3 Techniques of Integration 积分技巧

到目前为止，我们已经有了一套相当完整的程序来计算许多基本积分。然而，尽管我们可以通过换元 $u = x^{2}$ 来积分 $\int{x\mspace{2mu}\text{sin}(x^{2})dx}$，但像 $\int{x\mspace{2mu}\text{sin}\mspace{2mu} x\ dx}$ 这样看似简单的积分却难住了我们。许多学生想知道积分是否有乘积法则。并没有，但有一种基于求导乘积法则的技巧，它允许我们用一个积分替换另一个积分。我们称这种技巧为分部积分。

若 $h(x) = f(x)g(x)$，则由乘积法则可得 $h^{\prime}(x) = f^{\prime}(x)g(x) + g^{\prime}(x)f(x)$。尽管起初这看似适得其反，但现在我们对等式两边同时积分：$\int{h^{\prime}(x)dx = {\int{\left( {g(x)f^{\prime}(x) + f(x)g^{\prime}(x)} \right)dx.}}}$

这给出

$$h(x) = f(x)g(x) = {\int{g(x)f^{\prime}(x)dx + {\int{f(x)g^{\prime}(x)dx.}}}}$$

现在我们解出 $\int{f(x)g^{\prime}(x)dx:}$

$$\int{f(x)g^{\prime}(x)dx = f(x)g(x) - {\int{g(x)f^{\prime}(x)dx.}}}$$

通过换元 $u = f(x)$ 与 $v = g(x)$，进而有 $du = f^{\prime}(x)dx$ 与 $dv = g^{\prime}(x)dx$，我们得到更紧凑的形式

$${\int{u\ dv = uv -}}{\int{v\ du.}}$$

设 $u = f(x)$ 与 $v = g(x)$ 为具有连续导数的函数。则关于包含这两个函数的积分的分部积分公式为：

$${\int{u\ dv = uv - {\int{v\ du}}}}.$$ (3.1)

使用分部积分公式的优点在于，我们可以用它把一个积分换成另一个可能更简单的积分。下面的示例说明了它的用法。

取 $u = x$、$dv = \text{sin}\mspace{2mu} x\ dx$，用分部积分公式计算 $\int{x\mspace{2mu}\text{sin}\mspace{2mu} x\ dx.}$

取 $u = x$，则 $du = 1dx$。因为 $dv = \text{sin}\mspace{2mu} x\ dx$，得到 $v = {\int{\text{sin}\mspace{2mu} x\ dx = \text{−}\text{cos}\mspace{2mu} x.}}$。把这些值按如下方式记录较为方便：

$$\begin{array}{rllcrll} u & = & x & & {dv} & = & {\text{sin}\mspace{2mu} x\ dx} \\ {du} & = & {1dx} & & v & = & {\int{\text{sin}\mspace{2mu} x\ dx = \text{−}\text{cos}\mspace{2mu} x.}} \end{array}$$

应用分部积分公式得到

$$\begin{array}{clcl} {\int{x\mspace{2mu}\text{sin}\mspace{2mu} x\ dx}} & {= (x)(\text{−}\text{cos}\mspace{2mu} x) - {\int{(\text{−}\text{cos}\mspace{2mu} x)(1dx)}}} & & {\text{Substitute}.} \\ & {= \text{−}x\mspace{2mu}\text{cos}\mspace{2mu} x + {\int{\text{cos}\mspace{2mu} x\ dx}}} & & {\text{Simplify}.} \\ & {= \text{−}x\mspace{2mu}\text{cos}\mspace{2mu} x + \text{sin}\mspace{2mu} x + C.} & & {\text{Use}\ {\int{\text{cos}\mspace{2mu} x\ dx = \text{sin}\mspace{2mu} x + C.}}} \end{array}$$

此时，可能有几点需要澄清。首先，你或许会好奇：如果我们取 $u = \text{sin}\mspace{2mu} x$、$dv = x$ 会怎样。若这样做，则 $du = \text{cos}\mspace{2mu} x{dx}$、$v = \frac{1}{2}x^{2}$。于是应用分部积分后，有 $\int^{}x\mspace{2mu}\text{sin}\mspace{2mu} x\ dx = \frac{1}{2}x^{2}\text{sin}\mspace{2mu} x - \int^{}\frac{1}{2}x^{2}\text{cos}\mspace{2mu} x\ dx$。遗憾的是，对于新的积分，我们并没有比之前更好。重要的是要记住，应用分部积分时，在找到可行的 $u$ 与 $dv$ 之前，我们可能需要尝试若干种选择。

其次，你或许会想：当我们求出 $v = \int^{}\text{sin}\mspace{2mu} x\ dx = \text{−}\text{cos}\mspace{2mu} x$ 时，为什么不加 $v = \text{−}\text{cos}\mspace{2mu} x + K$。为了说明这没有区别，我们用 $v = \text{−}\text{cos}\mspace{2mu} x + K\text{:}$ 重做一遍：

$$\begin{array}{ll} {\int^{\ }x\mspace{2mu}\text{sin}\mspace{2mu} x\ dx} & {= (x)(\text{−}\text{cos}\mspace{2mu} x + K) - \int^{\ }(\text{−}\text{cos}\mspace{2mu} x + K)(1dx)} \\ & {= \text{−}x\mspace{2mu}\text{cos}\mspace{2mu} x + Kx + \int^{\ }\text{cos}\mspace{2mu} x\ dx - \int^{\ }Kdx} \\ & {= \text{−}x\mspace{2mu}\text{cos}\mspace{2mu} x\ + Kx + \text{sin}\mspace{2mu} x - Kx + C} \\ & {= \text{−}x\mspace{2mu}\text{cos}\mspace{2mu} x + \text{sin}\mspace{2mu} x + C.} \end{array}$$

如你所见，这并不影响最终结果。

最后，我们可以对 $\text{−}x\mspace{2mu}\text{cos}\mspace{2mu} x + \text{sin}\mspace{2mu} x + C\text{:}$ 求导来验证原函数是否正确：

$$\begin{array}{ll} {\frac{d}{dx}\left( {\text{−}x\mspace{2mu}\text{cos}\mspace{2mu} x + \text{sin}\mspace{2mu} x + C} \right)} & {= (-1)\text{cos}\mspace{2mu} x + (\text{−}x)(\text{−}\text{sin}\mspace{2mu} x) + \text{cos}\mspace{2mu} x} \\ & {= x\mspace{2mu}\text{sin}\mspace{2mu} x.} \end{array}$$

因此，该原函数验证无误。

观看该视频并访问该网站，以查看分部积分的示例。

用分部积分公式，取 $u = x$、$dv = e^{2x}dx$，计算 $\int^{}xe^{2x}dx$。

此时自然会问：我们如何知道怎样选择 $u$ 与 $dv$？有时这纯属试错；然而，首字母缩写 LIATE 常常能帮我们减少选择时的猜测。这个缩写代表 Logarithmic Functions（对数函数）、Inverse Trigonometric Functions（反三角函数）、Algebraic Functions（代数函数）、Trigonometric Functions（三角函数）与 Exponential Functions（指数函数）。这个助记符有助于我们为 $u$ 做出合适的选择。

积分中在列表中排在最前面的函数类型，应作为我们首选的 $u$。例如，若一个积分同时含有对数函数和代数函数，我们应取 $u$ 为对数函数，因为 LIATE 中 L 在 A 之前。示例 3.1 中的积分含有三角函数 $\text{(}{\text{sin}\mspace{2mu} x}\text{)}$ 与代数函数 $(x)$。由于 LIATE 中 A 在 T 之前，我们取 $u$ 为代数函数。选定 $u$ 后，$dv$ 取为待积分函数的剩余部分连同 $dx$。

这个助记符为何有效？记住，我们选作 $dv$ 的必须是能够积分的东西。由于我们并没有能直接积分简单对数函数和反三角函数的公式，它们不应被选作 $dv$ 的值是说得通的。因此，它们应排在列表最前面，作为 $u$ 的候选。于是我们把 LI 放在助记符的开头。（同样也可以从 IL 开始，因为这两类函数不会同时出现在一个分部积分问题中。）指数函数和三角函数位于列表末尾，因为它们相当容易积分，适合作为 $dv$。于是助记符末尾是 TE。（结尾用 ET 也可以，因为当这两类函数同时出现时，通常谁作 $u$、谁作 $dv$ 没多大关系。）代数函数通常既容易积分也容易求导，它们位于助记符中间。

计算 $\int{\frac{\text{ln}\mspace{2mu} x}{x^{3}}dx.}$

先将积分改写：

$${\int{\frac{\text{ln}\mspace{2mu} x}{x^{3}}dx}} = \int^{}x^{-3}\text{ln}\mspace{2mu} x\ dx.$$

由于该积分含有代数函数 $x^{-3}$ 与对数函数 $\text{ln}\mspace{2mu} x$，取 $u = \text{ln}\mspace{2mu} x$，因为 LIATE 中 L 在 A 之前。选定 $u = \text{ln}\mspace{2mu} x$ 后，必须取 $dv = x^{-3}dx$。

接着，由 $u = \text{ln}\mspace{2mu} x$ 得 $du = \frac{1}{x}dx$。又有 $v = \int^{}x^{-3}dx = - \frac{1}{2}x^{-2}$。归纳如下：

$$\begin{array}{rllcrll} u & = & {\text{ln}\mspace{2mu} x} & & {dv} & = & {x^{-3}dx} \\ {du} & = & {\frac{1}{x}dx} & & v & = & {\int^{}x^{-3}dx = - \frac{1}{2}x^{-2}.} \end{array}$$

代入分部积分公式（式 3.1）得到

$$\begin{array}{clccc} {\int{\frac{\text{ln}\mspace{2mu} x}{x^{3}}dx}} & {= \int^{}x^{-3}\text{ln}\mspace{2mu} x\ dx = \left( {\text{ln}\mspace{2mu} x)(\text{−}\ \frac{1}{2}x^{-2}} \right) - \int^{}\left( {\text{−}\ \frac{1}{2}x^{-2}} \right)(\frac{1}{x}dx)} & & & \\ & {= - \frac{1}{2}x^{-2}\text{ln}\mspace{2mu} x + \int^{}\frac{1}{2}x^{-3}dx} & & & {\text{Simplify}.} \\ & {= - \frac{1}{2}x^{-2}\text{ln}\mspace{2mu} x - \frac{1}{4}x^{-2} + C} & & & {\text{Integrate}.} \\ & {= - \frac{1}{2x^{2}}\text{ln}\mspace{2mu} x - \frac{1}{4x^{2}} + C.} & & & \text{Rewrite with positive integers.} \end{array}$$

计算 $\int^{}x\mspace{2mu}\text{ln}\mspace{2mu} x\ dx$。

在某些情况下，如下面两个示例所示，可能需要多次应用分部积分。

计算 $\int^{}x^{2}e^{3x}dx$。

用 LIATE，取 $u = x^{2}$、$dv = e^{3x}dx$。于是 $du = 2x\ dx$、$v = {\int{e^{3x}dx = \left( \frac{1}{3} \right)e^{3x}}}$。因此

$$\begin{array}{rllcrll} u & = & x^{2} & & {dv} & = & {e^{3x}dx} \\ {du} & = & {2x\ dx} & & v & = & {{\int{e^{3x}dx = \frac{1}{3}e^{3x}}}.} \end{array}$$

代入式 3.1 得

$${\int{x^{2}e^{3x}dx}} = \frac{1}{3}x^{2}e^{3x} - {\int{\frac{2}{3}xe^{3x}dx}}.$$

我们仍无法直接积分 $\int{\frac{2}{3}xe^{3x}dx}$，但该积分中 $x$ 的幂次降低了。我们可以再次使用分部积分来计算这个新积分。为此，取 $u = x$、$dv = \frac{2}{3}e^{3x}dx$。于是 $du = dx$、$v = {\int{\left( \frac{2}{3} \right)e^{3x}dx = \left( \frac{2}{9} \right)e^{3x}}}$。现在有

$$\begin{array}{rllcrll} u & = & x & & {dv} & = & {\frac{2}{3}e^{3x}dx} \\ {du} & = & {dx} & & v & = & {{\int{\frac{2}{3}e^{3x}dx = \frac{2}{9}e^{3x}}}.} \end{array}$$

代回前面的等式得

$$\int^{}x^{2}e^{3x}dx = \frac{1}{3}x^{2}e^{3x} - \left( {\frac{2}{9}xe^{3x} - \int^{}\frac{2}{9}e^{3x}dx} \right).$$

计算最后一个积分并化简后，得到

$${\int{x^{2}e^{3x}dx}} = \frac{1}{3}x^{2}e^{3x} - \frac{2}{9}xe^{3x} + \frac{2}{27}e^{3x} + C.$$

计算 $\int^{}t^{3}e^{t^{2}}dt$。

如果我们严格按照助记符 LIATE 来选择 $u$，会得到 $u = t^{3}$、$dv = e^{t^{2}}dt$。遗憾的是，这个选择行不通，因为我们无法计算 $\int^{}e^{t^{2}}dt$。不过，既然我们可以计算 $\int^{}te^{t^{2}}dt$，不妨尝试取 $u = t^{2}$、$dv = te^{t^{2}}dt$。这样选择后有

$$\begin{array}{rllcrll} u & = & t^{2} & & {dv} & = & {te^{t^{2}}dt} \\ {du} & = & {2t\ dt} & & v & = & {\int^{}te^{t^{2}}dt = \frac{1}{2}e^{t^{2}}.} \end{array}$$

于是我们得到

$$\begin{array}{ll} {\int^{\ }t^{3}e^{t^{2}}dt} & {= \frac{1}{2}t^{2}e^{t^{2}} - {\int{\frac{1}{2}e^{t^{2}}2tdt}}} \\ & {= \frac{1}{2}t^{2}e^{t^{2}} - \frac{1}{2}e^{t^{2}} + C.} \end{array}$$

计算 $\int^{}\text{sin}(\text{ln}\mspace{2mu} x)dx$。

这个积分看上去只有一个函数——即 $\text{sin}(\text{ln}\mspace{2mu} x)$——然而，我们总可以把常数函数 1 当作另一个函数。在此例中，取 $u = \text{sin}(\text{ln}\mspace{2mu} x)$、$dv = 1dx$。（取 $u = \text{sin}(\text{ln}\mspace{2mu} x)$ 很容易决定。我们不能取 $dv = \ \text{sin}(\text{ln}\mspace{2mu} x)dx$，因为如果真能积分它，我们一开始就不会用分部积分了！）因此 $du = (1\text{/}x)\text{cos}(\text{ln}\mspace{2mu} x)dx$、$v = \int^{}1dx = x$。对积分应用分部积分并化简后，有

$$\int^{}\text{sin}(\text{ln}\mspace{2mu} x)dx = x\mspace{2mu}\text{sin}\left( {\text{ln}\mspace{2mu} x} \right) - \int^{}\text{cos}\left( {\text{ln}\mspace{2mu} x} \right)dx.$$

遗憾的是，这个过程留给我们一个与原式非常相似的新积分。不过，我们看看再应用一次分部积分会怎样。这次取 $u = \text{cos}(\text{ln}\mspace{2mu} x)$、$dv = 1dx$，于是 $du = \text{−}(1\text{/}x)\text{sin}(\text{ln}\mspace{2mu} x)dx$、$v = \int^{}1dx = x$。代入后得

$$\int^{}\text{sin}(\text{ln}\mspace{2mu} x)dx = x\mspace{2mu}\text{sin}\left( {\text{ln}\mspace{2mu} x} \right) - \left( {x\mspace{2mu}\text{cos}(\text{ln}\mspace{2mu} x)—\int^{} - \text{sin}(\text{ln}\mspace{2mu} x)dx} \right).$$

化简后我们得到

$$\int^{}\text{sin}(\text{ln}\mspace{2mu} x)dx = x\mspace{2mu}\text{sin}(\text{ln}\mspace{2mu} x) - x\mspace{2mu}\text{cos}(\text{ln}\mspace{2mu} x) - \int^{}\text{sin}(\text{ln}\mspace{2mu} x)dx.$$

最后一个积分现在已经和原来相同。似乎我们只是在兜圈子，但现在实际上可以算出该积分了。为了更清楚地看出怎么做，令 $I = \int^{}\text{sin}(\text{ln}\mspace{2mu} x)dx$。于是等式变为

$$I = x\mspace{2mu}\text{sin}(\text{ln}\mspace{2mu} x) - x\mspace{2mu}\text{cos}(\text{ln}\mspace{2mu} x) - I.$$

首先，等式两边同时加上 $I$，得到

$$2I = x\mspace{2mu}\text{sin}(\text{ln}\mspace{2mu} x) - x\mspace{2mu}\text{cos}(\text{ln}\mspace{2mu} x).$$

接着，两边除以 2：

$$I = \frac{1}{2}x\mspace{2mu}\text{sin}(\text{ln}\mspace{2mu} x) - \frac{1}{2}x\mspace{2mu}\text{cos}(\text{ln}\mspace{2mu} x).$$

再把 $I = \int^{}\text{sin}(\text{ln}\mspace{2mu} x)dx$ 代回，得

$$\int^{}\text{sin}(\text{ln}\mspace{2mu} x)dx = \frac{1}{2}x\mspace{2mu}\text{sin}(\text{ln}\mspace{2mu} x) - \frac{1}{2}x\mspace{2mu}\text{cos}(\text{ln}\mspace{2mu} x).$$

由此可知，$(1\text{/}2)x\mspace{2mu}\text{sin}(\text{ln}\mspace{2mu} x) - (1\text{/}2)x\mspace{2mu}\text{cos}(\text{ln}\mspace{2mu} x)$ 是 $\text{sin}(\text{ln}\mspace{2mu} x)dx$ 的一个原函数。要得到最一般的不定积分，再加上 $+ C\text{:}$

$$\int^{}\text{sin}(\text{ln}\mspace{2mu} x)dx = \frac{1}{2}x\mspace{2mu}\text{sin}(\text{ln}\mspace{2mu} x) - \frac{1}{2}x\mspace{2mu}\text{cos}(\text{ln}\mspace{2mu} x) + C.$$

如果这个方法起初让你觉得有点奇怪，我们可以通过求导来验证答案：

$$\begin{array}{l} {\frac{d}{dx}\left( {\frac{1}{2}x\mspace{2mu}\text{sin}(\text{ln}\mspace{2mu} x) - \frac{1}{2}x\mspace{2mu}\text{cos}(\text{ln}\mspace{2mu} x)} \right)} \\ \\ {= \frac{1}{2}(\text{sin}(\text{ln}\mspace{2mu} x)) + \text{cos}(\text{ln}\mspace{2mu} x) \cdot \frac{1}{x} \cdot \frac{1}{2}x - \left( {\frac{1}{2}\text{cos}(\text{ln}\mspace{2mu} x) - \text{sin}(\text{ln}\mspace{2mu} x) \cdot \frac{1}{x} \cdot \frac{1}{2}x} \right)} \\ {= \text{sin}(\text{ln}\mspace{2mu} x).} \end{array}$$

计算 $\int^{}x^{2}\text{sin}\mspace{2mu} x\ dx$。

既然我们已经成功地运用分部积分法计算了不定积分，现在把注意力转向定积分。积分技巧其实相同，只是增加一步：在积分的上限与下限处计算该积分的值。

设 $u = f(x)$ 与 $v = g(x)$ 是在 $\lbrack a,b\rbrack$ 上具有连续导数的函数。则

$${\int_{a}^{b}{u\ dv}} = \left. {uv} \right|_{a}^{b} - {\int_{a}^{b}{v\ du}}.$$ (3.2)

求由 $y = \text{tan}^{-1}x$ 的图像为上界、以 $x$ 轴为下界、在区间 $\left\lbrack {0,1} \right\rbrack$ 上的区域面积。

该区域如图 3.2 所示。为求面积，我们必须计算 $\int\limits_{0}^{1}\text{tan}^{-1}x\ dx.$

对于这个积分，取 $u = \text{tan}^{-1}x$ 与 $dv = dx,$ 从而得到 $du = \frac{1}{x^{2} + 1}dx$ 与 $v = x.$ 应用分部积分公式（式 3.2）后，我们得到

$${\text{Area} = x\mspace{2mu}\text{tan}^{-1}\left. x \right|_{0}^{1}} - {\int\limits_{0}^{1}{\frac{x}{x^{2} + 1}dx.}}$$

利用 *u* 换元法可得

$$\int\limits_{0}^{1}{\frac{x}{x^{2} + 1}dx = \frac{1}{2}\text{ln}\left| x^{2} + 1|_{0}^{1}. \right.}$$

因此，

$$\left. \text{Area} = x\ \tan^{- 1}x|_{0}^{1} - \frac{1}{2}\ln\left| x^{2} \right. + 1 \middle| |_{0}^{1} = \frac{\pi}{4} - \frac{1}{2}\ln\ 2. \right.$$

此时，对我们的解答做一次“合理性检验”不失为明智之举。由于 $\frac{\pi}{4} - \frac{1}{2}\text{ln}\mspace{2mu} 2 \approx 0.4388,$ 而由图 3.2 我们预期面积略小于 0.5，因此该解答看来是合理的。

求由 $f(x) = e^{\text{−}x}$ 的图像、*x* 轴、*y* 轴以及直线 $x = 1$ 所围成区域绕 *y* 轴旋转所得立体的体积。

解决此问题的最佳选择是使用柱壳法。先从画出待旋转区域及一条典型矩形（见下图）开始。

为用柱壳法求体积，我们必须计算 $2\pi{\int_{0}^{1}{xe^{\text{−}x}dx}}.$ 为此取 $u = x$ 与 $dv = e^{\text{−}x}.$ 这些选择给出 $du = dx$ 与 $v = \int^{}e^{\text{−}x} = \text{−}e^{\text{−}x}.$ 代入式 3.2，我们得到

$$\begin{array}{clcl} \text{Volume} & {= 2\pi\int\limits_{0}^{1}xe^{\text{−}x}dx = 2\pi(\text{−}xe^{\text{−}x}|_{0}^{1} + \int\limits_{0}^{1}e^{\text{−}x}dx)} & & {\text{Use integration by parts}.} \\ & {{= -2\pi xe^{\text{−}x}|_{0}^{1}} - 2\pi e^{\text{−}x}|_{0}^{1}} & & {{\text{Evaluate}\ \int\limits_{0}^{1}e^{\text{−}x}dx = \text{−}e^{\text{−}x}|_{0}^{1}}.} \\ & {= 2\pi - \frac{4\pi}{e}.} & & {\text{Evaluate and simplify}.} \end{array}$$

再次强调，检验解答的合理性是个好主意。我们观察到该立体的体积略小于：半径为 $1$、高为 $1\text{/}e$ 的圆柱体积，加上底半径为 $1$、高为 $1 - \frac{1}{e}$ 的圆锥体积之和。因此，该立体的体积应略小于

$$\pi{(1)}^{2}\frac{1}{e} + \left( \frac{\pi}{3} \right)(1)^{2}\left( {1 - \frac{1}{e}} \right) = \frac{2\pi}{3e} + \frac{\pi}{3} \approx 1.8177.$$

由于 $2\pi - \frac{4\pi}{e} \approx 1.6603,$ 可见我们计算出的体积是合理的。

计算 $\int_{0}^{\pi\text{/}2}{x\mspace{2mu}\text{cos}\mspace{2mu} x\ dx.}$

在使用分部积分法时，必须谨慎选择哪个式子作为 *u*。对下列每个问题，运用本节中的准则来选择 *u*。不要计算这些积分。

1.

$\int{x^{3}e^{2x}dx}$

2\.

$\int{x^{3}\text{ln}(x)dx}$

3.

$\int{y^{3}\text{cos}\mspace{2mu} ydy}$

4\.

$\int{x^{2}\text{arctan}\mspace{2mu} x\ dx}$

5.

$\int{e^{3x}\text{sin}(2x)dx}$

用最简单的方法求积分。并非所有问题都需要分部积分法。

6\.

$\int{v\mspace{2mu}\text{sin}\mspace{2mu} vdv}$

7.

$\int{\text{ln}\mspace{2mu} x\ dx}$（*提示：* $\int{\text{ln}\mspace{2mu} x\ dx}$ 等价于 ${\int{1 \cdot \text{ln}(x)dx}}$.）

8\.

$\int{x\mspace{2mu}\text{cos}\mspace{2mu} x\ dx}$

9.

$\int{\text{tan}^{-1}x\ dx}$

10\.

${\int x^{2}}e^{x}dx$

11.

$\int{x\mspace{2mu}\text{sin}(2x)dx}$

12\.

$\int{xe^{4x}dx}$

13.

$\int{xe^{\text{−}x}dx}$

14\.

$\int{x\mspace{2mu}\text{cos}\mspace{2mu} 3x\ dx}$

15.

${\int x^{2}}\text{cos}\mspace{2mu} x\ dx$

16\.

$\int{x\mspace{2mu}\text{ln}\mspace{2mu} x\ dx}$

17.

$\int{\text{ln}(2x + 1)dx}$

18\.

${\int{x^{2}e^{4x}}}dx$

19.

$\int{e^{x}\text{sin}\mspace{2mu} x\ dx}$

20\.

$\int{e^{x}\text{cos}\mspace{2mu} x\ dx}$

21.

$\int{xe^{\text{−}x^{2}}dx}$

22\.

$\int{x^{2}e^{\text{−}x}dx}$

23.

$\int{\text{sin}(\text{ln}(2x))dx}$

24\.

$\int{cos(\text{ln}\mspace{2mu} x)dx}$

25.

$\int{{(\text{ln}\mspace{2mu} x)}^{2}dx}$

26\.

$\int{\text{ln}(x^{2})dx}$

27.

$\int{x^{2}\text{ln}\mspace{2mu} x\ dx}$

28\.

$\int{\text{sin}^{-1}x\ dx}$

29.

$\int{\text{cos}^{-1}(2x)dx}$

30\.

$\int{x\mspace{2mu}\text{arctan}\mspace{2mu} x\ dx}$

31.

$\int{x^{2}\text{sin}\mspace{2mu} x\ dx}$

32\.

$\int{x^{3}\text{cos}\mspace{2mu} x\ dx}$

33.

$\int{x^{3}\text{sin}\mspace{2mu} x\ dx}$

34\.

$\int{x^{3}e^{x}dx}$

35.

$\int{x\mspace{2mu}\text{sec}^{-1}x\ dx}$

36\.

$\int{x\mspace{2mu}\text{sec}^{2}x\ dx}$

37.

$\int{x\mspace{2mu}\text{cosh}\mspace{2mu} x\ dx}$

计算下列定积分。用绘图工具验证你的答案。

38\.

$\int_{1\text{/}e}^{1}{\text{ln}\mspace{2mu} x\ dx}$

39.

$\int_{0}^{1}{xe^{-2x}dx}$（答案用精确形式表示。）

40\.

$\int_{0}^{1}{e^{\sqrt{x}}dx\left( {\text{let}\ u = \sqrt{x}} \right)}$

41.

$\int_{1}^{e}{\text{ln}(x^{2})dx}$

42\.

$\int_{0}^{\pi}{x\mspace{2mu}\text{cos}\mspace{2mu} x\ dx}$

43.

$\int_{\text{−}\pi}^{\pi}{x\mspace{2mu}\text{sin}\mspace{2mu} x\ dx}$（答案用精确形式表示。）

44\.

$\int_{0}^{3}{\text{ln}(x^{2} + 1)dx}$（答案用精确形式表示。）

45.

$\int_{0}^{\pi\text{/}2}{x^{2}\text{sin}\mspace{2mu} x\ dx}$（答案用精确形式表示。）

46\.

$\int_{0}^{1}{x5^{x}dx}$（答案用五位有效数字表示。）

47.

计算 $\int{\text{cos}\mspace{2mu} x\mspace{2mu}\text{ln}(\text{sin}\mspace{2mu} x)dx}$

用分部积分法推导下列公式。假设 *n* 为正整数。这些公式称为*递推公式*，因为在每种情况下 *x* 项的指数都降低了一次。第二个积分比原积分更简单。

48\.

$\int{x^{n}e^{x}dx = x^{n}e^{x} - n{\int{x^{n - 1}e^{x}dx}}}$

49.

${\int{x^{n}\text{cos}\mspace{2mu} x\ dx}} = x^{n}\text{sin}\mspace{2mu} x - n{\int{x^{n - 1}\text{sin}\mspace{2mu} x\ dx}}$

50\.

${\int{x^{n}\text{sin}\mspace{2mu} x\ dx}} = \underset{\_\_\_\_\_\_}{}$

51.

用两种方法积分 ${\int{2x\sqrt{2x - 3}}}dx$：

1. 使用分部积分法，取 $dv = \sqrt{2x - 3}\mspace{2mu} dx$

2. 使用换元法，取 $u = 2x - 3$

说明你是否会使用分部积分法来计算该积分。若使用，指出 *u* 与 *dv*。若不使用，则描述所用的方法（无需实际计算）。

52\.

$\int{x\mspace{2mu}\text{ln}\mspace{2mu} x\ dx}$

53.

$\int{\frac{\text{ln}^{2}x}{x}dx}$

54\.

$\int{xe^{x}dx}$

55.

$\int{xe^{x^{2} - 3}dx}$

56\.

$\int{x^{2}\text{sin}\mspace{2mu} x\ dx}$

57.

$\int{x^{2}\text{sin}(3x^{3} + 2)dx}$

画出由上方的曲线、*x* 轴以及 $x = 1$ 所围成区域的图形，并求该区域的面积。答案用精确形式表示，或按所指位数四舍五入。

58\.

$y = 2xe^{\text{−}x}$（答案近似到四位小数。）

59.

$y = e^{\text{−}x}\text{sin}(\pi x)$（答案近似到五位小数。）

求由给定曲线围成区域绕指定直线旋转所产生的体积。答案用精确形式表示，或近似到所指的小数位数。

60\.

$y = \text{sin}\mspace{2mu} x,y = 0,x = 2\pi,x = 3\pi$ 绕 y 轴（答案用精确形式表示。）

61.

$y = e^{\text{−}x}$ $y = 0,x = -1x = 0;$ 绕 $x = 1$（答案用精确形式表示。）

62\.

沿直线运动的粒子在 *t* 秒后的速度为 $v(t) = t^{2}e^{\text{−}t}$。它在最初的 2 秒内行进了多远？（假设单位为英尺，答案用精确形式表示。）

63.

求 $y = \text{sec}^{3}x$ 的图像下方、从 $x = 0\mspace{2mu}\text{to}\ x = 1$ 之间的面积。（答案四舍五入到两位有效数字。）

64\.

求 $y = (x - 2)e^{x}$ 与 *x* 轴之间、从 $x = 2$ 到 $x = 5$ 的面积。（答案用精确形式表示。）

65.

求由曲线 $y = x\mspace{2mu}\text{cos}\mspace{2mu} x$ 与 *x* 轴所围成区域在如下范围内的面积

$\frac{11\pi}{2} \leq x \leq \frac{13\pi}{2}.$（答案用精确形式表示。）

66\.

求由曲线 $y = \text{ln}\mspace{2mu} x$、*x* 轴以及竖直直线 $x = e^{2}$ 所围成区域绕 *x* 轴旋转所得立体的体积。（答案用精确形式表示。）

67.

求由曲线 $y = 4\mspace{2mu}\text{cos}\mspace{2mu} x$ 与 *x* 轴所围成区域（其中 $\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}$）绕 *x* 轴旋转所得立体的体积。（答案用精确形式表示。）

68\.

求第一象限内由 $y = e^{x}$ 与 *x* 轴所围成区域（从 $x = 0$ 到 $x = \text{ln}(7)$）绕 *y* 轴旋转所得立体的体积。（答案用精确形式表示。）

在本节中，我们考察如何对各类三角函数的乘积进行积分。这些积分称为三角积分。它们是称为*三角换元法*的积分技巧的重要组成部分，该技巧将在“三角换元”中重点介绍。这一技巧使我们能够将原本可能无法积分的代数表达式转换为涉及三角函数的表达式，而后者可用本节所述技巧积分。此外，在学习极坐标、柱坐标和球坐标系统时，这类积分也会频繁出现。让我们从 $\text{sin}\mspace{2mu} x$ 与 $\text{cos}\mspace{2mu} x$ 的乘积开始研究。

用于积分 $\text{sin}\mspace{2mu} x$ 与 $\text{cos}\mspace{2mu} x$ 的乘积与幂的组合的策略，其背后的关键思想是将这些表达式改写为形如 $\int\text{sin}^{j}x\mspace{2mu}\text{cos}\mspace{2mu} x\ dx$ 或 $\int\text{cos}^{j}x\mspace{2mu}\text{sin}\mspace{2mu} x\ dx$ 的积分的和与差。重写这些积分后，我们使用 *u* 换元来计算它们。在详细描述一般过程之前，我们先来看下面的例子。

计算 $\int\text{cos}^{3}x\mspace{2mu}\text{sin}\mspace{2mu} x\ dx.$

使用 $u$ 换元，令 $u = \text{cos}\mspace{2mu} x.$ 此时 $du = \text{−}\text{sin}\mspace{2mu} x\ dx.$ 于是

$$\begin{array}{ll} {\int\text{cos}^{3}\ x\mspace{2mu}\text{sin}\mspace{2mu} x\ dx} & {= \text{−}\int u^{3}\ du} \\ & {= - \frac{1}{4}u^{4} + C} \\ & {= - \frac{1}{4}\text{cos}^{4}\ x + C.} \end{array}$$

计算 $\int\text{sin}^{4}x\mspace{2mu}\text{cos}\mspace{2mu} x\ dx.$

计算 $\int\text{cos}^{2}x\mspace{2mu}\text{sin}^{3}x\ dx.$

为了将该积分转化为形如 $\int\text{cos}^{j}x\mspace{2mu}\text{sin}\mspace{2mu} x\ dx$ 的积分，将 $\text{sin}^{3}x = \text{sin}^{2}x\mspace{2mu}\text{sin}\mspace{2mu} x$ 改写，并作换元 $\text{sin}^{2}x = 1 - \text{cos}^{2}x.$ 于是

$$\begin{matrix} {\int\text{cos}^{2}x\mspace{2mu}\text{sin}^{3}x\ dx} & {= \int\text{cos}^{2}x\left( 1 - \text{cos}^{2}x \right)\text{sin}\mspace{2mu} x\ dx} & {\text{Let}\ u = \text{cos}\mspace{2mu} x;\ \text{then}\ du = \text{−}\text{sin}\mspace{2mu} x\ dx.} \\ & {= \text{−}\int u^{2}\left( 1 - u^{2} \right)du} & \\ & {= \int\left( u^{4} - u^{2} \right)du} & \\ & {= \frac{1}{5}u^{5} - \frac{1}{3}u^{3} + C} & \\ & {= \frac{1}{5}\text{cos}^{5}x - \frac{1}{3}\text{cos}^{3}x + C.} & \end{matrix}$$

计算 $\int\text{cos}^{3}x\mspace{2mu}\text{sin}^{2}x\ dx.$

在下一个示例中，我们将看到当 $\text{sin}\mspace{2mu} x$ 与 $\text{cos}\mspace{2mu} x$ 都只有偶数幂时必须采用的策略。对于此类积分，下列恒等式

$$\text{sin}^{2}x = \frac{1}{2} - \frac{1}{2}\text{cos}(2x) = \frac{1 - \text{cos}\left( {2x} \right)}{2}$$

以及

$$\text{cos}^{2}x = \frac{1}{2} + \frac{1}{2}\text{cos}(2x) = \frac{1 + \text{cos}\left( {2x} \right)}{2}$$

非常重要。这些恒等式有时称为 *power-reducing identities*（降幂恒等式），它们可由二倍角公式 $\text{cos}\left( {2x} \right) = \text{cos}^{2}x - \text{sin}^{2}x$ 与勾股恒等式 $\text{cos}^{2}x + \text{sin}^{2}x = 1$ 推导得出。

计算 $\int\text{sin}^{2}x\ dx.$

为了计算该积分，我们使用三角恒等式 $\text{sin}^{2}x = \frac{1}{2} - \frac{1}{2}\text{cos}(2x).$ 于是

$$\begin{array}{ll} {\int\text{sin}^{2}x\ dx} & {= \int\left( {\frac{1}{2} - \frac{1}{2}\text{cos}(2x)} \right)dx} \\ & {= \frac{1}{2}x - \frac{1}{4}\text{sin}\left( {2x} \right) + C.} \end{array}$$

计算 $\int{\text{cos}^{2}x\ dx.}$

积分 $\text{sin}\mspace{2mu} x$ 与 $\text{cos}\mspace{2mu} x$ 的幂之乘积的一般过程，总结于下列准则中。

为积分 $\int\text{cos}^{j}x\mspace{2mu}\text{sin}^{k}x\ dx$，采用以下策略：

1. 若 $k$ 为奇数，将 $\text{sin}^{k}x = \text{sin}^{k - 1}x\mspace{2mu}\text{sin}\mspace{2mu} x$ 改写，并利用恒等式 $\text{sin}^{2}x = 1 - \text{cos}^{2}x$ 将 $\text{sin}^{k - 1}x$ 用 $\text{cos}\mspace{2mu} x$ 表示。使用换元 $u = \text{cos}\mspace{2mu} x$ 积分。该换元给出 $du = \text{−}\text{sin}\mspace{2mu} x\ dx.$

2. 若 $j$ 为奇数，将 $\text{cos}^{j}x = \text{cos}^{j - 1}x\mspace{2mu}\text{cos}\mspace{2mu} x$ 改写，并利用恒等式 $\text{cos}^{2}x = 1 - \text{sin}^{2}x$ 将 $\text{cos}^{j - 1}x$ 用 $\text{sin}\mspace{2mu} x$ 表示。使用换元 $u = \text{sin}\mspace{2mu} x$ 积分。该换元给出 $du = \text{cos}\mspace{2mu} x\ dx.$（*注*：若 $j$ 与 $k$ 均为奇数，则策略 1 或策略 2 均可使用。）

3. 若 $j$ 与 $k$ 均为偶数，使用 $\text{sin}^{2}x = (1\text{/}2) - (1\text{/}2)\text{cos}(2x)$ 与 $\text{cos}^{2}x = (1\text{/}2) + (1\text{/}2)\text{cos}(2x).$ 应用这些公式后，化简并按需重新应用策略 1 至 3。

计算 $\int\text{cos}^{8}x\mspace{2mu}\text{sin}^{5}x\ dx.$

由于 $\text{sin}\mspace{2mu} x$ 的幂为奇数，使用策略 1。于是

$$\begin{array}{clcl} {\int\text{cos}^{8}x\mspace{2mu}\text{sin}^{5}x\ dx} & {= \int\text{cos}^{8}x\mspace{2mu}\text{sin}^{4}x\mspace{2mu}\text{sin}\mspace{2mu} x\ dx} & & {\text{Break off}\ \text{sin}\mspace{2mu} x.} \\ & {= \int\text{cos}^{8}x{(\text{sin}^{2}x)}^{2}\text{sin}\mspace{2mu} x\ dx} & & {\text{Rewrite}\ \text{sin}^{4}x = {(\text{sin}^{2}x)}^{2}.} \\ & {= \int\text{cos}^{8}x{(1 - \text{cos}^{2}x)}^{2}\text{sin}\mspace{2mu} x\ dx} & & {\text{Substitute}\ \text{sin}^{2}x = 1 - \text{cos}^{2}x.} \\ & {= \int u^{8}{(1 - u^{2})}^{2}(\text{−}du)} & & {\text{Let}\ u = \text{cos}\mspace{2mu} x\ \text{and}\ du = \text{−}\text{sin}\mspace{2mu} x\ dx.} \\ & {= \int\left( {\text{−}u^{8} + 2u^{10} - u^{12}} \right)du} & & {\text{Expand}.} \\ & {= - \frac{1}{9}u^{9} + \frac{2}{11}u^{11} - \frac{1}{13}u^{13} + C} & & {\text{Evaluate the integral}.} \\ & {= - \frac{1}{9}\text{cos}^{9}x + \frac{2}{11}\text{cos}^{11}x - \frac{1}{13}\text{cos}^{13}x + C.} & & {\text{Substitute}\ u = \text{cos}\mspace{2mu} x.} \end{array}$$

计算 $\int\text{sin}^{4}x\ dx.$

由于 $\text{sin}\mspace{2mu} x$ 的幂为偶数 $\left( {k = 4} \right)$ 且 $\text{cos}\mspace{2mu} x$ 的幂为偶数 $\left( {j = 0} \right),$ 我们必须使用策略 3。于是

$$\begin{array}{clcc} {\int\text{sin}^{4}x\ dx} & {= \int\left( {\text{sin}^{2}x} \right)^{2}dx} & & {\text{Rewrite}\ \text{sin}^{4}x = \left( {\text{sin}^{2}x} \right)^{2}.} \\ & {= \int\left( {\frac{1}{2} - \frac{1}{2}\text{cos}(2x)} \right)^{2}dx} & & {\text{Substitute}\ \text{sin}^{2}x = \frac{1}{2} - \frac{1}{2}\text{cos}(2x).} \\ & {= \int\left( {\frac{1}{4} - \frac{1}{2}\text{cos}\left( {2x} \right) + \frac{1}{4}\text{cos}^{2}(2x)} \right)dx} & & {\text{Expand}\mspace{2mu}\left( {\frac{1}{2} - \frac{1}{2}\text{cos}(2x)} \right)^{2}.} \\ & {= \int\left( {\frac{1}{4} - \frac{1}{2}\text{cos}\left( {2x} \right) + \frac{1}{4}\left(\frac{1}{2} + \frac{1}{2}\text{cos}(4x)\right)} \right)dx.} & & \end{array}$$

由于 $\text{cos}^{2}(2x)$ 的指数为偶数，代入 $\text{cos}^{2}(2x) = \frac{1}{2} + \frac{1}{2}\text{cos}\left( {4x} \right)\text{:}$

$$\begin{array}{ll} {= \int\left( {\frac{3}{8} - \frac{1}{2}\text{cos}\left( {2x} \right) + \frac{1}{8}\text{cos}(4x)} \right)dx} & {\text{Simplify}.} \\ {= \frac{3}{8}x - \frac{1}{4}\text{sin}\left( {2x} \right) + \frac{1}{32}\mspace{2mu}\text{sin}\left( {4x} \right) + C} & {\text{Evaluate the integral}.} \end{array}$$

计算 $\int\text{cos}^{3}x\ dx.$

计算 $\int\text{cos}^{2}(3x)dx.$

在物理学的某些领域，例如量子力学、信号处理以及傅里叶级数的计算中，常常需要积分包含 $\text{sin}(ax),$ $\text{sin}(bx),$ $\text{cos}(ax),$ 与 $\text{cos}(bx)$ 的乘积。这些积分通过应用三角恒等式来计算，如下述法则所述。

为积分包含 $\text{sin}(ax),$ $\text{sin}(bx),$ $\text{cos}(ax),$ 与 $\text{cos}(bx)$ 的乘积，采用下列换元

$$\mspace{13mu}\text{sin}\left( {ax} \right)\text{sin}\left( {bx} \right) = \frac{1}{2}\text{cos}(\left( {a - b} \right)x) - \frac{1}{2}\text{cos}(\left( {a + b} \right)x)$$ (3.3) $$\text{sin}\left( {ax} \right)\text{cos}\left( {bx} \right) = \frac{1}{2}\text{sin}\left( {\left( {a - b} \right)x} \right) + \frac{1}{2}\text{sin}(\left( {a + b} \right)x)$$ (3.4) $$\mspace{9mu}\text{cos}\left( {ax} \right)\text{cos}\left( {bx} \right) = \frac{1}{2}\text{cos}(\left( {a - b} \right)x) + \frac{1}{2}\text{cos}(\left( {a + b} \right)x)$$ (3.5)

这些公式可由正弦与余弦的和角公式推导得出。

计算 $\int\text{sin}\left( {5x} \right)\text{cos}(3x)dx.$

应用恒等式 $\text{sin}(5x)\text{cos}(3x) = \frac{1}{2}\text{sin}(2x) + \frac{1}{2}\text{sin}(8x).$ 于是

$$\begin{array}{ll} {\int\text{sin}\left( {5x} \right)\text{cos}(3x)dx} & {= \int\frac{1}{2}\text{sin}\left( {\left. 2x \right){dx}} \right. + \int\frac{1}{2}\text{sin}\left( {\left. 8x \right){dx}} \right.} \\ & {= - \frac{1}{4}\text{cos}\left( {2x} \right) - \frac{1}{16}\mspace{2mu}\text{cos}\left( {8x} \right) + C.} \end{array}$$

计算 $\int\text{cos}\left( {6x} \right)\text{cos}(5x)dx.$

在讨论 tan *x* 与 sec *x* 的乘积与幂的积分之前，回顾一下我们已经学过的涉及 tan *x* 与 sec *x* 的积分是有帮助的：

1. $\int\text{sec}^{2}x\ dx = \text{tan}\mspace{2mu} x + C$

2. $\int\text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x\ dx = \text{sec}\mspace{2mu} x + C$

3. $\int\text{tan}\mspace{2mu} x\ dx = \text{ln}\left| {\text{sec}\mspace{2mu} x} \right| + C$

4. $\int\text{sec}\mspace{2mu} x\ dx = \text{ln}\left| {\text{sec}\mspace{2mu} x + \text{tan}\mspace{2mu} x} \right| + C.$

对于 tan *x* 与 sec *x* 的乘积与幂的大多数积分，我们把要积分的表达式改写为形如 $\int\text{tan}^{j}x\mspace{2mu}\text{sec}^{2}x\ dx$ 或 $\int\text{sec}^{j}x\mspace{2mu}\text{tan}\mspace{2mu} x\ dx$ 的积分之和或差。如下例所示，我们可以用 *u* 换元法来计算这些新积分。

计算 $\int\text{sec}^{5}x\mspace{2mu}\text{tan}\mspace{2mu} x\ dx.$

先把 $\text{sec}^{5}x\mspace{2mu}\text{tan}\mspace{2mu} x$ 改写为 $\text{sec}^{4}x\mspace{2mu}\text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x$。

$$\begin{array}{clcc} {\int\text{sec}^{5}x\mspace{2mu}\text{tan}\mspace{2mu} x\ dx} & {= \int\text{sec}^{4}x\mspace{2mu}\text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x\ dx} & & {\text{Let}\ u = \text{sec}\mspace{2mu} x;\ \text{then},\ du = \text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x\ dx.} \\ & {= \int u^{4}du} & & {\text{Evaluate the integral}.} \\ & {= \frac{1}{5}u^{5} + C} & & {\text{Substitute}\ \text{sec}\mspace{2mu} x = u.} \\ & {= \frac{1}{5}\text{sec}^{5}x + C} & & \end{array}$$

你可以在此网站（website）上阅读一些有趣的信息，了解一个涉及正割的常见积分。

计算 $\int\text{tan}^{5}x\mspace{2mu}\text{sec}^{2}x\ dx.$

我们现在来看对 sec *x* 与 tan *x* 的乘积与幂进行积分的各种策略。

要积分 $\int\text{tan}^{k}x\mspace{2mu}\text{sec}^{j}x\ dx$，使用以下策略：

1. 若 $j$ 为偶数且 $j \geq 2$，将 $\text{sec}^{j}x = \text{sec}^{j - 2}x\mspace{2mu}\text{sec}^{2}x$ 改写，并利用 $\text{sec}^{2}x = \text{tan}^{2}x + 1$ 把 $\text{sec}^{j - 2}x$ 用 $\text{tan}\mspace{2mu} x$ 表示。令 $u = \text{tan}\mspace{2mu} x$，且 $du = \text{sec}^{2}x{dx}.$

2. 若 $k$ 为奇数且 $j \geq 1$，将 $\text{tan}^{k}x\mspace{2mu}\text{sec}^{j}x = \text{tan}^{k - 1}x\mspace{2mu}\text{sec}^{j - 1}x\mspace{2mu}\text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x$ 改写，并利用 $\text{tan}^{2}x = \text{sec}^{2}x - 1$ 把 $\text{tan}^{k - 1}x$ 用 $\text{sec}\mspace{2mu} x$ 表示。令 $u = \text{sec}\mspace{2mu} x$，且 $du = \text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x\ dx$。（*注*：若 $j$ 为偶数且 $k$ 为奇数，则策略 1 或策略 2 均可使用。）

3. 若 $k$ 为奇数且 $k \geq 3$、$j = 0$，将 $\text{tan}^{k}x = \text{tan}^{k - 2}x\mspace{2mu}\text{tan}^{2}x = \text{tan}^{k - 2}x(\text{sec}^{2}x - 1) = \text{tan}^{k - 2}x\mspace{2mu}\text{sec}^{2}x - \text{tan}^{k - 2}x$ 改写。可能需要对 $\text{tan}^{k - 2}x$ 项重复这一过程。

4. 若 $k$ 为偶数且 $j$ 为奇数，则利用 $\text{tan}^{2}x = \text{sec}^{2}x - 1$ 把 $\text{tan}^{k}x$ 用 $\text{sec}\mspace{2mu} x$ 表示。用分部积分法积分 sec *x* 的奇次幂。

计算 $\int\text{tan}^{6}x\mspace{2mu}\text{sec}^{4}x\ dx.$

由于 sec *x* 的幂为偶数，将 $\text{sec}^{4}x = \text{sec}^{2}x\mspace{2mu}\text{sec}^{2}x$ 改写，并利用 $\text{sec}^{2}x = \text{tan}^{2}x + 1$ 把第一个 $\text{sec}^{2}x$ 用 $\text{tan}\mspace{2mu} x$ 表示。于是，

$$\begin{array}{clcl} {\int\text{tan}^{6}x\mspace{2mu}\text{sec}^{4}x\ dx} & {= \int\text{tan}^{6}x\left( {\text{tan}^{2}x + 1} \right)\text{sec}^{2}x\ dx} & & {\text{Let}\ u = \text{tan}\mspace{2mu} x\ \text{and}\ du = \text{sec}^{2}x{dx}.} \\ & {= \int u^{6}\left( {u^{2} + 1} \right)du} & & {\text{Expand}.} \\ & {= \int(u^{8} + u^{6})du} & & {\text{Evaluate the integral}.} \\ & {= \frac{1}{9}u^{9} + \frac{1}{7}u^{7} + C} & & {\text{Substitute}\ \text{tan}\mspace{2mu} x = u.} \\ & {= \frac{1}{9}\text{tan}^{9}x + \frac{1}{7}\text{tan}^{7}x + C.} & & \end{array}$$

计算 $\int\text{tan}^{5}x\mspace{2mu}\text{sec}^{3}x\ dx.$

由于 tan *x* 的幂为奇数，先把 $\text{tan}^{5}x\mspace{2mu}\text{sec}^{3}x = \text{tan}^{4}x\mspace{2mu}\text{sec}^{2}x\mspace{2mu}\text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x$ 改写。于是，

$$\begin{array}{cllccc} {\text{tan}^{5}x\mspace{2mu}\text{sec}^{3}x} & = & {\text{tan}^{4}x\mspace{2mu}\text{sec}^{2}x\mspace{2mu}\text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x.} & & & {\text{Write}\ \text{tan}^{4}x = {(\text{tan}^{2}x)}^{2}.} \\ {\int\text{tan}^{5}x\mspace{2mu}\text{sec}^{3}x\ dx} & = & {\int{(\text{tan}^{2}x)}^{2}\text{sec}^{2}x\mspace{2mu}\text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x\ dx} & & & {\text{Use}\ \text{tan}^{2}x = \text{sec}^{2}x - 1.} \\ & = & {\int{(\text{sec}^{2}x - 1)}^{2}\text{sec}^{2}x\mspace{2mu}\text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x\ dx} & & & {\text{Let}\ u = \text{sec}\mspace{2mu} x\ \text{and}\ du = \text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x\ dx.} \\ & = & {\int{(u^{2} - 1)}^{2}u^{2}du} & & & {\text{Expand}.} \\ & = & {\int\left( {u^{6} - 2u^{4} + u^{2}} \right)du} & & & {\text{Integrate}.} \\ & = & {\frac{1}{7}u^{7} - \frac{2}{5}u^{5} + \frac{1}{3}u^{3} + C} & & & {\text{Substitute}\ \text{sec}\mspace{2mu} x = u.} \\ & = & {\frac{1}{7}\text{sec}^{7}x - \frac{2}{5}\text{sec}^{5}x + \frac{1}{3}\text{sec}^{3}x + C.} & & & \end{array}$$

计算 $\int\text{tan}^{3}x\ dx.$

先把 $\text{tan}^{3}x = \text{tan}\mspace{2mu} x\mspace{2mu}\text{tan}^{2}x = \text{tan}\mspace{2mu} x\left( {\text{sec}^{2}x - 1} \right) = \text{tan}\mspace{2mu} x\mspace{2mu}\text{sec}^{2}x - \text{tan}\mspace{2mu} x$ 改写。于是，

$$\begin{array}{ll} {\int\text{tan}^{3}x\ dx} & {= \int\left( {\text{tan}\mspace{2mu} x\mspace{2mu}\text{sec}^{2}x - \text{tan}\mspace{2mu} x} \right)dx} \\ & {= \int\text{tan}\mspace{2mu} x\mspace{2mu}\text{sec}^{2}x\ dx - \int\text{tan}\mspace{2mu} x\ dx} \\ & {= \frac{1}{2}\text{tan}^{2}x - \text{ln}\left| {\text{sec}\mspace{2mu} x} \right| + C.} \end{array}$$

对第一个积分，用换元 $u = \text{tan}\mspace{2mu} x$。对第二个积分，用该公式。

计算 $\int\text{sec}^{3}x\ dx.$

该积分需要分部积分。开始时，令 $u = \text{sec}\mspace{2mu} x$ 且 $dv = \text{sec}^{2}x{dx}$。这些选择使得 $du = \text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x$ 且 $v = \text{tan}\mspace{2mu} x$。于是，

$$\begin{array}{clcc} {\int\text{sec}^{3}x\ dx} & {= \text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x - \int\text{tan}\mspace{2mu} x\mspace{2mu}\text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x\ dx} & & \\ & {= \text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x - \int\text{tan}^{2}x\mspace{2mu}\text{sec}\mspace{2mu} x\ dx} & & {\text{Simplify}.} \\ & {= \text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x - \int\left( {\text{sec}^{2}x - 1} \right)\text{sec}\mspace{2mu} x\ dx} & & {\text{Substitute}\ \text{tan}^{2}x = \text{sec}^{2}x - 1.} \\ & {= \text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x + \int\text{sec}\mspace{2mu} x\ dx - \int\text{sec}^{3}x\ dx} & & {\text{Rewrite}.} \\ & {= \text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x + \text{ln}\left| {\text{sec}\mspace{2mu} x + \text{tan}\mspace{2mu} x} \right| - \int\text{sec}^{3}x\ dx.} & & {\text{Evaluate}\int\text{sec}\mspace{2mu} x\ dx.} \end{array}$$

我们现在有

$$\int\text{sec}^{3}x\ dx = \text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x + \text{ln}\left| {\text{sec}\mspace{2mu} x + \text{tan}\mspace{2mu} x} \right| - \int\text{sec}^{3}x\ dx.$$

由于积分 $\int\text{sec}^{3}x\ dx$ 在右侧再次出现，我们可以通过把它加到等式两边来求解 $\int\text{sec}^{3}x\ dx$。这样做，我们得到

$$2\int\text{sec}^{3}x\ dx = \text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x + \text{ln}\left| {\text{sec}\mspace{2mu} x + \text{tan}\mspace{2mu} x} \right|.$$

两边除以 2，我们得到

$$\int\text{sec}^{3}x\ dx = \frac{1}{2}\text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x + \frac{1}{2}\text{ln}\left| {\text{sec}\mspace{2mu} x + \text{tan}\mspace{2mu} x} \right| + C.$$

计算 $\int\text{tan}^{3}x\mspace{2mu}\text{sec}^{7}x\ dx.$

对 $n$ 为奇数的 $\int\text{sec}^{n}x\ dx$ 的计算需要分部积分。此外，要计算 $\int\text{sec}^{n}x\ dx$，我们还必须知道 $\int\text{sec}^{n - 2}x\ dx$ 的值。$\int\text{tan}^{n}x\ dx$ 的计算也需要能够积分 $\int\text{tan}^{n - 2}x\ dx$。为了使过程更简单，我们可以推导并应用下列幂次降幂公式。这些法则使我们能够把 sec *x* 或 tan *x* 的某次幂的积分替换为 sec *x* 或 tan *x* 的较低次幂的积分。

$$\mspace{73mu}\int\text{sec}^{n}x\ dx = \frac{1}{n - 1}\text{sec}^{n - 2}x\mspace{2mu}\text{tan}\mspace{2mu} x + \frac{n - 2}{n - 1}\int\text{sec}^{n - 2}x\ dx$$ (3.6) $$\int\text{tan}^{n}x\ dx = \frac{1}{n - 1}\text{tan}^{n - 1}x - \int\text{tan}^{n - 2}x\ dx$$ (3.7)

第一条幂次降幂法则可通过应用分部积分来验证。第二条可通过按照积分 tan *x* 的奇次幂所概述的策略来验证。

应用降幂公式来计算 $\int\text{sec}^{3}x\ dx.$

应用第一条降幂公式，我们得到

$$\begin{array}{ll} {\int\text{sec}^{3}x\ dx} & {= \frac{1}{2}\text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x + \frac{1}{2}\int\text{sec}\mspace{2mu} x\ dx} \\ & {= \frac{1}{2}\text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x + \frac{1}{2}\text{ln}\left| {\text{sec}\mspace{2mu} x + \text{tan}\mspace{2mu} x} \right| + C.} \end{array}$$

计算 $\int\text{tan}^{4}x\ dx.$

对 $\int\text{tan}^{4}x\ dx$ 应用降幂公式，我们有

$$\begin{array}{clcc} {\int\text{tan}^{4}x\ dx} & {= \frac{1}{3}\text{tan}^{3}x - \int\text{tan}^{2}x\ dx} & & \\ & {= \frac{1}{3}\text{tan}^{3}x - (\text{tan}\mspace{2mu} x - \int\text{tan}^{0}x\ dx)} & & {\text{Apply the reduction formula to}\int\text{tan}^{2}x\ dx.} \\ & {= \frac{1}{3}\text{tan}^{3}x - \text{tan}\mspace{2mu} x + \int 1\ dx} & & {\text{Simplify}.} \\ & {= \frac{1}{3}\text{tan}^{3}x - \text{tan}\mspace{2mu} x + x + C.} & & {\text{Evaluate}\int 1dx.} \end{array}$$

对 $\int\text{sec}^{5}x\ dx$ 应用降幂公式。

填空使等式成立。

69.

$\text{sin}^{2}x + \operatorname{\_\_\_\_\_\_\_} = 1$

70\.

$\text{sec}^{2}x - 1 = \operatorname{\_\_\_\_\_\_\_}$

利用恒等式将三角函数的幂次降为一次三角函数。

71.

$\text{sin}^{2}x = \operatorname{\_\_\_\_\_\_\_}$

72\.

$\text{cos}^{2}x = \operatorname{\_\_\_\_\_\_\_}$

用 *u* 换元法计算下列各个积分。

73.

$\int{\text{sin}^{3}x\mspace{2mu}\text{cos}\mspace{2mu} x\ dx}$

74\.

$\int{\sqrt{\text{cos}\mspace{2mu} x}\mspace{2mu}\text{sin}\mspace{2mu} x\ dx}$

75.

$\int{\text{tan}^{5}(2x)\text{sec}^{2}(2x)dx}$

76\.

$\int{\text{sin}^{7}(2x)\text{cos}(2x)dx}$

77.

$\int{\text{tan}\left( \frac{x}{2} \right)\text{sec}^{2}\left( \frac{x}{2} \right)dx}$

78\.

$\int{\text{tan}^{2}x\mspace{2mu}\text{sec}^{2}x\ dx}$

使用三角幂函数积分的准则计算下列积分。可用计算机代数系统（CAS）检验解答。（*注*：部分题目可用之前学过的积分技巧完成。）

79.

$\int{\text{sin}^{3}x\ dx}$

80\.

$\int{\text{cos}^{3}x\ dx}$

81.

$\int{\text{sin}\mspace{2mu} x\mspace{2mu}\text{cos}\mspace{2mu} x\ dx}$

82\.

$\int{\text{cos}^{5}x\ dx}$

83.

$\int{\text{sin}^{5}x\mspace{2mu}\text{cos}^{2}x\ dx}$

84\.

$\int{\text{sin}^{3}x\mspace{2mu}\text{cos}^{3}x\ dx}$

85.

${\int\sqrt{\text{sin}\mspace{2mu} x}}\mspace{2mu}\text{cos}\mspace{2mu} x\ dx$

86\.

${\int\sqrt{\text{sin}\mspace{2mu} x}}\mspace{2mu}\text{cos}^{3}x\ dx$

87.

$\int{\text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x\ dx}$

88\.

$\int{\text{tan}(5x)dx}$

89.

$\int{\text{tan}^{2}x\mspace{2mu}\text{sec}\mspace{2mu} x\ dx}$

90\.

$\int{\text{tan}\mspace{2mu} x\mspace{2mu}\text{sec}^{3}x\ dx}$

91.

$\int{\text{sec}^{4}x\ dx}$

92\.

$\int{\text{cot}\mspace{2mu} x\ dx}$

93.

$\int{\text{csc}\mspace{2mu} x\ dx}$

94\.

$\int{\frac{\text{tan}^{3}x}{\sqrt{\text{sec}\mspace{2mu} x}}dx}$

对下列习题，求出这些积分的通式。

95.

$\int{\text{sin}^{2}ax\mspace{2mu}\text{cos}\mspace{2mu} ax\ dx}$

96\.

$\int{\text{sin}\mspace{2mu} ax\mspace{2mu}\text{cos}\mspace{2mu} ax\ dx.}$

利用倍角公式计算下列积分。

97.

$\int_{0}^{\pi}{\text{sin}^{2}x\ dx}$

98\.

$\int_{0}^{\pi}{\text{sin}^{4}x\ dx}$

99.

$\int{\text{cos}^{2}3x\ dx}$

100\.

$\int{\text{sin}^{2}x\mspace{2mu}\text{cos}^{2}x\ dx}$

101.

${\int{\text{sin}^{2}x\ dx}} + {\int{\text{cos}^{2}x\ dx}}$

102\.

$\int{\text{sin}^{2}x\mspace{2mu}\text{cos}^{2}(2x)dx}$

对下列习题，计算定积分。尽可能以精确形式给出答案。

103.

$\int_{0}^{2\pi}{\text{cos}\mspace{2mu} x\mspace{2mu}\text{sin}\mspace{2mu} 2x\ dx}$

104\.

$\int_{0}^{\pi}{\text{sin}\mspace{2mu} 3x\mspace{2mu}\text{sin}\mspace{2mu} 5x\ dx}$

105.

$\int{\text{cos}(99x)\text{sin}(101x)dx}$

106\.

$\int_{\text{−}\pi}^{\pi}{\text{cos}^{2}(3x)dx}$

107.

$\int_{0}^{2\pi}{\text{sin}\mspace{2mu} x\mspace{2mu}\text{sin}(2x)\text{sin}(3x)dx}$

108\.

${\int_{0}^{4\pi}{\text{cos}{{(x}\text{/}2})\text{sin}}}({x\text{/}2})dx$

109.

$\int_{\pi\text{/}6}^{\pi\text{/}3}{\frac{\text{cos}^{3}x}{\sqrt{\text{sin}\mspace{2mu} x}}dx}$ （将答案四舍五入到三位小数。）

110\.

$\int_{\text{−}\pi\text{/}3}^{\pi\text{/}3}{\sqrt{\text{sec}^{2}x - 1}\mspace{2mu} dx}$

111.

$\int_{0}^{\pi\text{/}2}{\sqrt{1 - \text{cos}(2x)}\mspace{2mu} dx}$

112\.

求由下列方程的图像所围成区域的面积 $y = \text{sin}\mspace{2mu} x,y = \text{sin}^{3}x,x = 0,\text{and}\ x = \frac{\pi}{2}.$

113.

求由下列方程的图像所围成区域的面积 $y = \text{cos}^{2}x,y = \text{sin}^{2}x,x = - \frac{\pi}{4},\text{and}\ x = \frac{\pi}{4}.$

114\.

一质点沿直线运动，其速度函数为 $v(t) = \text{sin}(\omega t)\text{cos}^{2}\left( {\omega t} \right).$ 若 $f(0) = 0$，求其位置函数 $x = f(t)$。

115.

求函数 $f(x) = \text{sin}^{2}x\mspace{2mu}\text{cos}^{3}x$ 在区间 $\left\lbrack {\text{−}\pi,\pi} \right\rbrack$ 上的平均值。

对下列习题，求解下列微分方程。

116\.

$\frac{dy}{dx} = \text{sin}^{2}x.$ 该曲线经过点 $(0,0).$

117.

$\frac{dy}{d\theta} = \text{sin}^{4}\left( {\pi\theta} \right)$

118\.

求曲线 $y = \text{ln}(\text{csc}\mspace{2mu} x),\frac{\pi}{4} \leq x \leq \frac{\pi}{2}.$ 的长度。

119.

求曲线 $y = \text{ln}(\text{sin}\mspace{2mu} x),\frac{\pi}{3} \leq x \leq \frac{\pi}{2}.$ 的长度。

120\.

求由曲线 $y = \text{cos}(3x)$ 绕 *x* 轴旋转所产生的体积，$0 \leq x \leq \frac{\pi}{36}$。

对下列习题，利用以下信息：两个函数 *f* 与 *g* 在区间 $\lbrack a,b\rbrack$ 上的内积定义为 $f(x) \cdot g(x) = \left\langle {f,g} \right\rangle = {\int_{a}^{b}{f \cdot gdx.}}$ 若 $\left\langle {f,g} \right\rangle = 0$，则称两个不同函数 *f* 与 *g* 正交。

121.

证明 $\left\\{ \text{sin}(2x),\text{cos}(3x) \right\\}$ 在区间 $\left\lbrack {\text{−}\pi,\pi} \right\rbrack$ 上正交。

122\.

Evaluate ${\int_{\text{−}\pi}^{\pi}{\text{sin}(mx)\text{cos}(nx)dx}}.$

123.

积分 $y^{\prime} = \sqrt{\text{tan}\mspace{2mu} x}\text{sec}^{4}x$。

对每一对积分，判断哪一个更难以计算，并说明理由。

124\.

$\int{\text{sin}^{456}x\mspace{2mu}\text{cos}\mspace{2mu} x\ dx}$ or $\int{\text{sin}^{2}x\mspace{2mu}\text{cos}^{2}x\ dx}$

125.

$\int{\text{tan}^{350}x\mspace{2mu}\text{sec}^{2}x\ dx}$ or $\int{\text{tan}^{350}x\mspace{2mu}\text{sec}\mspace{2mu} x\ dx}$

本节中，我们研究含有如下形式表达式的积分：$\sqrt{a^{2} - x^{2}}$、$\sqrt{a^{2} + x^{2}}$ 和 $\sqrt{x^{2} - a^{2}}$，其中 $a$ 取正值。我们已经遇到并计算过含有此类某些表达式的积分，但仍有许多无法用已有方法解决。三角换元法在计算这些积分时非常有用。这种方法利用换元将这些积分改写为三角积分。

在给出含 $\sqrt{a^{2} - x^{2}},$ 的积分的一般策略之前，先看积分 ${\int\sqrt{9 - x^{2}}}dx.$ 迄今讨论过的任何方法都算不出这个积分。但若作换元 $x = 3\mspace{2mu}\text{sin}\mspace{2mu}\theta,$ 则有 $dx = 3\mspace{2mu}\text{cos}\mspace{2mu}\theta d\theta.$ 代入积分后得到

$${\int{\sqrt{9 - x^{2}}\mspace{2mu} dx}} = \int^{}\sqrt{9 - {(3\mspace{2mu}\text{sin}\mspace{2mu}\theta)}^{2}}3\mspace{2mu}\text{cos}\mspace{2mu}\theta d\theta.$$

化简后得到

$$\int^{}\sqrt{9 - x^{2}}\mspace{2mu} dx = \int^{}9\sqrt{1 - \text{sin}^{2}\theta}\mspace{2mu}\text{cos}\mspace{2mu}\theta d\theta.$$

令 $1 - \text{sin}^{2}\theta = \text{cos}^{2}\theta,$ 现在有

$$\int^{}\sqrt{9 - x^{2}}\mspace{2mu} dx = \int^{}9\sqrt{\text{cos}^{2}\theta}\mspace{2mu}\text{cos}\mspace{2mu}\theta d\theta.$$

假设 $\text{cos}\mspace{2mu}\theta \geq 0,$ 则有

$$\int^{}\sqrt{9 - x^{2}}\mspace{2mu} dx = \int^{}9\mspace{2mu}\text{cos}^{2}\theta d\theta.$$

到这一步，就可以用前面为三角函数的幂与乘积的积分所建立的方法来计算该积分。在做完这个例子之前，先来看看这一想法背后的一般理论。

为计算含 $\sqrt{a^{2} - x^{2}},$ 的积分，我们作换元 $x = a\mspace{2mu}\text{sin}\mspace{2mu}\theta$ 与 $dx = a\mspace{2mu}\text{cos}\mspace{2mu}\theta.$ 为说明这样做确实合理，考虑如下论证：$\sqrt{a^{2} - x^{2}}$ 的定义域是 $\lbrack\text{−}a,a\rbrack.$ 因此 $\text{−}a \leq x \leq a.$ 于是 $-1 \leq \frac{x}{a} \leq 1.$ 由于 $\text{sin}\mspace{2mu} x$ 在 $\left\lbrack {\text{−}(\pi\text{/}2),\pi\text{/}2} \right\rbrack$ 上的值域为 $\lbrack-1,1\rbrack,$ 故存在唯一满足 $\text{−}(\pi\text{/}2) \leq \theta \leq \pi\text{/}2$ 的角 $\theta$ 使 $\text{sin}\mspace{2mu}\theta = x\text{/}a,$ 等价地使 $x = a\mspace{2mu}\text{sin}\mspace{2mu}\theta.$ 若把 $x = a\mspace{2mu}\text{sin}\mspace{2mu}\theta$ 代入 $\sqrt{a^{2} - x^{2}},$ 便得

$$\begin{array}{clccc} \sqrt{a^{2} - x^{2}} & {= \sqrt{a^{2} - {(a\mspace{2mu}\text{sin}\mspace{2mu}\theta)}^{2}}} & & & {\text{Let}\ x = a\mspace{2mu}\text{sin}\mspace{2mu}\theta\ \text{where}\ - \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}.\ \text{Simplify.}} \\ & {= \sqrt{a^{2} - a^{2}\text{sin}^{2}\theta}} & & & {\text{Factor out}\ a^{2}.} \\ & {= \sqrt{a^{2}(1 - \text{sin}^{2}\theta)}} & & & {\text{Substitute}\ 1 - \text{sin}^{2}x = \text{cos}^{2}x.} \\ & {= \sqrt{a^{2}\text{cos}^{2}\theta}} & & & \text{Take the square root.} \\ & {= \left| {a\mspace{2mu}\text{cos}\mspace{2mu}\theta} \right|} & & & \\ & {= a\mspace{2mu}\text{cos}\mspace{2mu}\theta.} & & & \end{array}$$

由于在 $- \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$ 上 $\text{cos}\mspace{2mu}\theta \geq 0$ 且 $a > 0,$ 故 $\left| {a\mspace{2mu}\text{cos}\mspace{2mu}\theta} \right| = a\mspace{2mu}\text{cos}\mspace{2mu}\theta.$ 由上述讨论可知，作换元 $x = a\mspace{2mu}\text{sin}\mspace{2mu}\theta,$ 就能把含根式的积分化为含三角函数的积分。算出该积分之后，再把结果化回含 $x.$ 的表达式。为说明具体做法，先假设 $0 < x < a.$ 此时 $0 < \theta < \frac{\pi}{2}.$ 由于 $\text{sin}\mspace{2mu}\theta = \frac{x}{a},$ 可以画出图 3.4 中的参考三角形，借助它把 $\text{cos}\mspace{2mu}\theta,$ $\text{tan}\mspace{2mu}\theta,$ 以及其余三角函数的值用 $x.$ 表示出来。可以证明：对一切满足 $- \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}.$ 的 $\theta$，这个三角形给出的三角函数在 $\theta$ 处的值都是正确的。值得注意的是，表达式 $\sqrt{a^{2} - x^{2}}$ 恰好就是三角形一条边的长度。最后，若 $\theta$ 单独出现，就用 $\theta = \text{sin}^{-1}\left( \frac{x}{a} \right).$

这一讨论的要点归纳为下面的解题策略。

1. 最好先确认该积分不能用别的方法轻易算出。例如，虽然本方法可用于形如 ${\int{\frac{1}{\sqrt{a^{2} - x^{2}}}dx}},$ ${\int{\frac{x}{\sqrt{a^{2} - x^{2}}}dx}},$ 与 $\int{x\sqrt{a^{2} - x^{2}}\mspace{2mu} dx,}$ 的积分，但它们各自都能直接用公式或简单的 *u* 换元求出。

2. 作换元 $x = a\mspace{2mu}\text{sin}\mspace{2mu}\theta$ 与 $dx = a\mspace{2mu}\text{cos}\mspace{2mu}\theta d\theta.$ *注*：该换元给出 $\sqrt{a^{2} - x^{2}} = a\mspace{2mu}\text{cos}\mspace{2mu}\theta.$

3. 化简表达式。

4. 用三角积分那一节的方法计算该积分。

5. 利用图 3.4 中的参考三角形把结果改写成含 $x.$ 的形式。这一步可能还要用到一些三角恒等式以及关系式 $\theta = \text{sin}^{-1}\left( \frac{x}{a} \right).$

下面的示例演示这一解题策略的应用。

计算 $\int^{}\sqrt{9 - x^{2}}\mspace{2mu} dx.$

先作换元 $x = 3\mspace{2mu}\text{sin}\mspace{2mu}\theta$ 与 $dx = 3\mspace{2mu}\text{cos}\mspace{2mu}\theta d\theta.$ 由于 $\text{sin}\mspace{2mu}\theta = \frac{x}{3},$ 可以作出下图所示的参考三角形。

于是，

$$\begin{array}{clccl} {\int^{}\sqrt{9 - x^{2}}\mspace{2mu} dx} & {= \int^{}\sqrt{9 - {(3\mspace{2mu}\text{sin}\mspace{2mu}\theta)}^{2}}3\mspace{2mu}\text{cos}\mspace{2mu}\theta d\theta} & & & {\text{Substitute}\ x = 3\mspace{2mu}\text{sin}\mspace{2mu}\theta\ \text{and}\ dx = 3\mspace{2mu}\text{cos}\mspace{2mu}\theta d\theta.} \\ & {= \int^{}\sqrt{9(1 - \text{sin}^{2}\theta)}3\mspace{2mu}\text{cos}\mspace{2mu}\theta d\theta} & & & \text{Simplify.} \\ & {= \int^{}\sqrt{9\mspace{2mu}\text{cos}^{2}\theta}3\mspace{2mu}\text{cos}\mspace{2mu}\theta d\theta} & & & {\text{Substitute}\ \text{cos}^{2}\theta = 1 - \text{sin}^{2}\theta.} \\ & {= \int^{}3\left| {\text{cos}\mspace{2mu}\theta} \right|3\mspace{2mu}\text{cos}\mspace{2mu}\theta d\theta} & & & \text{Take the square root.} \\ & {= \int^{}9\mspace{2mu}\text{cos}^{2}\theta d\theta} & & & \begin{array}{l} {\text{Simplify. Since}\ - \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2},\ \text{cos}\mspace{2mu}\theta \geq 0\ \text{and}} \\ {\left| {\text{cos}\mspace{2mu}\theta} \right| = \text{cos}\mspace{2mu}\theta.} \end{array} \\ & {= \int^{}9\left( {\frac{1}{2} + \frac{1}{2}\text{cos}(2\theta)} \right)d\theta} & & & \begin{array}{l} \text{Use the strategy for integrating an even power} \\ {\text{of}\ \text{cos}\mspace{2mu}\theta.} \end{array} \\ & {= \frac{9}{2}\theta + \frac{9}{4}\text{sin}(2\theta) + C} & & & \text{Evaluate the integral.} \\ & {= \frac{9}{2}\theta + \frac{9}{4}(2\mspace{2mu}\text{sin}\mspace{2mu}\theta\mspace{2mu}\text{cos}\mspace{2mu}\theta) + C} & & & {\text{Substitute}\ \text{sin}(2\theta) = 2\mspace{2mu}\text{sin}\mspace{2mu}\theta\mspace{2mu}\text{cos}\mspace{2mu}\theta.} \\ & {= \frac{9}{2}\text{sin}^{-1}\left( \frac{x}{3} \right) + \frac{9}{2} \cdot \frac{x}{3} \cdot \frac{\sqrt{9 - x^{2}}}{3} + C} & & & \begin{array}{l} {\text{Substitute}\ \text{sin}^{-1}\left( \frac{x}{3} \right) = \theta\ \text{and}\ \text{sin}\mspace{2mu}\theta = \frac{x}{3}.\ \text{Use}} \\ \text{the reference triangle to see that} \\ {\text{cos}\mspace{2mu}\theta = \frac{\sqrt{9 - x^{2}}}{3}\ \text{and make this substitution.}} \end{array} \\ & {= \frac{9}{2}\text{sin}^{-1}\left( \frac{x}{3} \right) + \frac{x\sqrt{9 - x^{2}}}{2} + C.} & & & \text{Simplify.} \end{array}$$

计算 $\int{\frac{\sqrt{4 - x^{2}}}{x}dx.}$

先作换元 $x = 2\mspace{2mu}\text{sin}\mspace{2mu}\theta$ 与 $dx = 2\mspace{2mu}\text{cos}\mspace{2mu}\theta d\theta.$ 由于 $\text{sin}\mspace{2mu}\theta = \frac{x}{2},$ 可以作出下图所示的参考三角形。

于是，

$$\begin{array}{clccl} {\int{\frac{\sqrt{4 - x^{2}}}{x}dx}} & {= {\int{\frac{\sqrt{4 - {(2\mspace{2mu}\text{sin}\mspace{2mu}\theta)}^{2}}}{2\mspace{2mu}\text{sin}\mspace{2mu}\theta}2\mspace{2mu}\text{cos}\mspace{2mu}\theta d\theta}}} & & & {\text{Substitute}\ x = 2\mspace{2mu}\text{sin}\mspace{2mu}\theta\ \text{and}\ {dx} = 2\mspace{2mu}\text{cos}\mspace{2mu}\theta d\theta.} \\ & {= {\int{\frac{2\mspace{2mu}\text{cos}^{2}\theta}{\text{sin}\mspace{2mu}\theta}d\theta}}} & & & {\text{Substitute}\ \text{cos}^{2}\theta = 1 - \text{sin}^{2}\theta\ \text{and simplify.}} \\ & {= {\int{\frac{2(1 - \text{sin}^{2}\theta)}{\text{sin}\mspace{2mu}\theta}d\theta}}} & & & {\text{Substitute}\ \text{sin}^{2}\theta = 1 - \text{cos}^{2}\theta.} \\ & {= \int^{}(2\mspace{2mu}\text{csc}\mspace{2mu}\theta - 2\mspace{2mu}\text{sin}\mspace{2mu}\theta)d\theta} & & & \begin{array}{l} \text{Separate the numerator, simplify, and use} \\ {\text{csc}\mspace{2mu}\theta = \frac{1}{\text{sin}\mspace{2mu}\theta}.} \end{array} \\ & {= 2\mspace{2mu}\text{ln}\left| {\text{csc}\mspace{2mu}\theta - \text{cot}\mspace{2mu}\theta} \right| + 2\mspace{2mu}\text{cos}\mspace{2mu}\theta + C} & & & \text{Evaluate the integral.} \\ & {= 2\mspace{2mu}\text{ln}\left| {\frac{2}{x} - \frac{\sqrt{4 - x^{2}}}{x}} \right| + \sqrt{4 - x^{2}} + C.} & & & \begin{array}{l} \text{Use the reference triangle to rewrite the} \\ {\text{expression in terms of}\ x\ \text{and simplify.}} \end{array} \end{array}$$

在下一个示例中可以看到，有时方法并不唯一，可以自行选择。

用两种方法计算 $\int^{}x^{3}\sqrt{1 - x^{2}}\mspace{2mu} dx$：先用换元 $u = 1 - x^{2}$，再用三角换元。

令 $u = 1 - x^{2}$，于是 $x^{2} = 1 - u.$ 因此 $du = -2x\mspace{2mu} dx.$ 这时积分化为

$$\begin{array}{clccc} {\int^{}x^{3}\sqrt{1 - x^{2}}\mspace{2mu} dx} & {= - \frac{1}{2}\int^{}x^{2}\sqrt{1 - x^{2}}(-2x\mspace{2mu} dx)} & & & \text{Make the substitution.} \\ & {= - \frac{1}{2}\int^{}(1 - u)\sqrt{u}\mspace{2mu} du} & & & \text{Expand the expression.} \\ & {= - \frac{1}{2}{\int{\left( {u^{1\text{/}2} - u^{3\text{/}2}} \right)du}}} & & & \text{Evaluate the integral.} \\ & {= - \frac{1}{2}\left( {\frac{2}{3}u^{3\text{/}2} - \frac{2}{5}u^{5\text{/}2}} \right) + C} & & & {\text{Rewrite in terms of}\ x.} \\ & {= - \frac{1}{3}\left( {1 - x^{2}} \right)^{3\text{/}2} + \frac{1}{5}\left( {1 - x^{2}} \right)^{5\text{/}2} + C.} & & & \end{array}$$

令 $x = \text{sin}\mspace{2mu}\theta.$ 这时 $dx = \text{cos}\mspace{2mu}\theta d\theta.$ 用这个换元，我们有

$$\begin{array}{clccc} {\int^{}x^{3}\sqrt{1 - x^{2}}\mspace{2mu} dx} & {= \int^{}\text{sin}^{3}\theta\mspace{2mu}\text{cos}^{2}\theta d\theta} & & & \\ & {= \int^{}\left( {1 - \text{cos}^{2}\theta} \right)\text{cos}^{2}\theta\mspace{2mu}\text{sin}\mspace{2mu}\theta d\theta} & & & {\text{Let}\ u = \text{cos}\mspace{2mu}\theta.\ \text{Thus,}\ du = \text{−}\text{sin}\mspace{2mu}\theta d\theta.} \\ & {= \int^{}\left( {u^{4} - u^{2}} \right)du} & & & \\ & {= \frac{1}{5}u^{5} - \frac{1}{3}u^{3} + C} & & & {\text{Substitute}\ \text{cos}\mspace{2mu}\theta = u.} \\ & {= \frac{1}{5}\text{cos}^{5}\theta - \frac{1}{3}\text{cos}^{3}\theta + C} & & & \begin{array}{l} \text{Use a reference triangle to see that} \\ {\text{cos}\mspace{2mu}\theta = \sqrt{1 - x^{2}}.} \end{array} \\ & {= \frac{1}{5}\left( {1 - x^{2}} \right)^{5\text{/}2} - \frac{1}{3}\left( {1 - x^{2}} \right)^{3\text{/}2} + C.} & & & \end{array}$$

用适当的三角换元改写积分 $\int{\frac{x^{3}}{\sqrt{25 - x^{2}}}dx}$（不必计算该积分）。

对于含有 $\sqrt{a^{2} + x^{2},}$ 的积分，我们先考虑该表达式的定义域。由于 $\sqrt{a^{2} + x^{2}}$ 对 $x$ 的所有实数值都有定义，我们将选择限制在值域为全体实数的那些三角函数上。因此，我们的选择局限于取 $x = a\mspace{2mu}\text{tan}\mspace{2mu}\theta$ 或 $x = a\mspace{2mu}\text{cot}\mspace{2mu}\theta.$ 这两种换元实际上都可行，但标准换元是 $x = a\mspace{2mu}\text{tan}\mspace{2mu}\theta$，或等价地 $\text{tan}\mspace{2mu}\theta = x\text{/}a.$ 采用这一换元时，我们假定 $\text{−}(\pi\text{/}2) < \theta < \pi\text{/}2,$ 从而还有 $\theta = \text{tan}^{-1}\left( {x\text{/}a} \right).$ 使用该换元的步骤在下述解题策略中列出。

1. 先检查该积分是否能用其他方法轻松计算。在某些情况下，使用替代方法更为方便。

2. 换元 $x = a\mspace{2mu}\text{tan}\mspace{2mu}\theta$ 并取 $dx = a\mspace{2mu}\text{sec}^{2}\theta d\theta.$ 这一换元得到

$\sqrt{a^{2} + x^{2}} = \sqrt{a^{2} + {(a\mspace{2mu}\text{tan}\mspace{2mu}\theta)}^{2}} = \sqrt{a^{2}(1 + \text{tan}^{2}\theta)} = \sqrt{a^{2}\text{sec}^{2}\theta} = \left| {a\mspace{2mu}\text{sec}\mspace{2mu}\theta} \right| = a\mspace{2mu}\text{sec}\mspace{2mu}\theta.$ （因为 $- \frac{\pi}{2} < \theta < \frac{\pi}{2}$ 且在此区间上 $\text{sec}\mspace{2mu}\theta > 0$，故 $\left| {a\mspace{2mu}\text{sec}\mspace{2mu}\theta} \right| = a\mspace{2mu}\text{sec}\mspace{2mu}\theta.$）

3. 化简表达式。

4. 利用三角积分一节中的技巧计算该积分。

5. 利用图 3.7 中的参考三角形将结果用 $x$ 表示。你可能还需要用到一些三角恒等式以及关系 $\theta = \text{tan}^{-1}\left( \frac{x}{a} \right).$ （*注*：参考三角形基于 $x > 0$ 的假设；不过，由参考三角形得到的三角比值与 $x \leq 0$ 时所得的比值相同。）

计算 $\int\frac{dx}{\sqrt{1 + x^{2}}}$ 并通过求导验证结果。

从换元 $x = \text{tan}\mspace{2mu}\theta$ 与 $dx = \text{sec}^{2}\theta d\theta$ 开始。由于 $\text{tan}\mspace{2mu}\theta = x$，在下图中画出参考三角形。

于是，

$$\begin{array}{clccc} {\int\frac{dx}{\sqrt{1 + x^{2}}}} & {= {\int{\frac{\text{sec}^{2}\theta}{\text{sec}\mspace{2mu}\theta}d\theta}}} & & & \begin{array}{l} {\text{Substitute}\ x = \text{tan}\mspace{2mu}\theta\ \text{and}\ dx = \text{sec}^{2}\theta d\theta.\ \text{This}} \\ {\text{substitution makes}\ \sqrt{1 + x^{2}} = \text{sec}\mspace{2mu}\theta.\ \text{Simplify.}} \end{array} \\ & {= \int^{}\text{sec}\mspace{2mu}\theta d\theta} & & & \text{Evaluate the integral.} \\ & {= \text{ln}\left| {\text{sec}\mspace{2mu}\theta + \text{tan}\mspace{2mu}\theta} \right| + C} & & & \begin{array}{l} \text{Use the reference triangle to express the result} \\ {\text{in terms of}\ x.} \end{array} \\ & {= \text{ln}\left| {\sqrt{1 + x^{2}} + x} \right| + C.} & & & \end{array}$$

为验证结果，求导：

$$\begin{array}{cl} {\frac{d}{dx}\left( {\text{ln}\left| {\sqrt{1 + x^{2}} + x} \right|} \right)} & {= \frac{1}{\sqrt{1 + x^{2}} + x} \cdot \left( {\frac{x}{\sqrt{1 + x^{2}}} + 1} \right)} \\ & {= \frac{1}{\sqrt{1 + x^{2}} + x} \cdot \frac{x + \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}} \\ & {= \frac{1}{\sqrt{1 + x^{2}}.}} \end{array}$$

由于对 $x$ 的所有取值都有 $\sqrt{1 + x^{2}} + x > 0$，若需要，可改写为 $\text{ln}\left| {\sqrt{1 + x^{2}} + x} \right| + C = \text{ln}\left( {\sqrt{1 + x^{2}} + x} \right) + C$。

用换元 $x = \text{sinh}\mspace{2mu}\theta$ 计算 $\int{\frac{dx}{\sqrt{1 + x^{2}}}.}$

由于 $\text{sinh}\mspace{2mu}\theta$ 的值域为全体实数，且 $1 + \text{sinh}^{2}\theta = \text{cosh}^{2}\theta$，我们也可以用换元 $x = \text{sinh}\mspace{2mu}\theta$ 来计算该积分。此时 $dx = \text{cosh}\mspace{2mu}\theta d\theta.$ 于是

$$\begin{array}{clccc} {\int\frac{dx}{\sqrt{1 + x^{2}}}} & {= {\int{\frac{\text{cosh}\mspace{2mu}\theta}{\sqrt{1 + \text{sinh}^{2}\theta}}d\theta}}} & & & \begin{array}{l} {\text{Substitute}\ x = \text{sinh}\mspace{2mu}\theta\ \text{and}\ dx = \text{cosh}\mspace{2mu}\theta d\theta.} \\ {\text{Substitute}\ 1 + \text{sinh}^{2}\theta = \text{cosh}^{2}\theta.} \end{array} \\ & {= {\int{\frac{\text{cosh}\mspace{2mu}\theta}{\sqrt{\text{cosh}^{2}\theta}}d\theta}}} & & & {\sqrt{\text{cosh}^{2}\theta} = \left| {\text{cosh}\mspace{2mu}\theta} \right|} \\ & {= {\int{\frac{\text{cosh}\mspace{2mu}\theta}{\left| {\text{cosh}\mspace{2mu}\theta} \right|}d\theta}}} & & & {\left| {\text{cosh}\mspace{2mu}\theta} \right| = \text{cosh}\mspace{2mu}\theta\ \text{since}\ \text{cosh}\mspace{2mu}\theta > 0\ \text{for all}\ \theta.} \\ & {= {\int{\frac{\text{cosh}\mspace{2mu}\theta}{\text{cosh}\mspace{2mu}\theta}d\theta}}} & & & \text{Simplify.} \\ & {= \int^{}1d\theta} & & & \text{Evaluate the integral.} \\ & {= \theta + C} & & & {\text{Since}\ x = \text{sinh}\mspace{2mu}\theta,\ \text{we know}\ \theta = \text{sinh}^{-1}x.} \\ & {= \text{sinh}^{-1}x + C.} & & & \end{array}$$

这个答案与用换元 $x = \text{tan}\mspace{2mu}\theta$ 所得答案看起来很不一样。为说明两种解法一致，令 $y = \text{sinh}^{-1}x.$ 于是 $\text{sinh}\mspace{2mu} y = x.$ 由该方程可得：

$$\frac{e^{y} - e^{\text{−}y}}{2} = x.$$

两边同乘 $2e^{y}$ 并改写后，该方程变为：

$$e^{2y} - 2xe^{y} - 1 = 0.$$

用二次方程解出 $e^{y}\text{:}$

$$e^{y} = \frac{2x \pm \sqrt{4x^{2} + 4}}{2}.$$

化简得：

$$e^{y} = x \pm \sqrt{x^{2} + 1}.$$

由于 $x - \sqrt{x^{2} + 1} < 0$，必定有 $e^{y} = x + \sqrt{x^{2} + 1}.$ 于是

$$y = \text{ln}\left( {x + \sqrt{x^{2} + 1}} \right).$$

最后，我们得到

$$\text{sinh}^{-1}x = \text{ln}\left( {x + \sqrt{x^{2} + 1}} \right).$$

最后注意到，由于 $x + \sqrt{x^{2} + 1} > 0$，

$$\text{ln}\left( {x + \sqrt{x^{2} + 1}} \right) = \text{ln}\left| {\sqrt{1 + x^{2}} + x} \right|,$$

可见两种方法给出了等价的解。

求曲线 $y = x^{2}$ 在区间 $\lbrack 0,\frac{1}{2}\rbrack$ 上的弧长。

由 $\frac{dy}{dx} = 2x$，弧长由下式给出

$${\int_{0}^{1\text{/}2}{\sqrt{1 + {(2x)}^{2}}\mspace{2mu} dx}} = {\int_{0}^{1\text{/}2}{\sqrt{1 + 4x^{2}}\mspace{2mu} dx}}.$$

为计算该积分，使用换元 $x = \frac{1}{2}\text{tan}\mspace{2mu}\theta$ 和 $dx = \frac{1}{2}\text{sec}^{2}\theta d\theta.$ 我们还需要改变积分限。若 $x = 0$，则 $\theta = 0$；若 $x = \frac{1}{2}$，则 $\theta = \frac{\pi}{4}.$ 于是

$$\begin{array}{clccc} {\int_{0}^{1\text{/}2}{\sqrt{1 + 4x^{2}}\mspace{2mu} dx}} & {= {\int_{0}^{\pi\text{/}4}{\sqrt{1 + \text{tan}^{2}\theta}\frac{1}{2}\text{sec}^{2}\theta d\theta}}} & & & \begin{array}{l} \text{After substitution,} \\ {\sqrt{1 + 4x^{2}} = \text{tan}\mspace{2mu}\theta.\ \text{Substitute}} \\ {1 + \text{tan}^{2}\theta = \text{sec}^{2}\theta\ \text{and simplify.}} \end{array} \\ & {= \frac{1}{2}{\int_{0}^{\pi\text{/}4}{\text{sec}^{3}\theta d\theta}}} & & & \begin{array}{l} \text{We derived this integral in the} \\ \text{previous section.} \end{array} \\ & {= \frac{1}{2}\left( {\frac{1}{2}\text{sec}\mspace{2mu}\theta\mspace{2mu}\text{tan}\mspace{2mu}\theta + \frac{1}{2}\text{ln}\left| {\text{sec}\mspace{2mu}\theta + \text{tan}\mspace{2mu}\theta} \right|} \right)\left| {}_{\begin{matrix} \\ 0 \end{matrix}}^{\begin{matrix} {\pi\text{/}4} \\ \end{matrix}} \right.} & & & \text{Evaluate and simplify.} \\ & {= \frac{1}{4}(\sqrt{2} + \text{ln}(\sqrt{2} + 1)).} & & & \end{array}$$

用含 $\text{tan}\mspace{2mu}\theta$ 的换元改写 $\int^{}x^{3}\sqrt{x^{2} + 4}\mspace{2mu} dx$。

表达式 $\sqrt{x^{2} - a^{2}}$ 的定义域为 $\left( {\text{−}\infty,\text{−}a} \right\rbrack \cup \left\lbrack {a,\text{+}\infty} \right).$ 因此，要么 $x \leq \text{−}a$，要么 $x \geq a.$ 从而 $\frac{x}{a} \leq - 1$ 或 $\frac{x}{a} \geq 1.$ 由于这些区间对应于 $\text{sec}\mspace{2mu}\theta$ 在集合 $\left\lbrack {0,\frac{\pi}{2}} \right) \cup \left( {\frac{\pi}{2},\pi} \right\rbrack$ 上的值域，采用换元 $\text{sec}\mspace{2mu}\theta = \frac{x}{a}$ 或等价地 $x = a\mspace{2mu}\text{sec}\mspace{2mu}\theta$ 是合理的，其中 $0 \leq \theta < \frac{\pi}{2}$ 或 $\frac{\pi}{2} < \theta \leq \pi.$ 对应的 $dx$ 换元为 $dx = a\mspace{2mu}\text{sec}\mspace{2mu}\theta\mspace{2mu}\text{tan}\mspace{2mu}\theta d\theta.$ 使用该换元的步骤在下述解题策略中列出。

1. 先检查该积分是否无法用其他方法计算。若是，我们或许应考虑采用一种替代技巧。

2. 换元 $x = a\mspace{2mu}\text{sec}\mspace{2mu}\theta$ 并取 $dx = a\mspace{2mu}\text{sec}\mspace{2mu}\theta\mspace{2mu}\text{tan}\mspace{2mu}\theta d\theta.$ 这一换元得到

$$\sqrt{x^{2} - a^{2}} = \sqrt{\left( {a\mspace{2mu}\text{sec}\mspace{2mu}\theta} \right)^{2} - a^{2}} = \sqrt{a^{2}(\text{sec}^{2}\theta –1)} = \sqrt{a^{2}\text{tan}^{2}\theta} = \left| {a\mspace{2mu}\text{tan}\mspace{2mu}\theta} \right|.$$

当 $x \geq a$ 时，$\left| {a\mspace{2mu}\text{tan}\mspace{2mu}\theta} \right| = a\mspace{2mu}\text{tan}\mspace{2mu}\theta$；当 $x \leq - a$ 时，$\left| {a\mspace{2mu}\text{tan}\mspace{2mu}\theta} \right| = \text{−}a\mspace{2mu}\text{tan}\mspace{2mu}\theta.$

3. 化简表达式。

4. 利用三角积分一节中的技巧计算该积分。

5. 利用图 3.9 中的参考三角形将结果用 $x$ 表示。你可能还需要用到一些三角恒等式以及关系 $\theta = \text{sec}^{-1}\left( \frac{x}{a} \right).$ （*注*：我们需要两个参考三角形，因为某些三角比的值在 $x \geq a$ 与 $x \leq \text{−}a$ 时不同。）

求图形 $f(x) = \sqrt{x^{2} - 9}$ 与 *x* 轴在区间 $\lbrack 3,5\rbrack$ 之间所围成区域的面积。

首先，如下列图形所示，粗略画出题目所描述区域的图像。

可见面积为 $A = {\int_{3}^{5}{\sqrt{x^{2} - 9}\mspace{2mu} dx.}}$ 为计算这一定积分，换元 $x = 3\mspace{2mu}\text{sec}\mspace{2mu}\theta$ 并取 $dx = 3\mspace{2mu}\text{sec}\mspace{2mu}\theta\mspace{2mu}\text{tan}\mspace{2mu}\theta d\theta.$ 我们还必须改变积分限。若 $x = 3$，则 $3 = 3\mspace{2mu}\text{sec}\mspace{2mu}\theta$，从而 $\theta = 0.$ 若 $x = 5$，则 $\theta = \text{sec}^{-1}\left( \frac{5}{3} \right).$ 进行这些换元并化简后，得到

$$\begin{array}{clccc} \text{Area} & {= {\int_{3}^{5}{\sqrt{x^{2} - 9}\mspace{2mu} dx}}} & & & \\ & {= {\int_{0}^{\text{sec}^{-1}{({5\text{/}3})}}{9\mspace{2mu}\text{tan}^{2}\theta\mspace{2mu}\text{sec}\mspace{2mu}\theta d\theta}}} & & & {\text{Use}\ \text{tan}^{2}\theta = 1 - \text{sec}^{2}\theta.} \\ & {= {\int_{0}^{\text{sec}^{-1}{({5\text{/}3})}}{9(\text{sec}^{2}\theta - 1)\text{sec}\mspace{2mu}\theta d\theta}}} & & & \text{Expand.} \\ & {= {\int_{0}^{\text{sec}^{-1}{({5\text{/}3})}}{9(\text{sec}^{3}\theta - \text{sec}\mspace{2mu}\theta)d\theta}}} & & & \text{Evaluate the integral.} \\ & {= \left( {\frac{9}{2}\text{ln}\left| {\text{sec}\mspace{2mu}\theta + \text{tan}\mspace{2mu}\theta} \right| + \frac{9}{2}\text{sec}\mspace{2mu}\theta\mspace{2mu}\text{tan}\mspace{2mu}\theta} \right) - 9\mspace{2mu}\text{ln}\left| {\text{sec}\mspace{2mu}\theta + \text{tan}\mspace{2mu}\theta} \right|\left| {}_{\begin{matrix} \\ 0 \end{matrix}}^{\begin{matrix} {\text{sec}^{-1}{({5\text{/}3})}} \\ \end{matrix}} \right.} & & & \text{Simplify.} \\ & {= \frac{9}{2}\text{sec}\mspace{2mu}\theta\mspace{2mu}\text{tan}\mspace{2mu}\theta - \frac{9}{2}\text{ln}\left| {\text{sec}\mspace{2mu}\theta + \text{tan}\mspace{2mu}\theta} \right|\left| {}_{\begin{matrix} \\ 0 \end{matrix}}^{\begin{matrix} {\text{sec}^{-1}{({5\text{/}3})}} \\ \end{matrix}} \right.} & & & \begin{array}{l} {\text{Evaluate. Use}\ \text{sec}\left( {\text{sec}^{-1}\frac{5}{3}} \right) = \frac{5}{3}} \\ {\text{and}\ \text{tan}\left( {\text{sec}^{-1}\frac{5}{3}} \right) = \frac{4}{3}.} \end{array} \\ & {= \frac{9}{2} \cdot \frac{5}{3} \cdot \frac{4}{3} - \frac{9}{2}\text{ln}\left| {\frac{5}{3} + \frac{4}{3}} \right| - \left( {\frac{9}{2} \cdot 1 \cdot 0 - \frac{9}{2}\text{ln}\left| {1 + 0} \right|} \right)} & & & \\ & {= 10 - \frac{9}{2}\text{ln}\mspace{2mu} 3.} & & & \end{array}$$

计算 ${\int\frac{dx}{\sqrt{x^{2} - 4}}}.$ 假定 $x > 2$。

通过用单个三角函数表示，化简下列各式。

126\.

$4 - 4\mspace{2mu}\text{sin}^{2}\theta$

127.

$9\mspace{2mu}\text{sec}^{2}\theta - 9$

128\.

$a^{2} + a^{2}\text{tan}^{2}\theta$

129.

$a^{2} + a^{2}\text{sinh}^{2}\theta$

130\.

$16\mspace{2mu}\text{cosh}^{2}\theta - 16$

使用配方法，将每个三项式表示为二项式的平方，或二项式的平方加上一个常数。

131.

$4x^{2} - 4x + 1$

132\.

$2x^{2} - 8x + 3$

133.

$\text{−}x^{2} - 2x + 4$

使用三角换元法积分。将最终答案用该变量表示。

134\.

$\int\frac{dx}{\sqrt{4 - x^{2}}}$

135.

$\int\frac{dx}{\sqrt{x^{2} - a^{2}}}$

136\.

$\int{\sqrt{4 - x^{2}}\mspace{2mu} dx}$

137.

$\int\frac{dx}{\sqrt{1 + 9x^{2}}}$

138\.

$\int\frac{x^{2}dx}{\sqrt{1 - x^{2}}}$

139.

$\int\frac{dx}{x^{2}\sqrt{1 - x^{2}}}$

140\.

$\int\frac{dx}{{(1 + x^{2})}^{2}}$

141.

${\int\sqrt{x^{2} + 9}}dx$

142\.

$\int{\frac{\sqrt{x^{2} - 25}}{x}dx}$

143.

$\int\frac{\theta^{3}d\theta}{\sqrt{9 - \theta^{2}}}$

144\.

$\int\frac{dx}{\sqrt{x^{6} - x^{2}}}$

145.

${\int\sqrt{x^{6} - x^{8}}}dx$

146\.

$\int\frac{dx}{\left( {1 + x^{2}} \right)^{3\text{/}2}}$

147.

$\int\frac{dx}{\left( {x^{2} - 9} \right)^{3\text{/}2}}$

148\.

$\int\frac{\sqrt{1 + x^{2}}\mspace{2mu} dx}{x}$

149.

$\int\frac{x^{2}dx}{\sqrt{x^{2} - 1}}$

150\.

$\int\frac{x^{2}dx}{x^{2} + 4}$

151.

$\int\frac{dx}{x^{2}\sqrt{x^{2} + 1}}$

152\.

$\int\frac{x^{2}dx}{\sqrt{1 + x^{2}}}$

153.

$\int\left( 1 - x^{2} \right)^{3\text{/}2}dx$

在以下习题中，使用换元 $x = \text{sinh}\mspace{2mu}\theta,\text{cosh}\mspace{2mu}\theta,$ 或 $\text{tanh}\mspace{2mu}\theta.$ 将最终答案用变量 *x.* 表示。

154\.

$\int\frac{dx}{\sqrt{x^{2} - 1}}$

155.

$\int\frac{dx}{x\sqrt{1 - x^{2}}}$

156\.

${\int\sqrt{x^{2} - 1}}dx$

157.

$\int{\frac{\sqrt{x^{2} - 1}}{x^{2}}dx}$

158\.

$\int\frac{dx}{1 - x^{2}}$

159.

$\int{\frac{\sqrt{1 + x^{2}}}{x^{2}}dx}$

使用配方法计算下列积分。

160\.

$\int{\frac{1}{x^{2} - 6x}dx}$

161.

$\int{\frac{1}{x^{2} + 2x + 1}dx}$

162\.

$\int{\frac{1}{\sqrt{\text{−}x^{2} + 2x + 8}}dx}$

163.

$\int{\frac{1}{\sqrt{\text{−}x^{2} + 10x}}dx}$

164\.

${\int\frac{1}{\sqrt{x^{2} + 4x - 12}}}dx$

165.

不用微积分计算该积分：$\int_{-3}^{3}{\sqrt{9 - x^{2}}\mspace{2mu} dx.}$

166\.

求椭圆 $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1.$ 所围成的面积。

167.

用两种不同的换元计算积分 $\int\frac{dx}{\sqrt{1 - x^{2}}}$。首先，令 $x = \text{cos}\mspace{2mu}\theta$，并用三角换元进行计算。其次，令 $x = \text{sin}\mspace{2mu}\theta$，并用三角换元。两个答案是否相同？

168\.

使用换元 $x = \text{sec}\mspace{2mu}\theta$ 计算积分 $\int\frac{dx}{x\sqrt{x^{2} - 1}}$。接着，使用换元 $x = \text{csc}\mspace{2mu}\theta$ 计算同一积分。证明两个结果等价。

169.

利用形式 ${\int{\frac{1}{u}du}}$ 计算积分 ${\int\frac{x}{x^{2} + 1}}dx$。接着，使用换元 $x = \text{tan}\mspace{2mu}\theta$ 计算同一积分。两个结果是否相同？

170\.

说明你将用于计算积分 ${\int{x\sqrt{x^{2} + 1}\mspace{2mu} dx}}$ 的积分方法。你为何选择该方法？

171.

说明你将用于计算积分 ${\int{x^{2}\sqrt{x^{2} - 1}\mspace{2mu} dx}}$ 的积分方法。你为何选择该方法？

172\.

计算 $\int_{-1}^{1}\frac{x\mspace{2mu} dx}{x^{2} + 1}$

173.

求曲线在指定区间上的弧长：$y = \text{ln}\mspace{2mu} x,\lbrack 1,5\rbrack.$ 答案保留三位小数。

174\.

求由曲线 $y = x^{2},y = 0,x = 0,\text{and}\ x = \sqrt{2}$ 所围成区域绕 *x*-轴旋转所生成立体的表面积。（答案保留三位小数）。

175.

由 $f(x) = \frac{1}{1 + x^{2}}$ 的图像与 *x*-轴在 $x = 0$ 和 $x = 1$ 之间所围成的区域绕 *x-*轴旋转。求所生成立体的体积。

求解初值问题，将 *y* 表示为 *x* 的函数。

176\.

$\left( {x^{2} + 36} \right)\frac{dy}{dx} = 1,y(6) = 0$

177.

$\left( {64 - x^{2}} \right)\frac{dy}{dx} = 1,y(0) = 3$

178\.

求由 $y = \frac{2}{\sqrt{64 - 4x^{2}}},x = 0,y = 0,\text{and}\ x = 2$ 所围成的面积。

179.

一个储油罐可描述为：由 $y = \frac{16}{\sqrt{64 + x^{2}}},x = 0,y = 0,x = 2$ 所围成的区域绕 *x*-轴旋转所生成的体积。求该油罐的体积（单位：立方米）。

180\.

在每个周期中，一台机器人焊接装置的速率 *v*（单位：英尺每秒）由 $v = 2t - \frac{14}{4 + t^{2}}$ 给出，其中 *t* 为时间（单位：秒）。若 $s = 0$ 当 $t = 0$，求位移 *s*（单位：英尺）关于 *t* 的表达式。

181.

求曲线 $y = \sqrt{16 - x^{2}}$ 在 $x = 0$ 与 $x = 2$ 之间的弧长。

3.4 Partial Fractions 3.4 部分分式

• 3.4.1 Integrate a rational function using the method of partial fractions.
• 3.4.2 Recognize simple linear factors in a rational function.
• 3.4.3 Recognize repeated linear factors in a rational function.
• 3.4.4 Recognize quadratic factors in a rational function.

• 3.4.1 使用部分分式法积分有理函数。
• 3.4.2 识别有理函数中的互异一次因式。
• 3.4.3 识别有理函数中的重复一次因式。
• 3.4.4 识别有理函数中的二次因式。

We have seen some techniques that allow us to integrate specific rational functions. For example, we know that

$${\int\frac{du}{u}} = \text{ln}|u| + C\ \text{and}\ {\int\frac{du}{u^{2} + a^{2}}} = \frac{1}{a}\text{tan}^{-1}\left( \frac{u}{a} \right) + C\text{.}$$

However, we do not yet have a technique that allows us to tackle arbitrary quotients of this type. Thus, it is not immediately obvious how to go about evaluating $\int{\frac{3x}{x^{2} - x - 2}dx.}$ However, we know from material previously developed that

$$\int{\left( {\frac{1}{x + 1} + \frac{2}{x - 2}} \right)dx = \text{ln}\left| {x + 1} \right| + 2\mspace{2mu}\text{ln}\left| {x - 2} \right| + C.}$$

In fact, by getting a common denominator, we see that

$$\frac{1}{x + 1} + \frac{2}{x - 2} = \frac{3x}{x^{2} - x - 2}.$$

Consequently,

$${\int{\frac{3x}{x^{2} - x - 2}dx}} = {\int{\left( {\frac{1}{x + 1} + \frac{2}{x - 2}} \right)dx}}.$$

In this section, we examine the method of partial fraction decomposition, which allows us to decompose rational functions into sums of simpler, more easily integrated rational functions. Using this method, we can rewrite an expression such as: $\frac{3x}{x^{2} - x - 2}$ as an expression such as $\frac{1}{x + 1} + \frac{2}{x - 2}.$

The key to the method of partial fraction decomposition is being able to anticipate the form that the decomposition of a rational function will take. As we shall see, this form is both predictable and highly dependent on the factorization of the denominator of the rational function. It is also extremely important to keep in mind that partial fraction decomposition can be applied to a rational function $\frac{P(x)}{Q(x)}$ only if $\text{deg}\left( {P(x)} \right) < \text{deg}\left( {Q(x)} \right).$ In the case when $\text{deg}\left( {P(x)} \right) \geq \text{deg}\left( {Q(x)} \right),$ we must first perform long division to rewrite the quotient $\frac{P(x)}{Q(x)}$ in the form $A(x) + \frac{R(x)}{Q(x)},$ where $\text{deg}\left( {R(x)} \right) < \text{deg}\left( {Q(x)} \right).$ We then do a partial fraction decomposition on $\frac{R(x)}{Q(x)}.$ The following example, although not requiring partial fraction decomposition, illustrates our approach to integrals of rational functions of the form $\int{\frac{P(x)}{Q(x)}dx,}$ where $\text{deg}\left( {P(x)} \right) \geq \text{deg}\left( {Q(x)} \right).$

Integrating $\int{\frac{P(x)}{Q(x)}dx,}$ where $\text{deg}\left( {P(x)} \right) \geq \text{deg}\left( {Q(x)} \right)$ 积分 $\int{\frac{P(x)}{Q(x)}dx,}$（其中 $\text{deg}\left( {P(x)} \right) \geq \text{deg}\left( {Q(x)} \right)$）

Evaluate ${\int{\frac{x^{2} + 3x + 5}{x + 1}dx}}.$

Solution 解答

Since $\text{deg}\left( {x^{2} + 3x + 5} \right) \geq \text{deg}\left( {x + 1} \right),$ we perform long division to obtain

$$\frac{x^{2} + 3x + 5}{x + 1} = x + 2 + \frac{3}{x + 1}.$$

Thus,

$$\begin{array}{cl} {\int{\frac{x^{2} + 3x + 5}{x + 1}dx}} & {= {\int{\left( {x + 2 + \frac{3}{x + 1}} \right)dx}}} \\ & {= \frac{1}{2}x^{2} + 2x + 3\mspace{2mu}\text{ln}\left| {x + 1} \right| + C.} \end{array}$$

Visit this website for a review of long division of polynomials.

Evaluate $\int{\frac{x - 3}{x + 2}dx.}$

To integrate $\int{\frac{P(x)}{Q(x)}dx,}$ where $\text{deg}\left( {P(x)} \right) < \text{deg}\left( {Q(x)} \right),$ we must begin by factoring $Q(x).$

Nonrepeated Linear Factors 互异一次因式

If $Q(x)$ can be factored as $\left( {a_{1}x + b_{1}} \right)\left( {a_{2}x + b_{2}} \right)\text{…}\left( {a_{n}x + b_{n}} \right),$ where each linear factor is distinct, then it is possible to find constants $A_{1},A_{2}\text{,…}\ A_{n}$ satisfying

$$\frac{P(x)}{Q(x)} = \frac{A_{1}}{a_{1}x + b_{1}} + \frac{A_{2}}{a_{2}x + b_{2}} + \cdots + \frac{A_{n}}{a_{n}x + b_{n}}.$$

The proof that such constants exist is beyond the scope of this course.

In this next example, we see how to use partial fractions to integrate a rational function of this type.

Partial Fractions with Nonrepeated Linear Factors 互异一次因式的部分分式

Evaluate $\int{\frac{3x + 2}{x^{3} - x^{2} - 2x}dx.}$

Solution 解答

Since $\text{deg}(3x + 2) < \text{deg}\left( {x^{3} - x^{2} - 2x} \right),$ we begin by factoring the denominator of $\frac{3x + 2}{x^{3} - x^{2} - 2x}.$ We can see that $x^{3} - x^{2} - 2x = x(x - 2)(x + 1).$ Thus, there are constants $A,$ $B,$ and $C$ satisfying

$$\frac{3x + 2}{x(x - 2)(x + 1)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 1}.$$

We must now find these constants. To do so, we begin by getting a common denominator on the right. Thus,

$$\frac{3x + 2}{x(x - 2)(x + 1)} = \frac{A\left( {x - 2} \right)(x + 1) + Bx\left( {x + 1} \right) + Cx(x - 2)}{x(x - 2)(x + 1)}.$$

Now, we set the numerators equal to each other, obtaining

$$3x + 2 = A\left( {x - 2} \right)\left( {x + 1} \right) + Bx\left( {x + 1} \right) + Cx\left( {x - 2} \right).$$ (3.8)

There are two different strategies for finding the coefficients $A,$ $B,$ and $C.$ We refer to these as the *method of equating coefficients* and the *method of strategic substitution*.

Rewrite Equation 3.8 in the form

$$3x + 2 = \left( {A + B + C} \right)x^{2} + \left( {\text{−}A + B - 2C} \right)x + \left( {-2A} \right).$$

Equating coefficients produces the system of equations

$$\begin{array}{rll} {A + B + C} & = & 0 \\ {- A + B - 2C} & = & 3 \\ {- 2A} & = & 2. \end{array}$$

To solve this system, we first observe that $-2A = 2\Rightarrow A = -1.$ Substituting this value into the first two equations gives us the system

$$\begin{array}{rll} {B + C} & = & 1 \\ {B - 2C} & = & 2. \end{array}$$

Multiplying the second equation by $-1$ and adding the resulting equation to the first produces

$$-3C = 1,$$

which in turn implies that $C = - \frac{1}{3}.$ Substituting this value into the equation $B + C = 1$ yields $B = \frac{4}{3}.$ Thus, solving these equations yields $A = -1,$ $B = \frac{4}{3},$ and $C = - \frac{1}{3}.$

It is important to note that the system produced by this method is consistent if and only if we have set up the decomposition correctly. If the system is inconsistent, there is an error in our decomposition.

The method of strategic substitution is based on the assumption that we have set up the decomposition correctly. If the decomposition is set up correctly, then there must be values of $A,$ $B,$ and $C$ that satisfy Equation 3.8 for *all* values of $x.$ That is, this equation must be true for any value of $x$ we care to substitute into it. Therefore, by choosing values of $x$ carefully and substituting them into the equation, we may find $A,$ $B,$ and $C$ easily. For example, if we substitute $x = 0,$ the equation reduces to $2 = A(-2)(1).$ Solving for $A$ yields $A = -1.$ Next, by substituting $x = 2,$ the equation reduces to $8 = B(2)(3),$ or equivalently $B = 4\text{/}3.$ Last, we substitute $x = -1$ into the equation and obtain $-1 = C(-1)(-3).$ Solving, we have $C = - \frac{1}{3}.$

It is important to keep in mind that if we attempt to use this method with a decomposition that has not been set up correctly, we are still able to find values for the constants, but these constants are meaningless. If we do opt to use the method of strategic substitution, then it is a good idea to check the result by recombining the terms algebraically.

Now that we have the values of $A,$ $B,$ and $C,$ we rewrite the original integral:

$${\int{\frac{3x + 2}{x^{3} - x^{2} - 2x}dx}} = {\int\left( {\text{−}\ \frac{1}{x} + \frac{4}{3} \cdot \frac{1}{\left( {x - 2} \right)} - \frac{1}{3} \cdot \frac{1}{(x + 1)}} \right)}dx.$$

Evaluating the integral gives us

$${\int{\frac{3x + 2}{x^{3} - x^{2} - 2x}dx}} = \text{−}\text{ln}|x| + \frac{4}{3}\text{ln}\left| {x - 2} \right| - \frac{1}{3}\text{ln}\left| {x + 1} \right| + C.$$

In the next example, we integrate a rational function in which the degree of the numerator is not less than the degree of the denominator.

Dividing before Applying Partial Fractions 应用部分分式前的除法

Evaluate $\int{\frac{x^{2} + 3x + 1}{x^{2} - 4}dx.}$

Solution 解答

Since $\text{degree}(x^{2} + 3x + 1) \geq \text{degree}(x^{2} - 4),$ we must perform long division of polynomials. This results in

$$\frac{x^{2} + 3x + 1}{x^{2} - 4} = 1 + \frac{3x + 5}{x^{2} - 4}.$$

Next, we perform partial fraction decomposition on $\frac{3x + 5}{x^{2} - 4} = \frac{3x + 5}{(x + 2)(x - 2)}.$ We have

$$\frac{3x + 5}{(x - 2)(x + 2)} = \frac{A}{x - 2} + \frac{B}{x + 2}.$$

Thus,

$$3x + 5 = A\left( {x + 2} \right) + B\left( {x - 2} \right).$$

Solving for $A$ and $B$ using either method, we obtain $A = 11\text{/}4$ and $B = 1\text{/}4.$

Rewriting the original integral, we have

$${\int{\frac{x^{2} + 3x + 1}{x^{2} - 4}dx}} = {\int{\left( {1 + \frac{11}{4} \cdot \frac{1}{x - 2} + \frac{1}{4} \cdot \frac{1}{x + 2}} \right)dx.}}$$

Evaluating the integral produces

$${\int{\frac{x^{2} + 3x + 1}{x^{2} - 4}dx}} = x + \frac{11}{4}\text{ln}\left| {x - 2} \right| + \frac{1}{4}\text{ln}\left| {x + 2} \right| + C.$$

As we see in the next example, it may be possible to apply the technique of partial fraction decomposition to a nonrational function. The trick is to convert the nonrational function to a rational function through a substitution.

Applying Partial Fractions after a Substitution 换元后应用部分分式

Evaluate ${\int{\frac{\text{cos}\mspace{2mu} x}{\text{sin}^{2}x - \text{sin}\mspace{2mu} x}dx}}.$

Solution 解答

Let's begin by letting $u = \text{sin}\mspace{2mu} x.$ Consequently, $du = \text{cos}\mspace{2mu} x\mspace{2mu} dx.$ After making these substitutions, we have

$${\int{\frac{\text{cos}\mspace{2mu} x}{\text{sin}^{2}x - \text{sin}\mspace{2mu} x}dx}} = {\int\frac{du}{u^{2} - u}} = {\int{\frac{du}{u(u - 1)}.}}$$

Applying partial fraction decomposition to $1\text{/}u(u - 1)$ gives $\frac{1}{u(u - 1)} = - \frac{1}{u} + \frac{1}{u - 1}.$

Thus,

$$\begin{array}{cl} {\int{\frac{\text{cos}\mspace{2mu} x}{\text{sin}^{2}x - \text{sin}\mspace{2mu} x}dx}} & {= \text{−}\text{ln}|u| + \text{ln}\left| {u - 1} \right| + C} \\ & {= \text{−}\text{ln}\left| {\text{sin}\mspace{2mu} x} \right| + \text{ln}\left| {\text{sin}\mspace{2mu} x - 1} \right| + C.} \end{array}$$

Evaluate ${\int{\frac{x + 1}{(x + 3)(x - 2)}dx}}.$

Repeated Linear Factors 重复的一次因式

For some applications, we need to integrate rational expressions that have denominators with repeated linear factors—that is, rational functions with at least one factor of the form ${(ax + b)}^{n},$ where $n$ is a positive integer greater than or equal to $2.$ If the denominator contains the repeated linear factor ${(ax + b)}^{n},$ then the decomposition must contain

$$\frac{A_{1}}{ax + b} + \frac{A_{2}}{{(ax + b)}^{2}} + \cdots + \frac{A_{n}}{{(ax + b)}^{n}}.$$

As we see in our next example, the basic technique used for solving for the coefficients is the same, but it requires more algebra to determine the numerators of the partial fractions.

Partial Fractions with Repeated Linear Factors 含重复一次因式的部分分式

Evaluate $\int{\frac{x - 2}{\left( {2x - 1} \right)^{2}(x - 1)}dx.}$

Solution 解答

We have $\text{degree}\left( {x - 2} \right) < \text{degree}\left( {\left( {2x - 1} \right)^{2}\left( {x - 1} \right)} \right),$ so we can proceed with the decomposition. Since ${(2x - 1)}^{2}$ is a repeated linear factor, include $\frac{A}{2x - 1} + \frac{B}{{(2x - 1)}^{2}}$ in the decomposition. Thus,

$$\frac{x - 2}{\left( {2x - 1} \right)^{2}(x - 1)} = \frac{A}{2x - 1} + \frac{B}{{(2x - 1)}^{2}} + \frac{C}{x - 1}.$$

After getting a common denominator and equating the numerators, we have

$$x - 2 = A\left( {2x - 1} \right)\left( {x - 1} \right) + B\left( {x - 1} \right) + C{(2x - 1)}^{2}.$$ (3.9)

We then use the method of equating coefficients to find the values of $A,$ $B,$ and $C.$

$$x - 2 = \left( {2A + 4C} \right)x^{2} + \left( {-3A + B - 4C} \right)x + \left( {A - B + C} \right).$$

Equating coefficients yields $2A + 4C = 0,$ $-3A + B - 4C = 1,$ and $A - B + C = -2.$ Solving this system yields $A = 2,$ $B = 3,$ and $C = -1.$

Alternatively, we can use the method of strategic substitution. In this case, substituting $x = 1$ and $x = 1\text{/}2$ into Equation 3.9 easily produces the values $B = 3$ and $C = -1.$ At this point, it may seem that we have run out of good choices for $x,$ however, since we already have values for $B$ and $C,$ we can substitute in these values and choose any value for $x$ not previously used. The value $x = 0$ is a good option. In this case, we obtain the equation $-2 = A(-1)(-1) + 3(-1) + (-1){(-1)}^{2}$ or, equivalently, $A = 2.$

Now that we have the values for $A,$ $B,$ and $C,$ we rewrite the original integral and evaluate it:

$$\begin{array}{cl} {\int{\frac{x - 2}{\left( {2x - 1} \right)^{2}(x - 1)}dx}} & {= {\int{\left( {\frac{2}{2x - 1} + \frac{3}{{(2x - 1)}^{2}} - \frac{1}{x - 1}} \right)dx}}} \\ & {= \text{ln}\left| {2x - 1} \right| - \frac{3}{2\left( {2x - 1} \right)} - \text{ln}\left| {x - 1} \right| + C.} \end{array}$$

Set up the partial fraction decomposition for $\int{\frac{x + 2}{\left( {x + 3} \right)^{3}{(x - 4)}^{2}}dx.}$ (Do not solve for the coefficients or complete the integration.)

The General Method 一般方法

Now that we are beginning to get the idea of how the technique of partial fraction decomposition works, let’s outline the basic method in the following problem-solving strategy.

Partial Fraction Decomposition 部分分式分解

To decompose the rational function $P(x)\text{/}Q(x),$ use the following steps:

1. Make sure that $\text{degree}\left( {P(x)} \right) < \text{degree}(Q(x)).$ If not, perform long division of polynomials.

2. Factor $Q(x)$ into the product of linear and irreducible quadratic factors. An irreducible quadratic is a quadratic that has no real zeros.

3. Assuming that $\text{deg}\left( {P(x)} \right) < \text{deg}(Q(x)),$ the factors of $Q(x)$ determine the form of the decomposition of $P(x)\text{/}Q(x).$

1. If $Q(x)$ can be factored as $\left( {a_{1}x + b_{1}} \right)\left( {a_{2}x + b_{2}} \right)\text{…}\left( {a_{n}x + b_{n}} \right),$ where each linear factor is distinct, then it is possible to find constants $A_{1},A_{2},...A_{n}$ satisfying

$$\frac{P(x)}{Q(x)} = \frac{A_{1}}{a_{1}x + b_{1}} + \frac{A_{2}}{a_{2}x + b_{2}} + \cdots + \frac{A_{n}}{a_{n}x + b_{n}}.$$

2. If $Q(x)$ contains the repeated linear factor ${(ax + b)}^{n},$ then the decomposition must contain

$$\frac{A_{1}}{ax + b} + \frac{A_{2}}{{(ax + b)}^{2}} + \cdots + \frac{A_{n}}{{(ax + b)}^{n}}.$$

3. For each irreducible quadratic factor $ax^{2} + bx + c$ that $Q(x)$ contains, the decomposition must include

$$\frac{Ax + B}{ax^{2} + bx + c}.$$

4. For each repeated irreducible quadratic factor $\left( {ax^{2} + bx + c} \right)^{n},$ the decomposition must include

$$\frac{A_{1}x + B_{1}}{ax^{2} + bx + c} + \frac{A_{2}x + B_{2}}{{(ax^{2} + bx + c)}^{2}} + \cdots + \frac{A_{n}x + B_{n}}{{(ax^{2} + bx + c)}^{n}}.$$

5. After the appropriate decomposition is determined, solve for the constants.

6. Last, rewrite the integral in its decomposed form and evaluate it using previously developed techniques or integration formulas.

Simple Quadratic Factors 简单二次因式

Now let’s look at integrating a rational expression in which the denominator contains an irreducible quadratic factor. Recall that the quadratic $ax^{2} + bx + c$ is irreducible if $ax^{2} + bx + c = 0$ has no real zeros—that is, if $b^{2} - 4ac < 0.$

Rational Expressions with an Irreducible Quadratic Factor 含不可约二次因式的有理表达式

Evaluate $\int{\frac{2x - 3}{x^{3} + x}dx.}$

Solution 解答

Since $\text{deg}(2x - 3) < \text{deg}(x^{3} + x),$ factor the denominator and proceed with partial fraction decomposition. Since $x^{3} + x = x(x^{2} + 1)$ contains the irreducible quadratic factor $x^{2} + 1,$ include $\frac{Ax + B}{x^{2} + 1}$ as part of the decomposition, along with $\frac{C}{x}$ for the linear term $x.$ Thus, the decomposition has the form

$$\frac{2x - 3}{x(x^{2} + 1)} = \frac{Ax + B}{x^{2} + 1} + \frac{C}{x}.$$

After getting a common denominator and equating the numerators, we obtain the equation

$$2x - 3 = \left( {Ax + B} \right)x + C\left( {x^{2} + 1} \right).$$

Solving for $A,B,$ and $C,$ we get $A = 3,$ $B = 2,$ and $C = -3.$

Thus,

$$\frac{2x - 3}{x^{3} + x} = \frac{3x + 2}{x^{2} + 1} - \frac{3}{x}.$$

Substituting back into the integral, we obtain

$$\begin{array}{clccc} {\int{\frac{2x - 3}{x^{3} + x}dx}} & {= {\int{\left( {\frac{3x + 2}{x^{2} + 1} - \frac{3}{x}} \right)dx}}} & & & \\ & {= 3{\int{\frac{x}{x^{2} + 1}dx}} + 2{\int{\frac{1}{x^{2} + 1}dx}} - 3{\int{\frac{1}{x}dx}}} & & & \text{Split up the integral.} \\ & {= \frac{3}{2}\text{ln}\left| {x^{2} + 1} \right| + 2\mspace{2mu}\text{tan}^{-1}x - 3\mspace{2mu}\text{ln}|x| + C.} & & & \text{Evaluate each integral.} \end{array}$$

*Note*: We may rewrite $\text{ln}\left| {x^{2} + 1} \right| = \text{ln}(x^{2} + 1),$ if we wish to do so, since $x^{2} + 1 > 0.$

Partial Fractions with an Irreducible Quadratic Factor 含不可约二次因式的部分分式

Evaluate $\int{\frac{dx}{x^{3} - 8}.}$

Solution 解答

We can start by factoring $x^{3} - 8 = (x - 2)(x^{2} + 2x + 4).$ We see that the quadratic factor $x^{2} + 2x + 4$ is irreducible since $2^{2} - 4(1)(4) = -12 < 0.$ Using the decomposition described in the problem-solving strategy, we get

$$\frac{1}{(x - 2)(x^{2} + 2x + 4)} = \frac{A}{x - 2} + \frac{Bx + C}{x^{2} + 2x + 4}.$$

After obtaining a common denominator and equating the numerators, this becomes

$$1 = A\left( {x^{2} + 2x + 4} \right) + \left( {Bx + C} \right)\left( {x - 2} \right).$$

Applying either method, we get $A = \frac{1}{12},B = - \frac{1}{12},\ \text{and}\ C = - \frac{1}{3}.$

Rewriting $\int{\frac{dx}{x^{3} - 8},}$ we have

$${\int\frac{dx}{x^{3} - 8}} = \frac{1}{12}{\int{\frac{1}{x - 2}dx}} - \frac{1}{12}{\int{\frac{x + 4}{x^{2} + 2x + 4}dx.}}$$

We can see that

${\int{\frac{1}{x - 2}dx}} = \text{ln}\left| {x - 2} \right| + C,$ but $\int{\frac{x + 4}{x^{2} + 2x + 4}dx}$ requires a bit more effort. Let’s begin by completing the square on $x^{2} + 2x + 4$ to obtain

$$x^{2} + 2x + 4 = {(x + 1)}^{2} + 3.$$

By letting $u = x + 1$ and consequently $du = dx,$ we see that

$$\begin{array}{clccl} {\int{\frac{x + 4}{x^{2} + 2x + 4}dx}} & {= {\int{\frac{x + 4}{{(x + 1)}^{2} + 3}dx}}} & & & \begin{array}{l} \text{Complete the square on the} \\ \text{denominator.} \end{array} \\ & {= {\int{\frac{u + 3}{u^{2} + 3}du}}} & & & \begin{array}{l} {\text{Substitute}\ u = x + 1,x = u - 1,} \\ {\text{and}\ du = dx.} \end{array} \\ & {= {\int{\frac{u}{u^{2} + 3}du}} + {\int{\frac{3}{u^{2} + 3}du}}} & & & \text{Split the numerator apart.} \\ & {= \frac{1}{2}\text{ln}\left| {u^{2} + 3} \right| + \frac{3}{\sqrt{3}}\text{tan}^{-1}\frac{u}{\sqrt{3}} + C} & & & \text{Evaluate each integral.} \\ & {= \frac{1}{2}\text{ln}\left| {x^{2} + 2x + 4} \right| + \sqrt{3}\text{tan}^{-1}\left( \frac{x + 1}{\sqrt{3}} \right) + C.} & & & \begin{array}{l} {\text{Rewrite in terms of}\ x\ \text{and}} \\ \text{simplify.} \end{array} \end{array}$$

Substituting back into the original integral and simplifying gives

$$\int^{}\frac{dx}{x^{3} - 8} = \frac{1}{12}\text{ln}\left| {x - 2} \right| - \frac{1}{24}\text{ln}\left| {x^{2} + 2x + 4} \right| - \frac{\sqrt{3}}{12}\text{tan}^{-1}\left( \frac{x + 1}{\sqrt{3}} \right) + C.$$

Here again, we can drop the absolute value if we wish to do so, since $x^{2} + 2x + 4 > 0$ for all $x.$

Finding a Volume 求体积

Find the volume of the solid of revolution obtained by revolving the region enclosed by the graph of $f(x) = \frac{x^{2}}{\left( {x^{2} + 1} \right)^{2}}$ and the *x*-axis over the interval $\left\lbrack {0,1} \right\rbrack$ about the *y*-axis.

Solution 解答

Let’s begin by sketching the region to be revolved (see Figure 3.11). From the sketch, we see that the shell method is a good choice for solving this problem.

The volume is given by

$$V = 2\pi{\int_{0}^{1}{x \cdot \frac{x^{2}}{\left( {x^{2} + 1} \right)^{2}}dx}} = 2\pi{\int_{0}^{1}{\frac{x^{3}}{{(x^{2} + 1)}^{2}}dx.}}$$

Since $\text{deg}\left( \left( {x^{2} + 1} \right)^{2} \right) = 4 > 3 = \text{deg}(x^{3}),$ we can proceed with partial fraction decomposition. Note that ${(x^{2} + 1)}^{2}$ is a repeated irreducible quadratic. Using the decomposition described in the problem-solving strategy, we get

$$\frac{x^{3}}{{(x^{2} + 1)}^{2}} = \frac{Ax + B}{x^{2} + 1} + \frac{Cx + D}{{(x^{2} + 1)}^{2}}.$$

Finding a common denominator and equating the numerators gives

$$x^{3} = \left( {Ax + B} \right)\left( {x^{2} + 1} \right) + Cx + D.$$

Solving, we obtain $A = 1,$ $B = 0,$ $C = -1,$ and $D = 0.$ Substituting back into the integral, we have

$$\begin{array}{cl} V & {= 2\pi{\int\limits_{0}^{1}{\frac{x^{3}}{{(x^{2} + 1)}^{2}}dx}}} \\ & {= 2\pi{\int\limits_{0}^{1}{\left( {\frac{x}{x^{2} + 1} - \frac{x}{{(x^{2} + 1)}^{2}}} \right)dx}}} \\ & {= 2\pi\left( {\frac{1}{2}\text{ln}(x^{2} + 1) + \frac{1}{2} \cdot \frac{1}{x^{2} + 1}} \right)\left| {}_{\begin{matrix} \\ 0 \end{matrix}}^{\begin{matrix} 1 \\ \end{matrix}} \right.} \\ & {= \pi\left( {\text{ln}\mspace{2mu} 2 - \frac{1}{2}} \right).} \end{array}$$

Set up the partial fraction decomposition for $\int{\frac{x^{2} + 3x + 1}{(x + 2){(x - 3)}^{2}{(x^{2} + 4)}^{2}}dx.}$

Section 3.4 Exercises 3.4 节习题

Express the rational function as a sum or difference of two simpler rational expressions.

182\.

$\frac{1}{(x - 3)(x - 2)}$

183.

$\frac{x^{2} + 1}{x(x + 1)(x + 2)}$

184\.

$\frac{1}{x^{3} - x}$

185.

$\frac{3x + 1}{x^{2}}$

186\.

$\frac{3x^{2}}{x^{2} + 1}$ (*Hint:* Use long division first.)

187.

$\frac{2x^{4}}{x^{2} - 2x}$

188\.

$\frac{1}{(x - 1)(x^{2} + 1)}$

189.

$\frac{1}{x^{2}(x - 1)}$

190\.

$\frac{x}{x^{2} - 4}$

191.

$\frac{1}{x(x - 1)(x - 2)(x - 3)}$

192\.

$\frac{1}{x^{4} - 1} = \frac{1}{(x + 1)(x - 1)\left( {x^{2} + 1} \right)}$

193.

$\frac{3x^{2}}{x^{3} - 1} = \frac{3x^{2}}{(x - 1)(x^{2} + x + 1)}$

194\.

$\frac{2x}{{(x + 2)}^{2}}$

195.

$\frac{3x^{4} + x^{3} + 20x^{2} + 3x + 31}{(x + 1)\left( {x^{2} + 4} \right)^{2}}$

Use the method of partial fractions to evaluate each of the following integrals.

196\.

$\int\frac{dx}{(x - 3)(x - 2)}$

197.

$\int{\frac{3x}{x^{2} + 2x - 8}dx}$

198\.

$\int\frac{dx}{x^{3} - x}$

199.

$\int{\frac{x}{x^{2} - 4}dx}$

200\.

$\int\frac{dx}{x(x - 1)(x - 2)(x - 3)}$

201.

$\int{\frac{2x^{2} + 4x + 22}{x^{2} + 2x + 10}dx}$

202\.

$\int\frac{dx}{x^{2} - 5x + 6}$

203.

$\int{\frac{2 - x}{x^{2} + x}dx}$

204\.

$\int{\frac{2}{x^{2} - x - 6}dx}$

205.

$\int\frac{dx}{x^{3} - 2x^{2} - 4x + 8}$

206\.

$\int\frac{dx}{x^{4} - 10x^{2} + 9}$

Evaluate the following integrals, which have irreducible quadratic factors.

207.

$\int{\frac{2}{(x - 4)\left( {x^{2} + 2x + 6} \right)}dx}$

208\.

$\int{\frac{x^{2}}{x^{3} - x^{2} + 4x - 4}dx}$

209.

$\int{\frac{x^{3} + 6x^{2} + 3x + 6}{x^{3} + 2x^{2}}dx}$

210\.

$\int{\frac{x}{(x - 1)\left( {x^{2} + 2x + 2} \right)^{2}}dx}$

Use the method of partial fractions to evaluate the following integrals.

211.

$\int{\frac{3x + 4}{\left( {x^{2} + 4} \right)(3 - x)}dx}$

212\.

$\int{\frac{2}{{(x + 2)}^{2}(2 - x)}dx}$

213.

$\int{\frac{3x + 4}{x^{3} - 2x - 4}dx}$ (*Hint:* Use the rational root theorem.)

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals.

214\.

$\int_{0}^{1}{\frac{e^{x}}{36 - e^{2x}}dx}$ (Give the exact answer and the decimal equivalent. Round to five decimal places.)

215.

$\int{\frac{e^{x}}{e^{2x} - e^{x}}dx}$

216\.

$\int\frac{\text{sin}\mspace{2mu} x\mspace{2mu} dx}{1 - \text{cos}^{2}x}$

217.

$\int{\frac{\text{sin}\mspace{2mu} x}{\text{cos}^{2}x + \text{cos}\mspace{2mu} x - 6}dx}$

218\.

$\int{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}dx}$

219.

$\int\frac{dt}{\left( {e^{t} - e^{\text{−}t}} \right)^{2}}$

220\.

$\int{\frac{1 + e^{x}}{1 - e^{x}}dx}$

221.

$\int\frac{dx}{1 + \sqrt{x + 1}}$

222\.

$\int\frac{dx}{\sqrt{x} + \sqrt[4]{x}}$

223.

$\int{\frac{\text{cos}\mspace{2mu} x}{\text{sin}\mspace{2mu} x(1 - \text{sin}\mspace{2mu} x)}dx}$

224\.

$\int{\frac{e^{x}}{\left( {e^{2x} - 4} \right)^{2}}dx}$

225.

$\int\limits_{1}^{2}{\frac{1}{x^{2}\sqrt{4 - x^{2}}}dx}$

226\.

$\int{\frac{1}{2 + e^{\text{−}x}}dx}$

227.

$\int{\frac{1}{1 + e^{x}}dx}$

Use the given substitution to convert the integral to an integral of a rational function, then evaluate.

228\.

${\int{\frac{1}{t - \sqrt[3]{t}}dt}}\mspace{2mu} t = x^{3}$

229.

${\int{\frac{1}{\sqrt{x} + \sqrt[3]{x}}dx}};x = u^{6}$

230\.

Graph the curve $y = \frac{x}{1 + x}$ over the interval $\left\lbrack {0,5} \right\rbrack.$ Then, find the area of the region bounded by the curve, the *x*-axis, and the line $x = 4.$

231.

Find the volume of the solid generated when the region bounded by $y = 1\text{/}\sqrt{x(3 - x)},$ $y = 0,$ $x = 1,$ and $x = 2$ is revolved about the *x-*axis.

232\.

The velocity of a particle moving along a line is a function of time given by $v(t) = \frac{88t^{2}}{t^{2} + 1}.$ Find the distance that the particle has traveled after $t = 5$ sec.

Solve the initial-value problem for *x* as a function of *t.*

233.

$\left( {t^{2} - 7t + 12} \right)\frac{dx}{dt} = 1,\left( {t > 4,x(5) = 0} \right)$

234\.

$(t + 5)\frac{dx}{dt} = x^{2} + 1,t > \text{−}5,x(1) = \text{tan}\mspace{2mu} 1$

235.

$\left( {2t^{3} - 2t^{2} + t - 1} \right)\frac{dx}{dt} = 3,x(2) = 0$

236\.

Find the *x*-coordinate of the centroid of the area bounded by

$y\left( {x^{2} - 9} \right) = 1,$ $y = 0,x = 4,\ \text{and}\ x = 5.$ (Round the answer to two decimal places.)

237.

Find the volume generated by revolving the area bounded by $y = \frac{1}{x^{3} + 7x^{2} + 6x}\text{,}\ x = 1,x = 7,\ \text{and}\ y = 0$ about the *y*-axis.

238\.

Find the area bounded by $y = \frac{x - 12}{x^{2} - 8x - 20},$ $y = 0,x = 2,\ \text{and}\ x = 4.$ (Round the answer to the nearest hundredth.)

239.

Evaluate the integral $\int{\frac{dx}{x^{3} + 1}.}$

For the following problems, use the substitutions $\text{tan}\left( \frac{x}{2} \right) = t,$ $dx = \frac{2}{1 + t^{2}}dt,$ $\text{sin}\mspace{2mu} x = \frac{2t}{1 + t^{2}},$ and $\text{cos}\mspace{2mu} x = \frac{1 - t^{2}}{1 + t^{2}}.$

240\.

$\int\frac{dx}{3 - 5\mspace{2mu}\text{sin}\mspace{2mu} x}$

241.

Find the area under the curve $y = \frac{1}{1 + \text{sin}\mspace{2mu} x}$ between $x = 0$ and $x = \pi.$ (Assume the dimensions are in inches.)

242\.

Given $\text{tan}\left( \frac{x}{2} \right) = t,$ derive the formulas $dx = \frac{2}{1 + t^{2}}dt,$ $\text{sin}\mspace{2mu} x = \frac{2t}{1 + t^{2}},$ and $\text{cos}\mspace{2mu} x = \frac{1 - t^{2}}{1 + t^{2}}.$

243.

Evaluate ${\int{\frac{\sqrt[3]{x - 8}}{x}dx}}.$

3.5 Other Strategies for Integration 3.5 其他的积分策略

• 3.5.1 Use a table of integrals to solve integration problems.
• 3.5.2 Use a computer algebra system (CAS) to solve integration problems.

• 3.5.1 使用积分表求解积分问题。
• 3.5.2 使用计算机代数系统（CAS）求解积分问题。

In addition to the techniques of integration we have already seen, several other tools are widely available to assist with the process of integration. Among these tools are integration tables, which are readily available in many books, including the appendices to this one. Also widely available are computer algebra systems (CAS), which are found on calculators and in many campus computer labs, and are free online.

Tables of Integrals 积分表

Integration tables, if used in the right manner, can be a handy way either to evaluate or check an integral quickly. Keep in mind that when using a table to check an answer, it is possible for two completely correct solutions to look very different. For example, in Trigonometric Substitution, we found that, by using the substitution $x = \text{tan}\mspace{2mu}\theta,$ we can arrive at

$${\int\frac{dx}{\sqrt{1 + x^{2}}}} = \text{ln}\left( {x + \sqrt{x^{2} + 1}} \right) + C.$$

However, using $x = \text{sinh}\mspace{2mu}\theta,$ we obtained a different solution—namely,

$${\int\frac{dx}{\sqrt{1 + x^{2}}}} = \text{sinh}^{-1}x + C.$$

We later showed algebraically that the two solutions are equivalent. That is, we showed that $\text{sinh}^{-1}x = \text{ln}\left( {x + \sqrt{x^{2} + 1}} \right).$ In this case, the two antiderivatives that we found were actually equal. This need not be the case. However, as long as the difference in the two antiderivatives is a constant, they are equivalent.

Using a Formula from a Table to Evaluate an Integral 利用积分表中的公式求积分

Use the table formula

$${\int{\frac{\sqrt{a^{2} - u^{2}}}{u^{2}}du}} = - \frac{\sqrt{a^{2} - u^{2}}}{u} - \text{sin}^{-1}\frac{u}{a} + C$$

to evaluate $\int{\frac{\sqrt{16 - e^{2x}}}{e^{x}}dx.}$

Solution 解答

If we look at integration tables, we see that several formulas contain expressions of the form $\sqrt{a^{2} - u^{2}}.$ This expression is actually similar to $\sqrt{16 - e^{2x}},$ where $a = 4$ and $u = e^{x}.$ Keep in mind that we must also have $du = e^{x}{dx}.$ Multiplying the numerator and the denominator of the given integral by $e^{x}$ should help to put this integral in a useful form. Thus, we now have

$${\int{\frac{\sqrt{16 - e^{2x}}}{e^{x}}dx}} = {\int{\frac{\sqrt{16 - e^{2x}}}{e^{2x}}e^{x}dx.}}$$

Substituting $u = e^{x}$ and $du = e^{x}{dx}$ produces $\int{\frac{\sqrt{a^{2} - u^{2}}}{u^{2}}du.}$ From the integration table (#88 in Appendix A),

$${\int{\frac{\sqrt{a^{2} - u^{2}}}{u^{2}}du}} = - \frac{\sqrt{a^{2} - u^{2}}}{u} - \text{sin}^{-1}\frac{u}{a} + C.$$

Thus,

$$\begin{array}{clccc} {\int{\frac{\sqrt{16 - e^{2x}}}{e^{x}}dx}} & {= {\int{\frac{\sqrt{16 - e^{2x}}}{e^{2x}}e^{x}dx}}} & & & {\text{Substitute}\ u = e^{x}\ \text{and}\ du = e^{x}dx.} \\ & {= {\int{\frac{\sqrt{4^{2} - u^{2}}}{u^{2}}du}}} & & & {\text{Apply the formula using}\ a = 4.} \\ & {= - \frac{\sqrt{4^{2} - u^{2}}}{u} - \text{sin}^{-1}\frac{u}{4} + C} & & & {\text{Substitute}\ u = e^{x}.} \\ & {= - \frac{\sqrt{16 - e^{2x}}}{e^{x}} - \text{sin}^{-1}\left( \frac{e^{x}}{4} \right) + C.} & & & \end{array}$$

Computer Algebra Systems 计算机代数系统

If available, a CAS is a faster alternative to a table for solving an integration problem. Many such systems are widely available and are, in general, quite easy to use.

Using a Computer Algebra System to Evaluate an Integral 使用计算机代数系统求积分

Use a computer algebra system to evaluate $\int{\frac{dx}{\sqrt{x^{2} - 4}}.}$ Compare this result with $\text{ln}\left| {\frac{\sqrt{x^{2} - 4}}{2} + \frac{x}{2}} \right| + C,$ a result we might have obtained if we had used trigonometric substitution.

Solution 解答

Using Wolfram Alpha, we obtain

$${\int\frac{dx}{\sqrt{x^{2} - 4}}} = \text{ln}\left| {\sqrt{x^{2} - 4} + x} \right| + C.$$

Notice that

$$\text{ln}\left| {\frac{\sqrt{x^{2} - 4}}{2} + \frac{x}{2}} \right| + C = \text{ln}\left| \frac{\sqrt{x^{2} - 4} + x}{2} \right| + C = \text{ln}\left| {\sqrt{x^{2} - 4} + x} \right| - \text{ln}\mspace{2mu} 2 + C.$$

Since these two antiderivatives differ by only a constant, the solutions are equivalent. We could have also demonstrated that each of these antiderivatives is correct by differentiating them.

You can access an integral calculator for more examples.

Using a CAS to Evaluate an Integral 使用 CAS 求积分

Evaluate $\int^{}\text{sin}^{3}x\mspace{2mu} dx$ using a CAS. Compare the result to $\frac{1}{3}\text{cos}^{3}x - \text{cos}\mspace{2mu} x + C,$ the result we might have obtained using the technique for integrating odd powers of $\text{sin}\mspace{2mu} x$ discussed earlier in this chapter.

Solution 解答

Using Wolfram Alpha, we obtain

$${\int{\text{sin}^{3}x\mspace{2mu} dx}} = \frac{1}{12}\left( {\text{cos}\left( {3x} \right) - 9\mspace{2mu}\text{cos}\mspace{2mu} x} \right) + C.$$

This looks quite different from $\frac{1}{3}\text{cos}^{3}x - \text{cos}\mspace{2mu} x + C.$ To see that these antiderivatives are equivalent, we can make use of a few trigonometric identities:

$$\begin{array}{cl} {\frac{1}{12}(\text{cos}\left( {3x} \right) - 9\mspace{2mu}\text{cos}\mspace{2mu} x)} & {= \frac{1}{12}(\text{cos}\left( {x + 2x} \right) - 9\mspace{2mu}\text{cos}\mspace{2mu} x)} \\ & {= \frac{1}{12}(\text{cos}(x)\text{cos}\left( {2x} \right) - \text{sin}(x)\text{sin}\left( {2x} \right) - 9\mspace{2mu}\text{cos}\mspace{2mu} x)} \\ & {= \frac{1}{12}(\text{cos}\mspace{2mu} x\left( {2\mspace{2mu}\text{cos}^{2}x - 1} \right) - \text{sin}\mspace{2mu} x\left( {2\mspace{2mu}\text{sin}\mspace{2mu} x\mspace{2mu}\text{cos}\mspace{2mu} x} \right) - 9\mspace{2mu}\text{cos}\mspace{2mu} x)} \\ & {= \frac{1}{12}(2\text{cos}^{3}x - \text{cos}\mspace{2mu} x - 2\mspace{2mu}\text{cos}\mspace{2mu} x\left( {1 - \text{cos}^{2}x} \right) - 9\mspace{2mu}\text{cos}\mspace{2mu} x)} \\ & {= \frac{1}{12}(4\text{cos}^{3}x - 12\mspace{2mu}\text{cos}\mspace{2mu} x)} \\ & {= \frac{1}{3}\text{cos}^{3}x - \text{cos}\mspace{2mu} x.} \end{array}$$

Thus, the two antiderivatives are identical.

We may also use a CAS to compare the graphs of the two functions, as shown in the following figure.

Use a CAS to evaluate $\int{\frac{dx}{\sqrt{x^{2} + 4}}.}$

Section 3.5 Exercises 3.5 节习题

Use a table of integrals to evaluate the following integrals.

244\.

$\int\limits_{0}^{4}{\frac{x}{\sqrt{1 + 2x}}dx}$

245.

$\int{\frac{x + 3}{x^{2} + 2x + 2}dx}$

246\.

$\int{x^{3}\sqrt{1 + 2x^{2}}\mspace{2mu} dx}$

247.

$\int{\frac{1}{\sqrt{x^{2} + 6x}}dx}$

248\.

$\int{\frac{x}{x + 1}dx}$

249.

$\int{x \cdot 2^{x^{2}}dx}$

250\.

$\int{\frac{1}{4x^{2} + 25}dx}$

251.

$\int\frac{dy}{\sqrt{4 - y^{2}}}$

252\.

$\int{\text{sin}^{3}(2x)\text{cos}(2x)dx}$

253.

$\int{\text{csc}(2w)\text{cot}(2w)dw}$

254\.

$\int{2^{y}dy}$

255.

$\int_{0}^{1}\frac{3x\mspace{2mu} dx}{\sqrt{x^{2} + 8}}$

256\.

$\int_{-1\text{/}4}^{1\text{/}4}{\text{sec}^{2}(\pi x)\text{tan}\left( {\pi x} \right)dx}$

257.

$\int_{0}^{\pi\text{/}2}{\text{tan}^{2}\left( \frac{x}{2} \right)dx}$

258\.

$\int{\text{cos}^{3}x\mspace{2mu} dx}$

259.

$\int{\text{tan}^{5}\left( {3x} \right)dx}$

260\.

$\int{\text{sin}^{2}y\mspace{2mu}\text{cos}^{3}ydy}$

Use a CAS to evaluate the following integrals. Tables can also be used to verify the answers.

261.

\[T\] $\int\frac{dw}{1 + \text{sec}\left( \frac{w}{2} \right)}$

262\.

\[T\] $\int\frac{dw}{1 - \text{cos}\left( {7w} \right)}$

263.

\[T\] $\int_{0}^{t}\frac{dt}{4\mspace{2mu}\text{cos}\mspace{2mu} t + 3\mspace{2mu}\text{sin}\mspace{2mu} t}$

264\.

\[T\] $\int{\frac{\sqrt{x^{2} - 9}}{3x}dx}$

265.

\[T\] $\int\frac{dx}{x^{1\text{/}2} + x^{1\text{/}3}}$

266\.

\[T\] $\int\frac{dx}{x\sqrt{x - 1}}$

267.

\[T\] $\int{x^{3}\text{sin}\mspace{2mu} x\mspace{2mu} dx}$

268\.

\[T\] $\int{x\sqrt{x^{4} - 9}\mspace{2mu} dx}$

269.

\[T\] $\int{\frac{x}{1 + e^{\text{−}x^{2}}}dx}$

270\.

\[T\] $\int{\frac{\sqrt{3 - 5x}}{2x}dx}$

271.

\[T\] $\int\frac{dx}{x\sqrt{x - 1}}$

272\.

\[T\] $\int{e^{x}\text{cos}^{-1}(e^{x})dx}$

Use a calculator or CAS to evaluate the following integrals.

273.

\[T\] $\int_{0}^{\pi\text{/}4}{\text{cos}(2x)dx}$

274\.

\[T\] $\int_{0}^{1}{x \cdot e^{\text{−}x^{2}}dx}$

275.

\[T\] $\int_{0}^{8}{\frac{2x}{\sqrt{x^{2} + 36}}dx}$

276\.

\[T\] $\int_{0}^{2\text{/}\sqrt{3}}{\frac{1}{4 + 9x^{2}}dx}$

277.

\[T\] $\int\frac{dx}{x^{2} + 4x + 13}$

278\.

\[T\] $\int\frac{dx}{1 + \text{sin}\mspace{2mu} x}$

Use tables to evaluate the integrals. You may need to complete the square or change variables to put the integral into a form given in the table.

279.

$\int\frac{dx}{x^{2} + 2x + 10}$

280\.

$\int\frac{dx}{\sqrt{x^{2} - 6x}}$

281.

$\int{\frac{e^{x}}{\sqrt{e^{2x} - 4}}dx}$

282\.

$\int{\frac{\text{cos}\mspace{2mu} x}{\text{sin}^{2}x + 2\mspace{2mu}\text{sin}\mspace{2mu} x}dx}$

283.

$\int{\frac{\text{arctan}\left( x^{3} \right)}{x^{4}}dx}$

284\.

$\int{\frac{\text{ln}|x|\text{arcsin}\left( {\text{ln}|x|} \right)}{x}dx}$

Use tables to perform the integration.

285.

$\int\frac{dx}{\sqrt{x^{2} + 16}}$

286\.

$\int{\frac{3x}{2x + 7}dx}$

287.

$\int\frac{dx}{1 - \text{cos}\left( {4x} \right)}$

288\.

$\int\frac{dx}{\sqrt{4x + 1}}$

289.

Find the area bounded by $y = \frac{5}{4 + 25x^{2}},x = 0,y = 0,\text{and}\ x = 4.$ Use a table of integrals or a CAS.

290\.

The region bounded between the curve $y = \frac{1}{\sqrt{1 + \text{cos}\mspace{2mu} x}},0.3 \leq x \leq 1.1,$ and the *x*-axis is revolved about the *x*-axis to generate a solid. Use a table of integrals to find the volume of the solid generated. (Round the answer to two decimal places.)

291.

Use substitution and a table of integrals to find the area of the surface generated by revolving the curve $y = e^{x},0 \leq x \leq 3,$ about the *x*-axis. (Round the answer to two decimal places.)

292\.

\[T\] Use an integral table and a calculator to find the area of the surface generated by revolving the curve $y = \frac{x^{2}}{2},0 \leq x \leq 1,$ about the *x*-axis. (Round the answer to two decimal places.)

293.

\[T\] Use a CAS or tables to find the area of the surface generated by revolving the curve $y = \text{cos}\mspace{2mu} x,0 \leq x \leq \frac{\pi}{2},$ about the *x*-axis. (Round the answer to two decimal places.)

294\.

Find the length of the curve $y = \frac{x^{2}}{4}$ over $\left\lbrack {0,8} \right\rbrack.$

295.

Find the length of the curve $y = e^{x}$ over $\left\lbrack {0,\text{ln}(2)} \right\rbrack.$

296\.

Find the area of the surface formed by revolving the graph of $y = 2\sqrt{x}$ over the interval $\left\lbrack {0,9} \right\rbrack$ about the *x*-axis.

297.

Find the average value of the function $f(x) = \frac{1}{x^{2} + 1}$ over the interval $\left\lbrack {-3,3} \right\rbrack.$

298\.

Approximate the arc length of the curve $y = \text{tan}\left( {\pi x} \right)$ over the interval $\left\lbrack {0,\frac{1}{4}} \right\rbrack.$ (Round the answer to three decimal places.)

3.6 Numerical Integration 3.6 数值积分

• 3.6.1 Approximate the value of a definite integral by using the midpoint and trapezoidal rules.
• 3.6.2 Determine the absolute and relative error in using a numerical integration technique.
• 3.6.3 Estimate the absolute and relative error using an error-bound formula.
• 3.6.4 Recognize when the midpoint and trapezoidal rules over- or underestimate the true value of an integral.
• 3.6.5 Use Simpson’s rule to approximate the value of a definite integral to a given accuracy.

• 3.6.1 使用中点法则与梯形法则近似计算定积分的值。
• 3.6.2 确定使用数值积分技术时的绝对误差与相对误差。
• 3.6.3 利用误差界公式估计绝对误差与相对误差。
• 3.6.4 识别中点法则与梯形法则何时高估或低估积分的真值。
• 3.6.5 使用辛普森法则将定积分的值近似到给定的精度。

The antiderivatives of many functions either cannot be expressed or cannot be expressed easily in closed form (that is, in terms of known functions). Consequently, rather than evaluate definite integrals of these functions directly, we resort to various techniques of numerical integration to approximate their values. In this section we explore several of these techniques. In addition, we examine the process of estimating the error in using these techniques.

The Midpoint Rule 中点法则

Earlier in this text we defined the definite integral of a function over an interval as the limit of Riemann sums. In general, any Riemann sum of a function $f(x)$ over an interval $\lbrack a,b\rbrack$ may be viewed as an estimate of ${\int_{a}^{b}{f(x)dx}}.$ Recall that a Riemann sum of a function $f(x)$ over an interval $\lbrack a,b\rbrack$ is obtained by selecting a partition

$$P = \left\\{ {x_{0},x_{1},x_{2}\text{,…},x_{n}} \right\\},\ \text{where}\ a = x_{0} < x_{1} < x_{2} < \cdots < x_{n} = b$$

and a set

$$S = \left\\{ {x_{1}^{*},x_{2}^{*}\text{,…},x_{n}^{*}} \right\\},\ \text{where}\ x_{i - 1} \leq x_{i}^{*} \leq x_{i}\ \text{for all}\ i.$$

The Riemann sum corresponding to the partition $P$ and the set $S$ is given by ${\sum\limits_{i = 1}^{n}{f(x_{i}^{*})\text{Δ}x_{i}}},$ where $\text{Δ}x_{i} = x_{i} - x_{i - 1},$ the length of the *i*th subinterval.

The midpoint rule for estimating a definite integral uses a Riemann sum with subintervals of equal width and the midpoints, $m_{i},$ of each subinterval in place of $x_{i}^{*}.$ Formally, we state a theorem regarding the convergence of the midpoint rule as follows.

The Midpoint Rule 中点法则

Assume that $f(x)$ is continuous on $\left\lbrack {a,b} \right\rbrack.$ Let *n* be a positive integer and $\text{Δ}x = \frac{b - a}{n}.$ If $\left\lbrack {a,b} \right\rbrack$ is divided into $n$ subintervals, each of length $\text{Δ}x,$ and $m_{i}$ is the midpoint of the *i*th subinterval, set

$$M_{n} = \sum\limits_{i = 1}^{n}f\left( m_{i} \right)\text{Δ}x.$$ (3.10)

Then $\underset{n\rightarrow\infty}{\text{lim}}M_{n} = {\int_{a}^{b}{f(x)dx}}.$

As we can see in Figure 3.13, if $f(x) \geq 0$ over $\lbrack a,b\rbrack,$ then $\sum\limits_{i = 1}^{n}f(m_{i})\text{Δ}x$ corresponds to the sum of the areas of rectangles approximating the area between the graph of $f(x)$ and the *x*-axis over $\left\lbrack {a,b} \right\rbrack.$ The graph shows the rectangles corresponding to $M_{4}$ for a nonnegative function over a closed interval $\lbrack a,b\rbrack.$

Using the Midpoint Rule with $M_{4}$ 使用 $M_{4}$ 的中点法则

Use the midpoint rule to estimate $\int_{0}^{1}{x^{2}dx}$ using four subintervals. Compare the result with the actual value of this integral.

Solution 解答

Each subinterval has length $\text{Δ}x = \frac{1 - 0}{4} = \frac{1}{4}.$ Therefore, the subintervals consist of

$$\left\lbrack {0,\frac{1}{4}} \right\rbrack,\left\lbrack {\frac{1}{4},\frac{1}{2}} \right\rbrack,\left\lbrack {\frac{1}{2},\frac{3}{4}} \right\rbrack,\ \text{and}\ \left\lbrack {\frac{3}{4},1} \right\rbrack.$$

The midpoints of these subintervals are $\left\\{ {\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8}} \right\\}.$ Thus,

$$M_{4} = \frac{1}{4}f\left( \frac{1}{8} \right) + \frac{1}{4}f\left( \frac{3}{8} \right) + \frac{1}{4}f\left( \frac{5}{8} \right) + \frac{1}{4}f\left( \frac{7}{8} \right) = \frac{1}{4} \cdot \frac{1}{64} + \frac{1}{4} \cdot \frac{9}{64} + \frac{1}{4} \cdot \frac{25}{64} + \frac{1}{4} \cdot \frac{49}{64} = \frac{21}{64}.$$

Since

$${\int_{0}^{1}{x^{2}dx = \frac{1}{3}}}\ \text{and}\ \left| {\frac{1}{3} - \frac{21}{64}} \right| = \frac{1}{192} \approx 0.0052,$$

we see that the midpoint rule produces an estimate that is somewhat close to the actual value of the definite integral.

Using the Midpoint Rule with $M_{6}$ 使用 $M_{6}$ 的中点法则

Use $M_{6}$ to estimate the length of the curve $y = \frac{1}{2}x^{2}$ on $\lbrack 1,4\rbrack.$

Solution 解答

The length of $y = \frac{1}{2}x^{2}$ on $\lbrack 1,4\rbrack$ is

$${\int_{1}^{4}{\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}\mspace{2mu} dx}}.$$

Since $\frac{dy}{dx} = x,$ this integral becomes ${\int_{1}^{4}{\sqrt{1 + x^{2}}\mspace{2mu} dx}}.$

If $\lbrack 1,4\rbrack$ is divided into six subintervals, then each subinterval has length $\text{Δ}x = \frac{4 - 1}{6} = \frac{1}{2}$ and the midpoints of the subintervals are $\left\\{ {\frac{5}{4},\frac{7}{4},\frac{9}{4},\frac{11}{4},\frac{13}{4},\frac{15}{4}} \right\\}.$ If we set $f(x) = \sqrt{1 + x^{2}},$

$$\begin{array}{cl} M_{6} & {= \frac{1}{2}f\left( \frac{5}{4} \right) + \frac{1}{2}f\left( \frac{7}{4} \right) + \frac{1}{2}f\left( \frac{9}{4} \right) + \frac{1}{2}f\left( \frac{11}{4} \right) + \frac{1}{2}f\left( \frac{13}{4} \right) + \frac{1}{2}f\left( \frac{15}{4} \right)} \\ & {\approx \frac{1}{2}\left( {1.6008 + 2.0156 + 2.4622 + 2.9262 + 3.4004 + 3.8810} \right) = 8.1431.} \end{array}$$

Use the midpoint rule with $n = 2$ to estimate ${\int_{1}^{2}{\frac{1}{x}dx}}.$

The Trapezoidal Rule 梯形法则

We can also approximate the value of a definite integral by using trapezoids rather than rectangles. In Figure 3.14, the area beneath the curve is approximated by trapezoids rather than by rectangles.

The trapezoidal rule for estimating definite integrals uses trapezoids rather than rectangles to approximate the area under a curve. To gain insight into the final form of the rule, consider the trapezoids shown in Figure 3.14. We assume that the length of each subinterval is given by $\text{Δ}x.$ First, recall that the area of a trapezoid with a height of *h* and bases of length $b_{1}$ and $b_{2}$ is given by $\text{Area} = \frac{1}{2}h(b_{1} + b_{2}).$ We see that the first trapezoid has a height $\text{Δ}x$ and parallel bases of length $f(x_{0})$ and $f(x_{1}).$ Thus, the area of the first trapezoid in Figure 3.14 is

$$\frac{1}{2}\text{Δ}x(f(x_{0}) + f(x_{1})).$$

The areas of the remaining three trapezoids are

$$\frac{1}{2}\text{Δ}x(f(x_{1}) + f(x_{2})),\frac{1}{2}\text{Δ}x(f(x_{2}) + f(x_{3})),\text{and}\ \frac{1}{2}\text{Δ}x(f(x_{3}) + f(x_{4})).$$

Consequently,

$${\int_{a}^{b}{f(x)dx \approx \frac{1}{2}\text{Δ}x(f(x_{0}) + f(x_{1})) + \frac{1}{2}\text{Δ}x(f(x_{1}) + f(x_{2})) + \frac{1}{2}\text{Δ}x(f(x_{2}) + f(x_{3})) + \frac{1}{2}\text{Δ}x(f(x_{3}) + f(x_{4}))}}.$$

After taking out a common factor of $\frac{1}{2}\text{Δ}x$ and combining like terms, we have

$$\int_{a}^{b}{f(x)dx \approx \frac{1}{2}\text{Δ}x\left( {f(x_{0}) + 2f\left( x_{1} \right) + 2f(x_{2}) + 2f\left( x_{3} \right) + f(x_{4})} \right).}$$

Generalizing, we formally state the following rule.

The Trapezoidal Rule 梯形法则

Assume that $f(x)$ is continuous over $\left\lbrack {a,b} \right\rbrack.$ Let *n* be a positive integer and $\text{Δ}x = \frac{b - a}{n}.$ Let $\left\lbrack {a,b} \right\rbrack$ be divided into $n$ subintervals, each of length $\text{Δ}x,$ with endpoints at $P = \left\\{ x_{0},x_{1},x_{2}\text{…},x_{n} \right\\}.$ Set

$$T_{n} = \frac{1}{2}\text{Δ}x\left( {f(x_{0}) + 2f\left( x_{1} \right) + 2f(x_{2}) + \cdots + 2f\left( x_{n - 1} \right) + f(x_{n})} \right).$$ (3.11)

Then, $\underset{n\rightarrow\text{+}\infty}{\text{lim}}T_{n} = {\int_{a}^{b}{f(x)dx}}.$

Before continuing, let’s make a few observations about the trapezoidal rule. First of all, it is useful to note that

$$T_{n} = \frac{1}{2}\left( {L_{n} + R_{n}} \right)\ \text{where}\ L_{n} = \sum\limits_{i = 1}^{n}f(x_{i - 1})\text{Δ}x\ \text{and}\ R_{n} = \sum\limits_{i = 1}^{n}f(x_{i})\text{Δ}x.$$

That is, $L_{n}$ and $R_{n}$ approximate the integral using the left-hand and right-hand endpoints of each subinterval, respectively. In addition, a careful examination of Figure 3.15 leads us to make the following observations about using the trapezoidal rules and midpoint rules to estimate the definite integral of a nonnegative function. The trapezoidal rule tends to overestimate the value of a definite integral systematically over intervals where the function is concave up and to underestimate the value of a definite integral systematically over intervals where the function is concave down. On the other hand, the midpoint rule tends to average out these errors somewhat by partially overestimating and partially underestimating the value of the definite integral over these same types of intervals. This leads us to hypothesize that, in general, the midpoint rule tends to be more accurate than the trapezoidal rule.

Using the Trapezoidal Rule 使用梯形法则

Use the trapezoidal rule to estimate $\int_{0}^{1}{x^{2}dx}$ using four subintervals.

Solution 解答

The endpoints of the subintervals consist of elements of the set $P = \left\\{ {0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1} \right\\}$ and $\text{Δ}x = \frac{1 - 0}{4} = \frac{1}{4}.$ Thus,

$$\begin{array}{cl} {\int_{0}^{1}{x^{2}dx}} & {\approx \frac{1}{2} \cdot \frac{1}{4}\left( {f(0) + 2f\left( \frac{1}{4} \right) + 2f\left( \frac{1}{2} \right) + 2f\left( \frac{3}{4} \right) + f(1)} \right)} \\ & {= \frac{1}{8}\left( {0 + 2 \cdot \frac{1}{16} + 2 \cdot \frac{1}{4} + 2 \cdot \frac{9}{16} + 1} \right)} \\ & {= \frac{11}{32}.} \end{array}$$

Use the trapezoidal rule with $n = 2$ to estimate ${\int_{1}^{2}{\frac{1}{x}dx}}.$

Absolute and Relative Error 绝对误差与相对误差

An important aspect of using these numerical approximation rules consists of calculating the error in using them for estimating the value of a definite integral. We first need to define absolute error and relative error.

If $B$ is our estimate of some quantity having an actual value of $A,$ then the absolute error is given by $\left| {A - B} \right|.$ The relative error is the error as a percentage of the absolute value and is given by $\left| \frac{A - B}{A} \right| = \left| \frac{A - B}{A} \right| \cdot 100\text{\%}.$

Calculating Error in the Midpoint Rule 计算中点法则的误差

Calculate the absolute and relative error in the estimate of $\int_{0}^{1}{x^{2}dx}$ using the midpoint rule, found in Example 3.39.

Solution 解答

The calculated value is $\int_{0}^{1}{x^{2}dx = \frac{1}{3}}$ and our estimate from the example is $M_{4} = \frac{21}{64}.$ Thus, the absolute error is given by $\left| {\left( \frac{1}{3} \right) - \left( \frac{21}{64} \right)} \right| = \frac{1}{192} \approx 0.0052.$ The relative error is

$$\frac{1\text{/}192}{1\text{/}3} = \frac{1}{64} \approx 0.015625 \approx 1.6\text{\%}.$$

Calculating Error in the Trapezoidal Rule 计算梯形法则的误差

Calculate the absolute and relative error in the estimate of $\int_{0}^{1}{x^{2}dx}$ using the trapezoidal rule, found in Example 3.41.

Solution 解答

The calculated value is $\int_{0}^{1}{x^{2}dx = \frac{1}{3}}$ and our estimate from the example is $T_{4} = \frac{11}{32}.$ Thus, the absolute error is given by $\left| {\frac{1}{3} - \frac{11}{32}} \right| = \frac{1}{96} \approx 0.0104.$ The relative error is given by

$$\frac{1\text{/}96}{1\text{/}3} = 0.03125 \approx 3.1\text{\%}.$$

In an earlier checkpoint, we estimated $\int_{1}^{2}{\frac{1}{x}dx}$ to be $\frac{24}{35}$ using $T_{2}.$ The actual value of this integral is $\text{ln}\mspace{2mu} 2.$ Using $\frac{24}{35} \approx 0.6857$ and $\text{ln}\mspace{2mu} 2 \approx 0.6931,$ calculate the absolute error and the relative error.

In the two previous examples, we were able to compare our estimate of an integral with the actual value of the integral; however, we do not typically have this luxury. In general, if we are approximating an integral, we are doing so because we cannot compute the exact value of the integral itself easily. Therefore, it is often helpful to be able to determine an upper bound for the error in an approximation of an integral. The following theorem provides error bounds for the midpoint and trapezoidal rules. The theorem is stated without proof.

Error Bounds for the Midpoint and Trapezoidal Rules 中点法则与梯形法则的误差界

Let $f(x)$ be a continuous function over $\left\lbrack {a,b} \right\rbrack,$ having a second derivative $f^{''}(x)$ over this interval. If $M$ is the maximum value of $\left| {f^{''}(x)} \right|$ over $\lbrack a,b\rbrack,$ then the upper bounds for the error in using $M_{n}$ and $T_{n}$ to estimate $\int_{a}^{b}{f(x)dx}$ are

$$\text{Error in}\ M_{n} \leq \frac{M\left( {b - a} \right)^{3}}{24n^{2}}$$ (3.12)

and

$$\text{Error in}\ T_{n} \leq \frac{M{(b - a)}^{3}}{12n^{2}}.$$ (3.13)

We can use these bounds to determine the value of $n$ necessary to guarantee that the error in an estimate is less than a specified value.

Determining the Number of Intervals to Use 确定所需区间数

What value of $n$ should be used to guarantee that an estimate of $\int_{0}^{1}{e^{x^{2}}dx}$ is accurate to within 0.01 if we use the midpoint rule?

Solution 解答

We begin by determining the value of $M,$ the maximum value of $\left| {f^{''}(x)} \right|$ over $\lbrack 0,1\rbrack$ for $f(x) = e^{x^{2}}.$ Since $f^{\prime}(x) = 2xe^{x^{2}},$ we have

$$f^{\text{''}}(x) = 2e^{x^{2}} + 4x^{2}e^{x^{2}}.$$

Thus,

$$\left| {f^{''}(x)} \right| = 2e^{x^{2}}\left( {1 + 2x^{2}} \right) \leq 2 \cdot e \cdot 3 = 6e.$$

From the error-bound Equation 3.12, we have

$$\text{Error in}\ M_{n} \leq \frac{M\left( {b - a} \right)^{3}}{24n^{2}} \leq \frac{6e{(1 - 0)}^{3}}{24n^{2}} = \frac{6e}{24n^{2}}.$$

Now we solve the following inequality for $n\text{:}$

$$\frac{6e}{24n^{2}} \leq 0.01.$$

Thus, $n \geq \sqrt{\frac{600e}{24}} \approx 8.24.$ Since $n$ must be an integer satisfying this inequality, a choice of $n = 9$ would guarantee that $\left| {\int_{0}^{1}{e^{x^{2}}dx - M_{n}}} \right| < 0.01.$

Analysis 分析

We might have been tempted to round $8.24$ down and choose $n = 8,$ but this would be incorrect because we must have an integer greater than or equal to $8.24.$ We need to keep in mind that the error estimates provide an upper bound only for the error. The actual estimate may, in fact, be a much better approximation than is indicated by the error bound.

Use Equation 3.13 to find an upper bound for the error in using $M_{4}$ to estimate ${\int_{0}^{1}{x^{2}dx}}.$

Simpson's Rule 辛普森法则

With the midpoint rule, we estimated areas of regions under curves by using rectangles. In a sense, we approximated the curve with piecewise constant functions. With the trapezoidal rule, we approximated the curve by using piecewise linear functions. What if we were, instead, to approximate a curve using piecewise quadratic functions? With Simpson's rule, we do just this. We partition the interval into an even number of subintervals, each of equal width. Over the first pair of subintervals we approximate $\int_{x_{0}}^{x_{2}}{f(x)dx}$ with $\int_{x_{0}}^{x_{2}}{p(x)dx,}$ where $p(x) = Ax^{2} + Bx + C$ is the quadratic function passing through $(x_{0},f\left( x_{0} \right)),$ $(x_{1},f\left( x_{1} \right)),$ and $(x_{2},f\left( x_{2} \right))$ (Figure 3.16). Over the next pair of subintervals we approximate $\int_{x_{2}}^{x_{4}}{f(x)dx}$ with the integral of another quadratic function passing through $(x_{2},f\left( x_{2} \right)),$ $(x_{3},f\left( x_{3} \right)),$ and $(x_{4},f\left( x_{4} \right)).$ This process is continued with each successive pair of subintervals.

To understand the formula that we obtain for Simpson's rule, we begin by deriving a formula for this approximation over the first two subintervals. As we go through the derivation, we need to keep in mind the following relationships:

$$\begin{array}{l} {f\left( x_{0} \right) = p\left( x_{0} \right) = Ax_{0}{}^{2} + Bx_{0} + C} \\ {f\left( x_{1} \right) = p\left( x_{1} \right) = Ax_{1}{}^{2} + Bx_{1} + C} \\ {f\left( x_{2} \right) = p\left( x_{2} \right) = Ax_{2}{}^{2} + Bx_{2} + C} \end{array}$$

$x_{2} - x_{0} = 2\text{Δ}x,$ where $\text{Δ}x$ is the length of a subinterval.

$$x_{2} + x_{0} = 2x_{1},\ \text{since}\ x_{1} = \frac{(x_{2} + x_{0})}{2}.$$

Thus,

$$\begin{matrix} {\int_{x_{0}}^{x_{2}}{f(x)dx}} & {\approx {\int_{x_{0}}^{x_{2}}{p(x)dx}}} & & & \\ & {= {\int_{x_{0}}^{x_{2}}{(Ax^{2} + Bx + C)dx}}} & & & \\ & {= \frac{A}{3}x^{3} + \frac{B}{2}x^{2} + Cx\left| \begin{matrix} {}^{x_{2}} \\ {}_{x_{0}} \end{matrix} \right.} & & & \text{Find the antiderivative.} \\ & {= \frac{A}{3}\left( {x_{2}{}^{3} - x_{0}{}^{3}} \right) + \frac{B}{2}\left( {x_{2}{}^{2} - x_{0}{}^{2}} \right) + C(x_{2} - x_{0})} & & & \text{Evaluate the antiderivative.} \\ & {= \frac{A}{3}\left( {x_{2} - x_{0}} \right)\left( {x_{2}{}^{2} + x_{2}x_{0} + x_{0}{}^{2}} \right)} & & & \\ & {+ \frac{B}{2}\left( {x_{2} - x_{0}} \right)\left( {x_{2} + x_{0}} \right) + C(x_{2} - x_{0})} & & & \\ & {= \frac{x_{2} - x_{0}}{6}\left( {2A\left( {x_{2}{}^{2} + x_{2}x_{0} + x_{0}{}^{2}} \right) + 3B(x_{2} + x_{0}) + 6C} \right)} & & & {\text{Factor out}\ \frac{x_{2} - x_{0}}{6}.} \\ & {= \frac{\text{Δ}x}{3}\left( {\left( {Ax_{2}{}^{2} + Bx_{2} + C} \right) + (Ax_{0}{}^{2} + Bx_{0} + C} \right)} & & & \\ & {+ A\left( {x_{2}{}^{2} + 2x_{2}x_{0} + x_{0}{}^{2}} \right) + 2B\left( {x_{2} + x_{0}} \right) + 4C)} & & & \\ & {= \frac{\text{Δ}x}{3}\left( {f\left( x_{2} \right) + f\left( x_{0} \right) + A\left( {x_{2} + x_{0}} \right)^{2} + 2B\left( {x_{2} + x_{0}} \right) + 4C} \right)} & & & \text{Rearrange the terms.} \\ & & & & \begin{array}{l} \text{Factor and substitute.} \\ {f\left( x_{2} \right) = Ax_{2}{}^{2} + Bx_{2} + C\ \text{and}} \\ {f\left( x_{0} \right) = Ax_{0}{}^{2} + Bx_{0} + C.} \end{array} \\ & {= \frac{\text{Δ}x}{3}\left( {f\left( x_{2} \right) + 4f\left( x_{1} \right) + f\left( x_{0} \right)} \right).} & & & {\text{Substitute}\ x_{2} + x_{0} = 2x_{1}.} \\ & {= \frac{\text{Δ}x}{3}\left( {f\left( x_{2} \right) + 4f\left( x_{1} \right) + f\left( x_{0} \right)} \right).} & & & \begin{array}{l} \text{Expand and substitute} \\ {f\left( x_{1} \right) = Ax_{1}{}^{2} + Bx_{1}\text{+}C.} \end{array} \end{matrix}$$

If we approximate $\int_{x_{2}}^{x_{4}}{f(x)dx}$ using the same method, we see that we have

$$\int_{x_{2}}^{x_{4}}{f(x)dx \approx \frac{\text{Δ}x}{3}\left( {f\left( x_{4} \right) + 4f\left( x_{3} \right) + f\left( x_{2} \right)} \right).}$$

Combining these two approximations, we get

$$\int_{x_{2}}^{x_{4}}{f(x)dx = \frac{\text{Δ}x}{3}\left( {f\left( x_{0} \right) + 4f\left( x_{1} \right) + 2f\left( x_{2} \right) + 4f\left( x_{3} \right) + f\left( x_{4} \right)} \right).}$$

The pattern continues as we add pairs of subintervals to our approximation. The general rule may be stated as follows.

Simpson's Rule

Assume that $f(x)$ is continuous over $\left\lbrack {a,b} \right\rbrack.$ Let *n* be a positive even integer and $\text{Δ}x = \frac{b - a}{n}.$ Let $\left\lbrack {a,b} \right\rbrack$ be divided into $n$ subintervals, each of length $\text{Δ}x,$ with endpoints at $P = \left\\{ x_{0},x_{1},x_{2}\text{,…},x_{n} \right\\}.$ Set

$$S_{n} = \frac{\text{Δ}x}{3}\left( {f\left( x_{0} \right) + 4f\left( x_{1} \right) + 2f\left( x_{2} \right) + 4f\left( x_{3} \right) + 2f\left( x_{4} \right) + \cdots + 2f\left( x_{n - 2} \right) + 4f\left( x_{n - 1} \right) + f\left( x_{n} \right)} \right).$$ (3.14)

Then,

$$\underset{n\rightarrow\text{+}\infty}{\text{lim}}S_{n} = {\int_{a}^{b}{f(x)dx.}}$$

Just as the trapezoidal rule is the average of the left-hand and right-hand rules for estimating definite integrals, Simpson's rule may be obtained from the midpoint and trapezoidal rules by using a weighted average. It can be shown that $S_{2n} = \left( \frac{2}{3} \right)M_{n} + \left( \frac{1}{3} \right)T_{n}.$

It is also possible to put a bound on the error when using Simpson's rule to approximate a definite integral. The bound in the error is given by the following rule:

Let $f(x)$ be a continuous function over $\lbrack a,b\rbrack$ having a fourth derivative, $f^{(4)}(x),$ over this interval. If $M$ is the maximum value of $\left| {f^{(4)}(x)} \right|$ over $\lbrack a,b\rbrack,$ then the upper bound for the error in using $S_{n}$ to estimate $\int_{a}^{b}{f(x)dx}$ is given by

$$\text{Error in}\ S_{n} \leq \frac{M{(b - a)}^{5}}{180n^{4}}.$$ (3.15)

Applying Simpson's Rule 1

Use $S_{2}$ to approximate ${\int_{0}^{1}{x^{3}dx}}.$ Estimate a bound for the error in $S_{2}.$

Solution

Since $\lbrack 0,1\rbrack$ is divided into two intervals, each subinterval has length $\text{Δ}x = \frac{1 - 0}{2} = \frac{1}{2}.$ The endpoints of these subintervals are $\left\\{ {0,\frac{1}{2},1} \right\\}.$ If we set $f(x) = x^{3},$ then

$S_{4} = \frac{1}{3} \cdot \frac{1}{2}\left( {f(0) + 4f\left( \frac{1}{2} \right) + f(1)} \right) = \frac{1}{6}\left( {0 + 4 \cdot \frac{1}{8} + 1} \right) = \frac{1}{4}.$ Since $f^{(4)}(x) = 0$ and consequently $M = 0,$ we see that

$$\text{Error in}\ S_{2} \leq \frac{0{(1)}^{5}}{180 \cdot 2^{4}} = 0.$$

This bound indicates that the value obtained through Simpson's rule is exact. A quick check will verify that, in fact, ${\int_{0}^{1}{x^{3}dx = \frac{1}{4}}}.$

Applying Simpson's Rule 2

Use $S_{6}$ to estimate the length of the curve $y = \frac{1}{2}x^{2}$ over $\lbrack 1,4\rbrack.$

Solution

The length of $y = \frac{1}{2}x^{2}$ over $\lbrack 1,4\rbrack$ is ${\int_{1}^{4}{\sqrt{1 + x^{2}}\mspace{2mu} dx}}.$ If we divide $\lbrack 1,4\rbrack$ into six subintervals, then each subinterval has length $\text{Δ}x = \frac{4 - 1}{6} = \frac{1}{2},$ and the endpoints of the subintervals are $\left\\{ {1,\frac{3}{2},2,\frac{5}{2},3,\frac{7}{2},4} \right\\}.$ Setting $f(x) = \sqrt{1 + x^{2}},$

$$S_{6} = \frac{1}{3} \cdot \frac{1}{2}\left( {f(1) + 4f\left( \frac{3}{2} \right) + 2f(2) + 4f\left( \frac{5}{2} \right) + 2f(3) + 4f\left( \frac{7}{2} \right) + f(4)} \right).$$

After substituting, we have

$$\begin{array}{cl} S_{6} & {= \frac{1}{6}\left( {1.4142 + 4 \cdot 1.80278 + 2 \cdot 2.23607 + 4 \cdot 2.69258 + 2 \cdot 3.16228 + 4 \cdot 3.64005 + 4.12311} \right)} \\ & {\approx 8.14594.} \end{array}$$

Use $S_{2}$ to estimate ${\int_{1}^{2}{\frac{1}{x}dx}}.$

Section 3.6 Exercises 3.6 节习题

Approximate the following integrals using either the midpoint rule, trapezoidal rule, or Simpson's rule as indicated. (Round answers to three decimal places.)

299.

${\int_{1}^{2}\frac{dx}{x}};$ trapezoidal rule; $n = 5$

300\.

${\int_{0}^{3}{\sqrt{4 + x^{3}}\mspace{2mu} dx}};$ trapezoidal rule; $n = 6$

301.

${\int_{0}^{3}{\sqrt{4 + x^{3}}\mspace{2mu} dx}};$ trapezoidal rule; $n = 3$

302\.

${\int_{0}^{12}{x^{2}dx}};$ midpoint rule; $n = 6$

303.

${\int_{0}^{1}{\text{sin}^{2}\left( {\pi x} \right)dx}};$ midpoint rule; $n = 3$

304\.

Use the midpoint rule with eight subdivisions to estimate ${\int_{2}^{4}{x^{2}dx}}.$

305.

Use the trapezoidal rule with four subdivisions to estimate ${\int_{2}^{4}{x^{2}dx}}.$

306\.

Find the exact value of ${\int_{2}^{4}{x^{2}dx}}.$ Find the error of approximation between the exact value and the value calculated using the trapezoidal rule with four subdivisions. Draw a graph to illustrate.

Approximate the integral to three decimal places using the indicated rule.

307.

${\int_{0}^{1}{\text{sin}^{2}\left( {\pi x} \right)dx}};$ trapezoidal rule; $n = 6$

308\.

${\int_{0}^{3}{\frac{1}{1 + x^{3}}dx}};$ trapezoidal rule; $n = 6$

309.

${\int_{0}^{3}{\frac{1}{1 + x^{3}}dx}};$ trapezoidal rule; $n = 3$

310\.

${\int_{0}^{0.8}{e^{\text{−}x^{2}}dx}};$ trapezoidal rule; $n = 4$

311.

${\int_{0}^{0.8}{e^{\text{−}x^{2}}dx}};$ Simpson's rule; $n = 4$

312\.

${\int_{0}^{0.4}{\text{sin}(x^{2})dx}};$ trapezoidal rule; $n = 4$

313.

${\int_{0}^{0.4}{\text{sin}(x^{2})dx}};$ Simpson's rule; $n = 4$

314\.

${\int_{0.1}^{0.5}{\frac{\text{cos}\mspace{2mu} x}{x}dx}};$ trapezoidal rule; $n = 4$

315.

${\int_{0.1}^{0.5}{\frac{\text{cos}\mspace{2mu} x}{x}dx}};$ Simpson's rule; $n = 4$

316\.

Evaluate $\int_{0}^{1}\frac{dx}{1 + x^{2}}$ exactly and show that the result is ${\pi\text{/}4}.$ Then, find the approximate value of the integral using the trapezoidal rule with $n = 4$ subdivisions. Use the result to approximate the value of $\pi.$

317.

Approximate $\int_{2}^{4}{\frac{1}{\text{ln}\mspace{2mu} x}dx}$ using the midpoint rule with four subdivisions to four decimal places.

318\.

Approximate $\int_{2}^{4}{\frac{1}{\text{ln}\mspace{2mu} x}dx}$ using the trapezoidal rule with eight subdivisions to four decimal places.

319.

Use the trapezoidal rule with four subdivisions to estimate $\int_{0}^{0.8}{x^{3}dx}$ to four decimal places.

320\.

Use the trapezoidal rule with four subdivisions to estimate ${\int_{0}^{0.8}{x^{3}dx}}.$ Compare this value with the exact value and find the error estimate.

321.

Using Simpson's rule with four subdivisions, find ${\int_{0}^{\pi\text{/}2}{\text{cos}(x)dx}}.$

322\.

Show that the exact value of $\int_{0}^{1}{xe^{\text{−}x}dx = 1 - \frac{2}{e}.}$ Find the absolute error if you approximate the integral using the midpoint rule with 16 subdivisions.

323.

Given $\int_{0}^{1}{xe^{\text{−}x}dx = 1 - \frac{2}{e},}$ use the trapezoidal rule with 16 subdivisions to approximate the integral and find the absolute error.

324\.

Find an upper bound for the error in estimating $\int_{0}^{3}{(5x + 4)dx}$ using the trapezoidal rule with six steps.

325.

Find an upper bound for the error in estimating $\int_{4}^{5}{\frac{1}{\left( {x - 1} \right)^{2}}dx}$ using the trapezoidal rule with seven subdivisions.

326\.

Find an upper bound for the error in estimating $\int_{0}^{3}{(6x^{2} - 1)dx}$ using Simpson's rule with $n = 10$ steps.

327.

Find an upper bound for the error in estimating $\int_{2}^{5}{\frac{1}{x - 1}dx}$ using Simpson's rule with $n = 10$ steps.

328\.

Find an upper bound for the error in estimating $\int_{0}^{\pi}{2x\mspace{2mu}\text{cos}(x)dx}$ using Simpson's rule with four steps.

329.

Estimate the minimum number of subintervals needed to approximate the integral $\int_{1}^{4}{\left( {5x^{2} + 8} \right)dx}$ with an error magnitude of less than 0.0001 using the trapezoidal rule.

330\.

Determine a value of *n* such that the trapezoidal rule will approximate $\int_{0}^{1}{\sqrt{1 + x^{2}}\mspace{2mu} dx}$ with an error of no more than 0.01.

331.

Estimate the minimum number of subintervals needed to approximate the integral $\int_{2}^{3}{\left( {2x^{3} + 4x} \right)dx}$ with an error of magnitude less than 0.0001 using the trapezoidal rule.

332\.

Estimate the minimum number of subintervals needed to approximate the integral $\int_{3}^{4}{\frac{1}{{(x - 1)}^{2}}dx}$ with an error magnitude of less than 0.0001 using the trapezoidal rule.

333.

Use Simpson's rule with four subdivisions to approximate the area under the probability density function $y = \frac{1}{\sqrt{2\pi}}e^{{\text{−}x^{2}}\text{/}2}$ from $x = 0$ to $x = 0.4.$

334\.

Use Simpson's rule with $n = 14$ to approximate (to three decimal places) the area of the region bounded by the graphs of $y = 0,$ $x = 0,$ and $x = {\pi\text{/}2}.$

335.

The length of one arch of the curve $y = 3\mspace{2mu}\text{sin}(2x)$ is given by $L = {\int_{0}^{\pi\text{/}2}{\sqrt{1 + 36\mspace{2mu}\text{cos}^{2}(2x)}\mspace{2mu} dx}}.$ Estimate *L* using the trapezoidal rule with $n = 6.$

336\.

The length of the ellipse $x = a\mspace{2mu}\text{cos}(t),y = b\mspace{2mu}\text{sin}(t),0 \leq t \leq 2\pi$ is given by $L = 4a{\int_{0}^{\pi\text{/}2}{\sqrt{1 - e^{2}\text{cos}^{2}(t)}\mspace{2mu} dt}},$ where *e* is the eccentricity of the ellipse. Use Simpson's rule with $n = 6$ subdivisions to estimate the length of the ellipse when $a = 2$ and $e = 1\text{/}3.$

337.

Estimate the area of the surface generated by revolving the curve $y = \text{cos}(2x),0 \leq x \leq \frac{\pi}{4}$ about the *x*-axis. Use the trapezoidal rule with six subdivisions.

338\.

Estimate the area of the surface generated by revolving the curve $y = 2x^{2},$ $0 \leq x \leq 3$ about the *x-*axis. Use Simpson's rule with $n = 6.$

339.

The growth rate of a certain tree (in feet) is given by $y = \frac{2}{t + 1} + e^{\text{−}t^{2}\text{/}2},$ where *t* is time in years. Estimate the growth of the tree through the end of the second year by using Simpson's rule, using two subintervals. (Round the answer to the nearest hundredth.)

340\.

\[T\] Use a calculator to approximate $\int_{0}^{1}{\text{sin}\left( {\pi x} \right)dx}$ using the midpoint rule with 25 subdivisions. Compute the relative error of approximation.

341.

\[T\] Given ${\int_{1}^{5}{\left( {3x^{2} - 2x} \right)dx}} = 100,$ approximate the value of this integral using the trapezoidal rule with 16 subdivisions and determine the absolute error.

342\.

Given that we know the Fundamental Theorem of Calculus, why would we want to develop numerical methods for definite integrals?

343.

The table represents the coordinates $\left( {x, y} \right)$ that give the boundary of a lot. The units of measurement are meters. Use the trapezoidal rule to estimate the number of square meters of land that is in this lot.

*x*	*y*	*x*	*y*
0	125	600	95
100	125	700	88
200	120	800	75
300	112	900	35
400	90	1000	0
500	90

*x*	*y*	*x*	*y*
0	125	600	95
100	125	700	88
200	120	800	75
300	112	900	35
400	90	1000	0
500	90

344\.

Choose the correct answer. When Simpson's rule is used to approximate the definite integral, it is necessary that the number of partitions be\_\_\_\_

1. an even number

2. odd number

3. either an even or an odd number

4. a multiple of 4

345.

The "Simpson" sum is based on the area under a \_\_\_\_.

346\.

The error formula for Simpson's rule depends on\_\_\_.

1. $f(x)$

2. $f^{\prime}(x)$

3. $f^{(4)}(x)$

4. the number of steps

3.7 Improper Integrals 3.7 反常积分

• 3.7.1 Evaluate an integral over an infinite interval.
• 3.7.2 Evaluate an integral over a closed interval with an infinite discontinuity within the interval.
• 3.7.3 Use the comparison theorem to determine whether a definite integral is convergent.

• 3.7.1 计算无穷区间上的积分。
• 3.7.2 计算在区间内部含有无穷间断点的闭区间上的积分。
• 3.7.3 使用比较定理判断定积分是否收敛。

Is the area between the graph of $f(x) = \frac{1}{x}$ and the *x*-axis over the interval $\lbrack 1,\text{+}\infty)$ finite or infinite? If this same region is revolved about the *x*-axis, is the volume finite or infinite? Surprisingly, the area of the region described is infinite, but the volume of the solid obtained by revolving this region about the *x*-axis is finite.

In this section, we define integrals over an infinite interval as well as integrals of functions containing a discontinuity on the interval. Integrals of these types are called improper integrals. We examine several techniques for evaluating improper integrals, all of which involve taking limits.

Integrating over an Infinite Interval 在无穷区间上积分

How should we go about defining an integral of the type ${\int_{a}^{+ \infty}{f(x)dx}}?$ We can integrate $\int_{a}^{t}{f(x)dx}$ for any value of $t,$ so it is reasonable to look at the behavior of this integral as we substitute larger values of $t.$ Figure 3.17 shows that $\int_{a}^{t}{f(x)dx}$ may be interpreted as area for various values of $t.$ In other words, we may define an improper integral as a limit, taken as one of the limits of integration increases or decreases without bound.

1. Let $f(x)$ be continuous over an interval of the form $\lbrack a,\text{+}\infty).$ Then

$$\int_{a}^{+ \infty}{f(x)dx = \underset{t\rightarrow\text{+}\infty}{\text{lim}}{\int_{a}^{t}{f(x)dx,}}}$$ (3.16)

provided this limit exists.

2. Let $f(x)$ be continuous over an interval of the form $(\text{−}\infty,b\rbrack.$ Then

$$\int_{\text{−}\infty}^{b}{f(x)dx = \underset{t\rightarrow\text{−}\infty}{\text{lim}}{\int_{t}^{b}{f(x)dx,}}}$$ (3.17)

provided this limit exists.

In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.

3. Let $f(x)$ be continuous over $(\text{−}\infty,\text{+}\infty).$ Then

$$\int_{\text{−}\infty}^{+ \infty}{f(x)dx = {\int_{\text{−}\infty}^{0}{f(x)dx + {\int_{0}^{+ \infty}{f(x)dx,}}}}}$$ (3.18)

provided that $\int_{\text{−}\infty}^{0}{f(x)dx}$ and $\int_{0}^{+ \infty}{f(x)dx}$ both converge. If either one or both of these two integrals diverge, then $\int_{\text{−}\infty}^{+ \infty}{f(x)dx}$ diverges. (It can be shown that, in fact, $\int_{\text{−}\infty}^{+ \infty}{f(x)dx = {\int_{\text{−}\infty}^{a}{f(x)dx + {\int_{a}^{+ \infty}{f(x)dx}}}}}$ for any value of $a.)$

In our first example, we return to the question we posed at the start of this section: Is the area between the graph of $f(x) = \frac{1}{x}$ and the $x$-axis over the interval $\lbrack 1,\text{+}\infty)$ finite or infinite?

Finding an Area 求面积

Determine whether the area between the graph of $f(x) = \frac{1}{x}$ and the *x*-axis over the interval $\lbrack 1,\text{+}\infty)$ is finite or infinite.

Solution 解答

We first do a quick sketch of the region in question, as shown in the following graph.

We can see that the area of this region is given by $A = {\int_{1}^{\infty}{\frac{1}{x}dx}}.$ Then we have

$$\begin{array}{clccc} A & {= {\int_{1}^{\infty}{\frac{1}{x}dx}}} & & & \\ & {= \underset{t\rightarrow\text{+}\infty}{\text{lim}}{\int_{1}^{t}{\frac{1}{x}dx}}} & & & \text{Rewrite the improper integral as a limit.} \\ & {= \underset{t\rightarrow\text{+}\infty}{\text{lim}}\text{ln}|x|\left| {}_{\begin{matrix} \\ 1 \end{matrix}}^{\begin{matrix} t \\ \end{matrix}} \right.} & & & \text{Find the antiderivative.} \\ & {= \underset{t\rightarrow\text{+}\infty}{\text{lim}}(\text{ln}|t| - \text{ln}\mspace{2mu} 1)} & & & \text{Evaluate the antiderivative.} \\ & {= \text{+}\infty.} & & & \text{Evaluate the limit.} \end{array}$$

Since the improper integral diverges to $+ \infty,$ the area of the region is infinite.

Finding a Volume 求体积

Find the volume of the solid obtained by revolving the region bounded by the graph of $f(x) = \frac{1}{x}$ and the *x*-axis over the interval $\lbrack 1,\text{+}\infty)$ about the $x$-axis.

Solution 解答

The solid is shown in Figure 3.19. Using the disk method, we see that the volume *V* is

$$V = \pi{\int_{1}^{+ \infty}{\frac{1}{x^{2}}dx}}.$$

Then we have

$$\begin{array}{clccc} V & {= \pi{\int_{1}^{+ \infty}{\frac{1}{x^{2}}dx}}} & & & \\ & {= \pi\underset{t\rightarrow\text{+}\infty}{\text{lim}}{\int_{1}^{t}{\frac{1}{x^{2}}dx}}} & & & \text{Rewrite as a limit.} \\ & {= \pi\underset{t\rightarrow\text{+}\infty}{\text{lim}} - \frac{1}{x}\left| {}_{\begin{matrix} \\ 1 \end{matrix}}^{\begin{matrix} t \\ \end{matrix}} \right.} & & & \text{Find the antiderivative.} \\ & {= \pi\underset{t\rightarrow\text{+}\infty}{\text{lim}}\left( {\text{−}\ \frac{1}{t} + 1} \right)} & & & \text{Evaluate the antiderivative.} \\ & {= \pi.} & & & \end{array}$$

The improper integral converges to $\pi.$ Therefore, the volume of the solid of revolution is $\pi.$

In conclusion, although the area of the region between the *x*-axis and the graph of $f(x) = 1\text{/}x$ over the interval $\lbrack 1,\text{+}\infty)$ is infinite, the volume of the solid generated by revolving this region about the *x*-axis is finite. The solid generated is known as *Gabriel’s Horn*.

Visit this website to read more about Gabriel’s Horn.

Chapter Opener: Traffic Accidents in a City 章节开篇：城市中的交通事故

In the chapter opener, we stated the following problem: Suppose that at a busy intersection, traffic accidents occur at an average rate of one every three months. After residents complained, changes were made to the traffic lights at the intersection. It has now been ten months since the changes were made and there have been no accidents. Were the changes effective or is the 10-month interval without an accident a result of chance?

Solution 解答

Revise to: Let $x$ represent the amount of time it takes for the next accident to occur. We want to know how likely it is that $x > 10$. Define the rate parameter $\lambda$ to be the average number of accidents per month. According to probability theory, for $a > 0$,

$$P\left( {x > a} \right) = {\int_{a}^{\infty}\lambda e^{- \lambda x}}dx,$$

In this example, since one accident happens every three months, on average, $\lambda = \frac{1}{3}$. The desired probability is:

$$\begin{array}{l} {P(x > 10) = \int_{10}^{\infty}\frac{1}{3}e^{- \frac{1}{3}x}dx} \\ {= \lim\limits_{t\rightarrow\infty}\int_{10}^{t}\frac{1}{3}e^{- \frac{1}{3}x}dx} \\ {= \lim\limits_{t\rightarrow\infty} - e^{- \frac{1}{3}x}|_{10}^{t}} \\ {= \lim\limits_{t\rightarrow\infty}\left( - e^{- \frac{t}{3}} + e^{- \frac{10}{3}} \right)} \\ {\approx 0.0357} \end{array}$$

The value $3.8\ \times\ 10^{-11}$ represents the probability of no accidents in 8 months under the initial conditions. Since this value is very, very small, it is reasonable to conclude the changes were effective.

Evaluating an Improper Integral over an Infinite Interval 在无穷区间上计算反常积分

Evaluate ${\int_{\text{−}\infty}^{0}{\frac{1}{x^{2} + 4}dx}}.$ State whether the improper integral converges or diverges.

Solution 解答

Begin by rewriting $\int_{\text{−}\infty}^{0}{\frac{1}{x^{2} + 4}dx}$ as a limit using Equation 3.17 from the definition. Thus,

$$\begin{array}{clccl} {\int_{\text{−}\infty}^{0}{\frac{1}{x^{2} + 4}dx}} & {= \underset{x\rightarrow\text{−}\infty}{\text{lim}}{\int_{t}^{0}{\frac{1}{x^{2} + 4}dx}}} & & & \text{Rewrite as a limit.} \\ & {= \underset{t\rightarrow\text{−}\infty}{\text{lim}}\frac{1}{2}\text{tan}^{-1}\frac{x}{2}\left| {}_{\begin{matrix} \\ t \end{matrix}}^{\begin{matrix} 0 \\ \end{matrix}} \right.} & & & \text{Find the antiderivative.} \\ & {= \frac{1}{2}\underset{t\rightarrow\text{−}\infty}{\text{lim}}(\text{tan}^{-1}0 - \text{tan}^{-1}\frac{t}{2})} & & & \text{Evaluate the antiderivative.} \\ & {= \frac{\pi}{4}.} & & & \text{Evaluate the limit and simplify.} \end{array}$$

The improper integral converges to $\frac{\pi}{4}.$

Evaluating an Improper Integral on $(\text{−}\infty,\text{+}\infty)$ 在 $(\text{−}\infty,\text{+}\infty)$ 上计算反常积分

Evaluate ${\int_{\text{−}\infty}^{+ \infty}{xe^{x}dx}}.$ State whether the improper integral converges or diverges.

Solution 解答

Start by splitting up the integral:

$${\int_{\text{−}\infty}^{+ \infty}{xe^{x}dx = {\int_{\text{−}\infty}^{0}{xe^{x}dx + {\int_{0}^{+ \infty}{xe^{x}dx}}}}}}.$$

If either $\int_{\text{−}\infty}^{0}{xe^{x}dx}$ or $\int_{0}^{+ \infty}{xe^{x}dx}$ diverges, then $\int_{\text{−}\infty}^{+ \infty}{xe^{x}dx}$ diverges. Compute each integral separately. For the first integral,

$$\begin{array}{clccl} {\int_{\text{−}\infty}^{0}{xe^{x}dx}} & {= \underset{t\rightarrow\text{−}\infty}{\text{lim}}{\int_{t}^{0}{xe^{x}dx}}} & & & \text{Rewrite as a limit.} \\ & {= \underset{t\rightarrow\text{−}\infty}{\text{lim}}\left( {xe^{x} - e^{x}} \right)\left| {}_{\begin{matrix} \\ t \end{matrix}}^{\begin{matrix} 0 \\ \end{matrix}} \right.} & & & \begin{array}{l} \text{Use integration by parts to find the} \\ {\text{antiderivative. (Here}\ u = x\ \text{and}\ dv = e^{x}{dx}.)} \end{array} \\ & {= \underset{t\rightarrow\text{−}\infty}{\text{lim}}\left( {-1 - te^{t} + e^{t}} \right)} & & & \text{Evaluate the antiderivative.} \\ & {= -1.} & & & \begin{array}{l} {\text{Evaluate the limit.}\ \textit{Note:}\ \underset{t\rightarrow\text{−}\infty}{\text{lim}}te^{t}\ \text{is}} \\ {\text{indeterminate of the form}\ 0 \cdot \infty.\ \text{Thus,}} \\ {\underset{t\rightarrow\text{−}\infty}{\text{lim}}te^{t} = \underset{t\rightarrow\text{−}\infty}{\text{lim}}\frac{t}{e^{\text{−}t}} = \underset{t\rightarrow\text{−}\infty}{\text{lim}}\frac{-1}{e^{\text{−}t}} = \underset{t\rightarrow\text{−}\infty}{\text{lim}} - e^{t} = 0\ \text{by}} \\ \text{L’Hôpital’s Rule.} \end{array} \end{array}$$

The first improper integral converges. For the second integral,

$$\begin{array}{clccl} {\int_{0}^{+ \infty}{xe^{x}dx}} & {= \underset{t\rightarrow\text{+}\infty}{\text{lim}}{\int_{0}^{t}{xe^{x}dx}}} & & & \text{Rewrite as a limit.} \\ & {= \underset{t\rightarrow\text{+}\infty}{\text{lim}}\left( {xe^{x} - e^{x}} \right)\left| {}_{\begin{matrix} \\ 0 \end{matrix}}^{\begin{matrix} t \\ \end{matrix}} \right.} & & & \text{Find the antiderivative.} \\ & {= \underset{t\rightarrow\text{+}\infty}{\text{lim}}\left( {te^{t} - e^{t} + 1} \right)} & & & \text{Evaluate the antiderivative.} \\ & {= \underset{t\rightarrow\text{+}\infty}{\text{lim}}\left( {\left( {t - 1} \right)e^{t} + 1} \right)} & & & {\text{Rewrite.}\ (te^{t} - e^{t}\ \text{is indeterminate.)}} \\ & {= \text{+}\infty.} & & & \text{Evaluate the limit.} \end{array}$$

Thus, $\int_{0}^{+ \infty}{xe^{x}dx}$ diverges. Since this integral diverges, $\int_{\text{−}\infty}^{+ \infty}{xe^{x}dx}$ diverges as well.

Evaluate ${\int_{-3}^{+ \infty}{e^{\text{−}x}dx}}.$ State whether the improper integral converges or diverges.

Integrating a Discontinuous Integrand 对不连续被积函数积分

Now let’s examine integrals of functions containing an infinite discontinuity in the interval over which the integration occurs. Consider an integral of the form $\int_{a}^{b}{f(x)dx,}$ where $f(x)$ is continuous over $\lbrack a,b)$ and discontinuous at $b.$ Since the function $f(x)$ is continuous over $\lbrack a,t\rbrack$ for all values of $t$ satisfying $a < t < b,$ the integral $\int_{a}^{t}{f(x)dx}$ is defined for all such values of $t.$ Thus, it makes sense to consider the values of $\int_{a}^{t}{f(x)dx}$ as $t$ approaches $b$ for $a < t < b.$ That is, we define ${\int_{a}^{b}{f(x)dx = \underset{t\rightarrow b^{-}}{\text{lim}}{\int_{a}^{t}{f(x)dx}}}},$ provided this limit exists. Figure 3.21 illustrates $\int_{a}^{t}{f(x)dx}$ as areas of regions for values of $t$ approaching $b.$

We use a similar approach to define $\int_{a}^{b}{f(x)dx,}$ where $f(x)$ is continuous over $(a,b\rbrack$ and discontinuous at $a.$ We now proceed with a formal definition.

1. Let $f(x)$ be continuous over $\lbrack a,b).$ Then,

$${\int_{a}^{b}{f(x)dx = \underset{t\rightarrow b^{-}}{\text{lim}}{\int_{a}^{t}{f(x)dx}}}}.$$ (3.19)

2. Let $f(x)$ be continuous over $(a,b\rbrack.$ Then,

$${\int_{a}^{b}{f(x)dx = \underset{t\rightarrow a^{+}}{\text{lim}}{\int_{t}^{b}{f(x)dx}}}}.$$ (3.20)

In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.

3. If $f(x)$ is continuous over $\lbrack a,b\rbrack$ except at a point $c$ in $(a,b),$ then

$$\int_{a}^{b}{f(x)dx = {\int_{a}^{c}{f(x)dx + {\int_{c}^{b}{f(x)dx,}}}}}$$ (3.21)

provided both $\int_{a}^{c}{f(x)dx}$ and $\int_{c}^{b}{f(x)dx}$ converge. If either of these integrals diverges, then $\int_{a}^{b}{f(x)dx}$ diverges.

The following examples demonstrate the application of this definition.

Integrating a Discontinuous Integrand 对不连续被积函数积分

Evaluate ${\int_{0}^{4}{\frac{1}{\sqrt{4 - x}}dx}},$ if possible. State whether the integral converges or diverges.

Solution 解答

The function $f(x) = \frac{1}{\sqrt{4 - x}}$ is continuous over $\lbrack 0,4)$ and discontinuous at 4. Using Equation 3.19 from the definition, rewrite $\int_{0}^{4}{\frac{1}{\sqrt{4 - x}}dx}$ as a limit:

$$\begin{array}{clccl} {\int_{0}^{4}{\frac{1}{\sqrt{4 - x}}dx}} & {= \underset{t\rightarrow 4^{-}}{\text{lim}}{\int_{0}^{t}{\frac{1}{\sqrt{4 - x}}dx}}} & & & \text{Rewrite as a limit.} \\ & {= \underset{t\rightarrow 4^{-}}{\text{lim}}\left( {-2\sqrt{4 - x}} \right)\left| {}_{\begin{matrix} \\ 0 \end{matrix}}^{\begin{matrix} t \\ \end{matrix}} \right.} & & & \text{Find the antiderivative.} \\ & {= \underset{t\rightarrow 4^{-}}{\text{lim}}\left( {-2\sqrt{4 - t} + 4} \right)} & & & \text{Evaluate the antiderivative.} \\ & {= 4.} & & & \text{Evaluate the limit.} \end{array}$$

The improper integral converges.

Integrating a Discontinuous Integrand 对不连续被积函数积分

Evaluate ${\int_{0}^{2}{x\mspace{2mu}\text{ln}\mspace{2mu} x\mspace{2mu} dx}}.$ State whether the integral converges or diverges.

Solution 解答

Since $f(x) = x\mspace{2mu}\text{ln}\mspace{2mu} x$ is continuous over $(0,2\rbrack$ and is discontinuous at zero, we can rewrite the integral in limit form using Equation 3.20:

$$\begin{array}{clccl} {\int_{0}^{2}{x\mspace{2mu}\text{ln}\mspace{2mu} x\mspace{2mu} dx}} & {= \underset{t\rightarrow 0^{+}}{\text{lim}}{\int_{t}^{2}{x\mspace{2mu}\text{ln}\mspace{2mu} x\mspace{2mu} dx}}} & & & \text{Rewrite as a limit.} \\ & {= \underset{t\rightarrow 0^{+}}{\text{lim}}\left( {\frac{1}{2}x^{2}\text{ln}\mspace{2mu} x - \frac{1}{4}x^{2}} \right)\left| {}_{\begin{matrix} \\ t \end{matrix}}^{\begin{matrix} 2 \\ \end{matrix}} \right.} & & & \begin{array}{l} {\text{Evaluate}\ \int^{}x\mspace{2mu}\text{ln}\mspace{2mu} x\mspace{2mu} dx\ \text{using integration by parts}} \\ {\text{with}\ u = \text{ln}\mspace{2mu} x\ \text{and}\ dv = x{dx}.} \end{array} \\ & {= \underset{t\rightarrow 0^{+}}{\text{lim}}\left( {2\mspace{2mu}\text{ln}\mspace{2mu} 2 - 1 - \frac{1}{2}t^{2}\text{ln}\mspace{2mu} t + \frac{1}{4}t^{2}} \right).} & & & \text{Evaluate the antiderivative.} \\ & {= 2\mspace{2mu}\text{ln}\mspace{2mu} 2 - 1.} & & & \begin{array}{l} {\text{Evaluate the limit.}\ \underset{t\rightarrow 0^{+}}{\text{lim}}t^{2}\text{ln}\mspace{2mu} t\ \text{is indeterminate.}} \\ \text{To evaluate it, rewrite as a quotient and apply} \\ \text{L’Hôpital’s rule.} \end{array} \end{array}$$

The improper integral converges.

Integrating a Discontinuous Integrand 对不连续被积函数积分

Evaluate ${\int_{-1}^{1}{\frac{1}{x^{3}}dx}}.$ State whether the improper integral converges or diverges.

Solution 解答

Since $f(x) = 1\text{/}x^{3}$ is discontinuous at zero, using Equation 3.21, we can write

$${\int_{-1}^{1}{\frac{1}{x^{3}}dx = {\int_{-1}^{0}{\frac{1}{x^{3}}dx + {\int_{0}^{1}{\frac{1}{x^{3}}dx}}}}}}.$$

If either of the two integrals diverges, then the original integral diverges. Begin with ${\int_{-1}^{0}{\frac{1}{x^{3}}dx}}:$

$$\begin{array}{clccl} {\int_{-1}^{0}{\frac{1}{x^{3}}dx}} & {= \underset{t\rightarrow 0^{-}}{\text{lim}}{\int_{-1}^{t}{\frac{1}{x^{3}}dx}}} & & & \text{Rewrite as a limit.} \\ & {= \underset{t\rightarrow 0^{-}}{\text{lim}}\left( {- \frac{1}{2x^{2}}} \right)\left| {}_{\begin{matrix} \\ -1 \end{matrix}}^{\begin{matrix} t \\ \end{matrix}} \right.} & & & \text{Find the antiderivative.} \\ & {= \underset{t\rightarrow 0^{-}}{\text{lim}}\left( {- \frac{1}{2t^{2}} + \frac{1}{2}} \right)} & & & \text{Evaluate the antiderivative.} \\ & {= \text{−}\infty.} & & & \text{Evaluate the limit.} \end{array}$$

Therefore, $\int_{-1}^{0}{\frac{1}{x^{3}}dx}$ diverges. Since $\int_{-1}^{0}{\frac{1}{x^{3}}dx}$ diverges, $\int_{-1}^{1}{\frac{1}{x^{3}}dx}$ diverges.

Evaluate ${\int_{0}^{2}{\frac{1}{x}dx}}.$ State whether the integral converges or diverges.

A Comparison Theorem 比较定理

It is not always easy or even possible to evaluate an improper integral directly; however, by comparing it with another carefully chosen integral, it may be possible to determine its convergence or divergence. To see this, consider two continuous functions $f(x)$ and $g(x)$ satisfying $0 \leq f(x) \leq g(x)$ for $x \geq a$ (Figure 3.22). In this case, we may view integrals of these functions over intervals of the form $\lbrack a,t\rbrack$ as areas, so we have the relationship

$$0 \leq {\int_{a}^{t}{f(x)dx \leq {\int_{a}^{t}{g(x)dx}}}}\ \text{for}\ t \geq a.$$

Thus, if

$$\int_{a}^{+ \infty}{f(x)dx = \underset{t\rightarrow\text{+}\infty}{\text{lim}}{\int_{a}^{t}{f(x)dx = \text{+}\infty,}}}$$

then

$\int_{a}^{+ \infty}{g(x)dx = \underset{t\rightarrow\text{+}\infty}{\text{lim}}{\int_{a}^{t}{g(x)dx = \text{+}\infty}}}$ as well. That is, if the area of the region between the graph of $f(x)$ and the *x*-axis over $\lbrack a,\text{+}\infty)$ is infinite, then the area of the region between the graph of $g(x)$ and the *x*-axis over $\lbrack a,\text{+}\infty)$ is infinite too.

On the other hand, if

$\int_{a}^{+ \infty}{g(x)dx = \underset{t\rightarrow\text{+}\infty}{\text{lim}}{\int_{a}^{t}{g(x)dx = L}}}$ for some real number $L,$ then

$\int_{a}^{+ \infty}{f(x)dx = \underset{t\rightarrow\text{+}\infty}{\text{lim}}{\int_{a}^{t}{f(x)dx}}}$ must converge to some value less than or equal to $L,$ since $\int_{a}^{t}{f(x)dx}$ increases as $t$ increases and $\int_{a}^{t}{f(x)dx \leq L}$ for all $t \geq a.$

If the area of the region between the graph of $g(x)$ and the *x*-axis over $\lbrack a,\text{+}\infty)$ is finite, then the area of the region between the graph of $f(x)$ and the *x*-axis over $\lbrack a,\text{+}\infty)$ is also finite.

These conclusions are summarized in the following theorem.

A Comparison Theorem 比较定理

Let $f(x)$ and $g(x)$ be continuous over $\lbrack a,\text{+}\infty).$ Assume that $0 \leq f(x) \leq g(x)$ for $x \geq a.$

1. If ${\int_{a}^{+ \infty}{f(x)dx = \underset{t\rightarrow\text{+}\infty}{\text{lim}}{\int_{a}^{t}{f(x)dx = \text{+}\infty}}}},$ then ${\int_{a}^{+ \infty}{g(x)dx = \underset{t\rightarrow\text{+}\infty}{\text{lim}}{\int_{a}^{t}{g(x)dx = \text{+}\infty}}}}.$

2. If $\int_{a}^{+ \infty}{g(x)dx = \underset{t\rightarrow\text{+}\infty}{\text{lim}}{\int_{a}^{t}{g(x)dx = L,}}}$ where $L$ is a real number, then $\int_{a}^{+ \infty}{f(x)dx = \underset{t\rightarrow\text{+}\infty}{\text{lim}}{\int_{a}^{t}{f(x)dx = M}}}$ for some real number $M \leq L.$

Applying the Comparison Theorem 应用比较定理

Use a comparison to show that $\int_{1}^{+ \infty}{\frac{1}{xe^{x}}dx}$ converges.

Solution 解答

We can see that

$$0 \leq \frac{1}{xe^{x}} \leq \frac{1}{e^{x}} = e^{\text{−}x},$$

so if $\int_{1}^{+ \infty}{e^{\text{−}x}dx}$ converges, then so does ${\int_{1}^{+ \infty}{\frac{1}{xe^{x}}dx}}.$ To evaluate ${\int_{1}^{+ \infty}{e^{\text{−}x}dx}},$ first rewrite it as a limit:

$$\begin{array}{cl} {{\int_{1}^{+ \infty}e^{\text{−}x}}dx} & {= \underset{t\rightarrow\text{+}\infty}{\text{lim}}{\int_{1}^{t}{e^{\text{−}x}dx}}} \\ & {= \underset{t\rightarrow\text{+}\infty}{\text{lim}}\left( {\text{−}e^{\text{−}x}} \right)\left| \begin{matrix} t \\ 1 \end{matrix} \right.} \\ & {= \underset{t\rightarrow\text{+}\infty}{\text{lim}}\left( {\text{−}e^{\text{−}t} + e^{–1}} \right)} \\ & {= e^{–1}.} \end{array}$$

Since $\int_{1}^{+ \infty}{e^{\text{−}x}dx}$ converges, so does ${\int_{1}^{+ \infty}{\frac{1}{xe^{x}}dx}}.$

Applying the Comparison Theorem 应用比较定理

Use the comparison theorem to show that $\int_{1}^{+ \infty}{\frac{1}{x^{p}}dx}$ diverges for all $p < 1.$

Solution 解答

For $p < 1,$ $1\text{/}x \leq 1\text{/}(x^{p})$ over $\lbrack 1,\text{+}\infty).$ In Example 3.47, we showed that ${\int_{1}^{+ \infty}{\frac{1}{x}dx = \text{+}\infty}}.$ Therefore, $\int_{1}^{+ \infty}{\frac{1}{x^{p}}dx}$ diverges for all $p < 1.$

Use a comparison to show that $\int_{e}^{+ \infty}{\frac{\text{ln}\mspace{2mu} x}{x}dx}$ diverges.

Laplace Transforms 拉普拉斯变换

In the last few chapters, we have looked at several ways to use integration for solving real-world problems. For this next project, we are going to explore a more advanced application of integration: integral transforms. Specifically, we describe the Laplace transform and some of its properties. The Laplace transform is used in engineering and physics to simplify the computations needed to solve some problems. It takes functions expressed in terms of time and *transforms* them to functions expressed in terms of frequency. It turns out that, in many cases, the computations needed to solve problems in the frequency domain are much simpler than those required in the time domain.

The Laplace transform is defined in terms of an integral as

$$L\left\\{ {f(t)} \right\\} = F(s) = {\int_{0}^{\infty}{e^{\text{−}st}f(t)dt}}.$$

Note that the input to a Laplace transform is a function of time, $f(t),$ and the output is a function of frequency, $F(s).$ Although many real-world examples require the use of complex numbers (involving the imaginary number $i = \sqrt{-1}),$ in this project we limit ourselves to functions of real numbers.

Let’s start with a simple example. Here we calculate the Laplace transform of $f(t) = t$. We have

$$L\left\\{ t \right\\} = {\int_{0}^{\infty}{te^{\text{−}st}dt}}.$$

This is an improper integral, so we express it in terms of a limit, which gives

$$L\left\\{ t \right\\} = {\int_{0}^{\infty}{te^{\text{−}st}dt}} = \underset{z\rightarrow\infty}{\text{lim}}{\int_{0}^{z}{te^{\text{−}st}dt}}.$$

Now we use integration by parts to evaluate the integral. Note that we are integrating with respect to *t*, so we treat the variable *s* as a constant. We have

$$\begin{array}{rllccrll} u & = & t & & & {dv} & = & {e^{\text{−}st}dt} \\ {du} & = & {dt} & & & v & = & {- \frac{1}{s}e^{\text{−}st}.} \end{array}$$

Then we obtain

$$\begin{array}{cl} {\underset{z\rightarrow\infty}{\text{lim}}{\int_{0}^{z}{te^{\text{−}st}dt}}} & {= \underset{z\rightarrow\infty}{\text{lim}}\left\lbrack {\left. \left\lbrack {- \frac{t}{s}e^{\text{−}st}} \right\rbrack \right|_{0}^{z} + \frac{1}{s}{\int_{0}^{z}{e^{\text{−}st}dt}}} \right\rbrack} \\ & {= \underset{z\rightarrow\infty}{\text{lim}}\left\lbrack {\left\lbrack {- \frac{z}{s}e^{\text{−}sz} + \frac{0}{s}e^{-0s}} \right\rbrack + \frac{1}{s}{\int_{0}^{z}{e^{\text{−}st}dt}}} \right\rbrack} \\ & {= \underset{z\rightarrow\infty}{\text{lim}}\left\lbrack {\left\lbrack {- \frac{z}{s}e^{\text{−}sz} + 0} \right\rbrack - \frac{1}{s}\left. \left\lbrack \frac{e^{\text{−}st}}{s} \right\rbrack \right|_{0}^{z}} \right\rbrack} \\ & {= \underset{z\rightarrow\infty}{\text{lim}}\left\lbrack {\left\lbrack {- \frac{z}{s}e^{\text{−}sz}} \right\rbrack - \frac{1}{s^{2}}\left\lbrack {e^{\text{−}sz} - 1} \right\rbrack} \right\rbrack} \\ & {= \underset{z\rightarrow\infty}{\text{lim}}\left\lbrack {- \frac{z}{se^{sz}}} \right\rbrack - \underset{z\rightarrow\infty}{\text{lim}}\left\lbrack \frac{1}{s^{2}e^{sz}} \right\rbrack + \underset{z\rightarrow\infty}{\text{lim}}\frac{1}{s^{2}}} \\ & {= 0 - 0 + \frac{1}{s^{2}}} \\ & {= \frac{1}{s^{2}}.} \end{array}$$

1. Calculate the Laplace transform of $f(t) = 1.$

2. Calculate the Laplace transform of $f(t) = e^{-3t}.$

3. Calculate the Laplace transform of $f(t) = t^{2}.$ (Note, you will have to integrate by parts twice.)

Laplace transforms are often used to solve differential equations. Differential equations are not covered in detail until later in this book; but, for now, let’s look at the relationship between the Laplace transform of a function and the Laplace transform of its derivative.

Let’s start with the definition of the Laplace transform. We have

$$L\left\\{ {f(t)} \right\\} = {\int_{0}^{\infty}{e^{\text{−}st}f(t)dt}} = \underset{z\rightarrow\infty}{\text{lim}}{\int_{0}^{z}{e^{\text{−}st}f(t)dt}}.$$

4. Use integration by parts to evaluate $\underset{z\rightarrow\infty}{\text{lim}}{\int_{0}^{z}{e^{\text{−}st}f(t)dt}}.$ (Let $u = f(t)$ and $dv = e^{\text{−}st}dt.)$

After integrating by parts and evaluating the limit, you should see that

$$L\left\\{ {f(t)} \right\\} = \frac{f(0)}{s} + \frac{1}{s}\left\lbrack {L\left\\{ {f^{\prime}(t)} \right\\}} \right\rbrack.$$

Then,

$$L\left\\{ {f^{\prime}(t)} \right\\} = sL\left\\{ {f(t)} \right\\} - f(0).$$

Thus, differentiation in the time domain simplifies to multiplication by *s* in the frequency domain.

The final thing we look at in this project is how the Laplace transforms of $f(t)$ and its antiderivative are related. Let $g(t) = {\int_{0}^{t}f}(u)du.$ Then,

$$L\left\\{ {g(t)} \right\\} = {\int_{0}^{\infty}{e^{\text{−}st}g(t)dt}} = \underset{z\rightarrow\infty}{\text{lim}}{\int_{0}^{z}{e^{\text{−}st}g(t)dt}}.$$

5. Use integration by parts to evaluate $\underset{z\rightarrow\infty}{\text{lim}}{\int_{0}^{z}{e^{\text{−}st}g(t)dt}}.$ (Let $u = g(t)$ and $dv = e^{\text{−}st}dt.$ Note, by the way, that we have defined $g(t),$ $du = f(t)dt.)$

As you might expect, you should see that

$$L\left\\{ {g(t)} \right\\} = \frac{1}{s} \cdot L\left\\{ {f(t)} \right\\}.$$

Integration in the time domain simplifies to division by *s* in the frequency domain.

Section 3.7 Exercises 3.7 节习题

Evaluate the following integrals. If the integral is not convergent, answer “divergent.”

347.

$\int_{2}^{4}\frac{dx}{{(x - 3)}^{2}}$

348\.

${\int_{0}^{\infty}\frac{1}{4 + x^{2}}}dx$

349.

$\int_{0}^{2}{\frac{1}{\sqrt{4 - x^{2}}}dx}$

350\.

$\int_{1}^{\infty}{\frac{1}{x\mspace{2mu}\text{ln}\mspace{2mu} x}dx}$

351.

$\int_{1}^{\infty}{xe^{\text{−}x}dx}$

352\.

$\int_{\text{−}\infty}^{\infty}{\frac{x}{x^{2} + 1}dx}$

353.

Without integrating, determine whether the integral $\int_{1}^{\infty}{\frac{1}{\sqrt{x^{3} + 1}}dx}$ converges or diverges by comparing the function $f(x) = \frac{1}{\sqrt{x^{3} + 1}}$ with $g(x) = \frac{1}{\sqrt{x^{3}}}.$

354\.

Without integrating, determine whether the integral $\int_{1}^{\infty}{\frac{1}{\sqrt{x + 1}}dx}$ converges or diverges.

Determine whether the improper integrals converge or diverge. If possible, determine the value of the integrals that converge.

355.

$\int_{0}^{\infty}{e^{\text{−}x}\text{cos}\mspace{2mu} x\mspace{2mu} dx}$

356\.

$\int_{1}^{\infty}{\frac{\text{ln}\mspace{2mu} x}{x}dx}$

357.

$\int_{0}^{1}{\frac{\text{ln}\mspace{2mu} x}{\sqrt{x}}dx}$

358\.

$\int_{0}^{1}{\text{ln}\mspace{2mu} x\mspace{2mu} dx}$

359.

$\int_{\text{−}\infty}^{\infty}{\frac{1}{x^{2} + 1}dx}$

360\.

$\int_{1}^{5}\frac{dx}{\sqrt{x - 1}}$

361.

$\int_{-2}^{2}\frac{dx}{\left( {1 + x} \right)^{2}}$

362\.

$\int_{0}^{\infty}{e^{\text{−}x}dx}$

363.

$\int_{0}^{\infty}{\text{sin}\mspace{2mu} x\mspace{2mu} dx}$

364\.

$\int_{\text{−}\infty}^{\infty}{\frac{e^{x}}{1 + e^{2x}}dx}$

365.

$\int_{0}^{1}\frac{dx}{\sqrt[3]{x}}$

366\.

$\int_{0}^{2}\frac{dx}{x^{3}}$

367.

$\int_{-1}^{2}\frac{dx}{x^{3}}$

368\.

$\int_{0}^{1}\frac{dx}{\sqrt{1 - x^{2}}}$

369.

$\int_{0}^{3}{\frac{1}{x - 1}dx}$

370\.

$\int_{1}^{\infty}{\frac{5}{x^{3}}dx}$

371.

$\int_{3}^{5}{\frac{5}{{(x - 4)}^{2}}dx}$

Determine the convergence of each of the following integrals by comparison with the given integral. If the integral converges, find the number to which it converges.

372\.

${\int_{1}^{\infty}\frac{dx}{x^{2} + 4x}};$ compare with ${\int_{1}^{\infty}\frac{dx}{x^{2}}}.$

373.

${\int_{1}^{\infty}\frac{dx}{\sqrt{x} + 1}};$ compare with ${\int_{1}^{\infty}\frac{dx}{2\sqrt{x}}}.$

Evaluate the integrals. If the integral diverges, answer “diverges.”

374\.

$\int_{1}^{\infty}\frac{dx}{x^{e}}$

375.

$\int_{0}^{1}\frac{dx}{x^{\pi}}$

376\.

$\int_{0}^{1}\frac{dx}{\sqrt{1 - x}}$

377.

$\int_{0}^{1}\frac{dx}{1 - x}$

378\.

$\int_{\text{−}\infty}^{0}\frac{dx}{x^{2} + 1}$

379.

$\int_{-1}^{1}\frac{dx}{\sqrt{1 - x^{2}}}$

380\.

$\int_{0}^{1}{\frac{\text{ln}\mspace{2mu} x}{x}dx}$

381.

$\int_{0}^{e}{\text{ln}(x)dx}$

382\.

$\int_{0}^{\infty}{xe^{\text{−}x}dx}$

383.

$\int_{\text{−}\infty}^{\infty}{\frac{x}{\left( {x^{2} + 1} \right)^{2}}dx}$

384\.

$\int_{0}^{\infty}{e^{x}dx}$

Evaluate the improper integrals. Each of these integrals has an infinite discontinuity either at an endpoint or at an interior point of the interval.

385.

$\int_{0}^{9}\frac{dx}{\sqrt{9 - x}}$

386\.

$\int_{-27}^{1}\frac{dx}{x^{2\text{/}3}}$

387.

$\int_{0}^{3}\frac{dx}{\sqrt{9 - x^{2}}}$

388\.

$\int_{6}^{24}\frac{dt}{t\sqrt{t^{2} - 36}}$

389.

$\int_{0}^{4}{x\mspace{2mu}\text{ln}(4x)dx}$

390\.

$\int_{0}^{3}{\frac{x}{\sqrt{9 - x^{2}}}dx}$

391.

Evaluate ${\int_{.5}^{1}\frac{dx}{\sqrt{1 - x^{2}}}}.$ (Be careful!) (Express your answer using three decimal places.)

392\.

Evaluate ${\int_{1}^{4}\frac{dx}{\sqrt{x^{2} - 1}}}.$ (Express the answer in exact form.)

393.

Evaluate ${\int_{2}^{\infty}\frac{dx}{{(x^{2} - 1)}^{3\text{/}2}}}.$

394\.

Find the area of the region in the first quadrant between the curve $y = e^{-6x}$ and the *x*-axis.

395.

Find the area of the region bounded by the curve $y = \frac{7}{x^{2}},$ the *x*-axis, and on the left by $x = 1.$

396\.

Find the area under the curve $y = \frac{1}{\left( {x + 1} \right)^{3\text{/}2}},$ bounded on the left by $x = 3.$

397.

Find the area under $y = \frac{5}{1 + x^{2}}$ in the first quadrant.

398\.

Find the volume of the solid generated by revolving about the *x*-axis the region under the curve $y = \frac{3}{x}$ from $x = 1$ to $x = \infty.$

399.

Find the volume of the solid generated by revolving about the *y*-axis the region under the curve $y = 6e^{-2x}$ in the first quadrant.

400\.

Find the volume of the solid generated by revolving about the *x*-axis the area under the curve $y = 3e^{\text{−}x}$ in the first quadrant.

The Laplace transform of a continuous function over the interval $\lbrack 0,\infty)$ is defined by $F(s) = {\int_{0}^{\infty}{e^{\text{−}sx}f(x)dx}}$ (see the Student Project). This definition is used to solve some important initial-value problems in differential equations, as discussed later. The domain of *F* is the set of all real numbers *s* such that the improper integral converges. Find the Laplace transform *F* of each of the following functions and give the domain of *F*.

401.

$f(x) = 1$

402\.

$f(x) = x$

403.

$f(x) = \text{cos}(2x)$

404\.

$f(x) = e^{ax}$

405.

Use the formula for arc length to show that the circumference of the circle $x^{2} + y^{2} = 1$ is $2\pi.$

A non-negative function is a probability density function if it satisfies the following definition: ${\int_{\text{−}\infty}^{\infty}{f(t)dt = 1}}.$ The probability that a random variable *x* lies between *a* and *b* is given by $P(a \leq x \leq b) = {\int_{a}^{b}{f(t)dt}}.$

406\.

Show that $f(x) = \left\{ \begin{matrix} {0\mspace{2mu}\text{if}\mspace{2mu} x < 0} \\ {7e^{-7x}\text{if}\mspace{2mu} x \geq 0} \end{matrix} \right.$ is a probability density function.

407.

Find the probability that *x* is between 0 and 0.3. (Use the function defined in the preceding problem.) Use four-place decimal accuracy.

Key Terms 关键术语

absolute error

if $B$ is an estimate of some quantity having an actual value of $A,$ then the absolute error is given by $\left| {A - B} \right|$

computer algebra system (CAS)

technology used to perform many mathematical tasks, including integration

improper integral

an integral over an infinite interval or an integral of a function containing an infinite discontinuity on the interval; an improper integral is defined in terms of a limit. The improper integral converges if this limit is a finite real number; otherwise, the improper integral diverges

integration by parts

a technique of integration that allows the exchange of one integral for another using the formula $\int^{}u\ dv = uv - \int^{}v\ du$

integration table

a table that lists integration formulas

midpoint rule

a rule that uses a Riemann sum of the form $M_{n} = \sum\limits_{i = 1}^{n}f(m_{i})\text{Δ}x,$ where $m_{i}$ is the midpoint of the *i*th subinterval to approximate $\int_{a}^{b}{f(x)dx}$

numerical integration

the variety of numerical methods used to estimate the value of a definite integral, including the midpoint rule, trapezoidal rule, and Simpson’s rule

partial fraction decomposition

a technique used to break down a rational function into the sum of simple rational functions

power reduction formula

a rule that allows an integral of a power of a trigonometric function to be exchanged for an integral involving a lower power

relative error

error as a percentage of the absolute value, given by $\left| \frac{A - B}{A} \right| = \left| \frac{A - B}{A} \right| \cdot 100\text{\%}$

Simpson’s rule

a rule that approximates $\int_{a}^{b}{f(x)dx}$ using the integrals of a piecewise quadratic function. The approximation $S_{n}$ to $\int_{a}^{b}{f(x)dx}$ is given by $S_{n} = \frac{\text{Δ}x}{3}\begin{pmatrix}{f\left( x_{0} \right) + 4f\left( x_{1} \right) + 2f\left( x_{2} \right) + 4f\left( x_{3} \right) + 2f\left( x_{4} \right) + 4f\left( x_{5} \right)} \\ {+ \cdots + 2f\left( x_{n - 2} \right) + 4f\left( x_{n - 1} \right) + f\left( x_{n} \right)}\end{pmatrix}$

trapezoidal rule

a rule that approximates $\int_{a}^{b}{f(x)dx}$ using trapezoids

trigonometric integral

an integral involving powers and products of trigonometric functions

trigonometric substitution

an integration technique that converts an algebraic integral containing expressions of the form $\sqrt{a^{2} - x^{2}},$ $\sqrt{a^{2} + x^{2}},$ or $\sqrt{x^{2} - a^{2}}$ into a trigonometric integral

绝对误差

若 $B$ 是某个实际值为 $A$ 的量的估计值，则绝对误差由 $\left| {A - B} \right|$ 给出。

计算机代数系统（CAS）

用于执行许多数学任务（包括积分）的技术。

反常积分（广义积分）

在无穷区间上的积分，或被积函数在该区间中含有无穷间断点的积分；反常积分用极限来定义。若该极限为有限实数，则反常积分收敛；否则反常积分发散。

分部积分（法）

一种积分技巧，利用公式 $\int^{}u\ dv = uv - \int^{}v\ du$ 将一个积分替换为另一个积分。

积分表

列出积分公式的表。

中点法则（中点法）

一种利用形如 $M_{n} = \sum\limits_{i = 1}^{n}f(m_{i})\text{Δ}x$ 的黎曼和来近似 $\int_{a}^{b}{f(x)dx}$ 的法则，其中 $m_{i}$ 是第 *i* 个子区间的中点。

数值积分

用于估计定积分值的各种数值方法，包括中点法则、梯形法则和辛普森法则。

部分分式分解

一种将有理函数分解为若干简单有理函数之和的技巧。

降幂公式（递推公式）

一种将三角函数的幂次积分替换为较低幂次积分的法则。

相对误差

作为绝对值的百分比的误差，由 $\left| \frac{A - B}{A} \right| = \left| \frac{A - B}{A} \right| \cdot 100\text{\%}$ 给出。

辛普森法则

一种利用分段二次函数的积分来近似 $\int_{a}^{b}{f(x)dx}$ 的法则。对 $\int_{a}^{b}{f(x)dx}$ 的近似 $S_{n}$ 由 $S_{n} = \frac{\text{Δ}x}{3}\begin{pmatrix}{f\left( x_{0} \right) + 4f\left( x_{1} \right) + 2f\left( x_{2} \right) + 4f\left( x_{3} \right) + 2f\left( x_{4} \right) + 4f\left( x_{5} \right)} \\ {+ \cdots + 2f\left( x_{n - 2} \right) + 4f\left( x_{n - 1} \right) + f\left( x_{n} \right)}\end{pmatrix}$ 给出。

梯形法则（梯形法）

一种利用梯形来近似 $\int_{a}^{b}{f(x)dx}$ 的法则。

三角积分

含三角函数的幂与乘积的积分。

三角换元（法）

一种将含有 $\sqrt{a^{2} - x^{2}},$ $\sqrt{a^{2} + x^{2}},$ 或 $\sqrt{x^{2} - a^{2}}$ 形式的代数积分转化为三角积分的积分技巧。

Key Equations 关键公式

Integration by parts formula	$\int{u\ dv = uv - {\int{v\ du}}}$
Integration by parts for definite integrals	${\int_{a}^{b}{u\ dv}} = \left. {uv} \right|_{a}^{b} - {\int_{a}^{b}{v\ du}}$

Integration by parts formula	$\int{u\ dv = uv - {\int{v\ du}}}$
Integration by parts for definite integrals	${\int_{a}^{b}{u\ dv}} = \left. {uv} \right|_{a}^{b} - {\int_{a}^{b}{v\ du}}$

To integrate products involving $\text{sin}(ax),$ $\text{sin}(bx),$ $\text{cos}(ax),$ and $\text{cos}(bx),$ use the substitutions.

Sine Products	$\text{sin}\left( {ax} \right)\text{sin}\left( {bx} \right) = \frac{1}{2}\text{cos}(\left( {a - b} \right)x) - \frac{1}{2}\text{cos}(\left( {a + b} \right)x)$
Sine and Cosine Products	$\text{sin}\left( {ax} \right)\text{cos}\left( {bx} \right) = \frac{1}{2}\text{sin}\left( {\left( {a - b} \right)x} \right) + \frac{1}{2}\text{sin}(\left( {a + b} \right)x)$
Cosine Products	$\text{cos}\left( {ax} \right)\text{cos}\left( {bx} \right) = \frac{1}{2}\text{cos}(\left( {a - b} \right)x) + \frac{1}{2}\text{cos}(\left( {a + b} \right)x)$
Power Reduction Formula	$\int\text{sec}^{n}x~dx = \frac{\sec^{n - 2}x~\tan~x}{n - 1} + \frac{n - 2}{n - 1}\int\text{sec}^{n - 2}x\ dx;n \neq 1$
Power Reduction Formula	$\int\text{tan}^{n}x\ dx = \frac{1}{n - 1}\text{tan}^{n - 1}x - \int\text{tan}^{n - 2}x\ dx$

Sine Products	$\text{sin}\left( {ax} \right)\text{sin}\left( {bx} \right) = \frac{1}{2}\text{cos}(\left( {a - b} \right)x) - \frac{1}{2}\text{cos}(\left( {a + b} \right)x)$
Sine and Cosine Products	$\text{sin}\left( {ax} \right)\text{cos}\left( {bx} \right) = \frac{1}{2}\text{sin}\left( {\left( {a - b} \right)x} \right) + \frac{1}{2}\text{sin}(\left( {a + b} \right)x)$
Cosine Products	$\text{cos}\left( {ax} \right)\text{cos}\left( {bx} \right) = \frac{1}{2}\text{cos}(\left( {a - b} \right)x) + \frac{1}{2}\text{cos}(\left( {a + b} \right)x)$
Power Reduction Formula	$\int\text{sec}^{n}x~dx = \frac{\sec^{n - 2}x~\tan~x}{n - 1} + \frac{n - 2}{n - 1}\int\text{sec}^{n - 2}x\ dx;n \neq 1$
Power Reduction Formula	$\int\text{tan}^{n}x\ dx = \frac{1}{n - 1}\text{tan}^{n - 1}x - \int\text{tan}^{n - 2}x\ dx$

Midpoint rule	$M_{n} = \sum\limits_{i = 1}^{n}f\left( m_{i} \right)\text{Δ}x$
Trapezoidal rule	$T_{n} = \frac{1}{2}\text{Δ}x\left( {f(x_{0}) + 2f\left( x_{1} \right) + 2f(x_{2}) + \cdots + 2f\left( x_{n - 1} \right) + f(x_{n})} \right)$
Simpson’s rule	$S_{n} = \frac{\text{Δ}x}{3}\left( {f\left( x_{0} \right) + 4f\left( x_{1} \right) + 2f\left( x_{2} \right) + 4f\left( x_{3} \right) + 2f\left( x_{4} \right) + 4f\left( x_{5} \right) + \cdots + 2f\left( x_{n - 2} \right) + 4f\left( x_{n - 1} \right) + f\left( x_{n} \right)} \right)$
Error bound for midpoint rule	$\text{Error in}\ M_{n} \leq \frac{M\left( {b - a} \right)^{3}}{24n^{2}}$
Error bound for trapezoidal rule	$\text{Error in}\ T_{n} \leq \frac{M{(b - a)}^{3}}{12n^{2}}$
Error bound for Simpson’s rule	$\text{Error in}\ S_{n} \leq \frac{M{(b - a)}^{5}}{180n^{4}}$

Midpoint rule	$M_{n} = \sum\limits_{i = 1}^{n}f\left( m_{i} \right)\text{Δ}x$
Trapezoidal rule	$T_{n} = \frac{1}{2}\text{Δ}x\left( {f(x_{0}) + 2f\left( x_{1} \right) + 2f(x_{2}) + \cdots + 2f\left( x_{n - 1} \right) + f(x_{n})} \right)$
Simpson’s rule	$S_{n} = \frac{\text{Δ}x}{3}\left( {f\left( x_{0} \right) + 4f\left( x_{1} \right) + 2f\left( x_{2} \right) + 4f\left( x_{3} \right) + 2f\left( x_{4} \right) + 4f\left( x_{5} \right) + \cdots + 2f\left( x_{n - 2} \right) + 4f\left( x_{n - 1} \right) + f\left( x_{n} \right)} \right)$
Error bound for midpoint rule	$\text{Error in}\ M_{n} \leq \frac{M\left( {b - a} \right)^{3}}{24n^{2}}$
Error bound for trapezoidal rule	$\text{Error in}\ T_{n} \leq \frac{M{(b - a)}^{3}}{12n^{2}}$
Error bound for Simpson’s rule	$\text{Error in}\ S_{n} \leq \frac{M{(b - a)}^{5}}{180n^{4}}$

Improper integrals

$\begin{array}{l} {{\int_{a}^{+ \infty}{f(x)dx}} = \underset{t\rightarrow\text{+}\infty}{\text{lim}}{\int_{a}^{t}{f(x)dx}}} \\ {{\int_{\text{−}\infty}^{b}{f(x)dx}} = \underset{t\rightarrow\text{−}\infty}{\text{lim}}{\int_{t}^{b}{f(x)dx}}} \\ {{\int_{\text{−}\infty}^{+ \infty}{f(x)dx}} = {\int_{\text{−}\infty}^{0}{f(x)dx + {\int_{0}^{+ \infty}{f(x)dx}}}}} \end{array}$

Improper integrals

$\begin{array}{l} {{\int_{a}^{+ \infty}{f(x)dx}} = \underset{t\rightarrow\text{+}\infty}{\text{lim}}{\int_{a}^{t}{f(x)dx}}} \\ {{\int_{\text{−}\infty}^{b}{f(x)dx}} = \underset{t\rightarrow\text{−}\infty}{\text{lim}}{\int_{t}^{b}{f(x)dx}}} \\ {{\int_{\text{−}\infty}^{+ \infty}{f(x)dx}} = {\int_{\text{−}\infty}^{0}{f(x)dx + {\int_{0}^{+ \infty}{f(x)dx}}}}} \end{array}$

Key Concepts 关键概念

3.1 Integration by Parts 3.1 分部积分

• The integration-by-parts formula allows the exchange of one integral for another, possibly easier, integral.

• 分部积分公式可用一个积分替换另一个（可能更简单的）积分。

• Integration by parts applies to both definite and indefinite integrals.

• 分部积分既适用于定积分，也适用于不定积分。

3.2 Trigonometric Integrals 3.2 三角积分

• Integrals of trigonometric functions can be evaluated by the use of various strategies. These strategies include

• 三角函数的积分可通过多种策略来计算。这些策略包括

1. Applying trigonometric identities to rewrite the integral so that it may be evaluated by *u*-substitution

2. Using integration by parts

3. Applying trigonometric identities to rewrite products of sines and cosines with different arguments as the sum of individual sine and cosine functions

4. Applying reduction formulas

3.3 Trigonometric Substitution 3.3 三角换元

• For integrals involving $\sqrt{a^{2} - x^{2}},$ use the substitution $x = a\mspace{2mu}\text{sin}\mspace{2mu}\theta$ and $dx = a\mspace{2mu}\text{cos}\mspace{2mu}\theta d\theta.$

• 对于含 $\sqrt{a^{2} - x^{2}},$ 的积分，使用换元 $x = a\mspace{2mu}\text{sin}\mspace{2mu}\theta$ 与 $dx = a\mspace{2mu}\text{cos}\mspace{2mu}\theta d\theta.$

• For integrals involving $\sqrt{a^{2} + x^{2}},$ use the substitution $x = a\mspace{2mu}\text{tan}\mspace{2mu}\theta$ and $dx = a\mspace{2mu}\text{sec}^{2}\theta d\theta.$

• 对于含 $\sqrt{a^{2} + x^{2}},$ 的积分，使用换元 $x = a\mspace{2mu}\text{tan}\mspace{2mu}\theta$ 与 $dx = a\mspace{2mu}\text{sec}^{2}\theta d\theta.$

• For integrals involving $\sqrt{x^{2} - a^{2}},$ substitute $x = a\mspace{2mu}\text{sec}\mspace{2mu}\theta$ and $dx = a\mspace{2mu}\text{sec}\mspace{2mu}\theta\mspace{2mu}\text{tan}\mspace{2mu}\theta d\theta.$

• 对于含 $\sqrt{x^{2} - a^{2}},$ 的积分，令 $x = a\mspace{2mu}\text{sec}\mspace{2mu}\theta$ 与 $dx = a\mspace{2mu}\text{sec}\mspace{2mu}\theta\mspace{2mu}\text{tan}\mspace{2mu}\theta d\theta.$

3.4 Partial Fractions 3.4 部分分式

• Partial fraction decomposition is a technique used to break down a rational function into a sum of simple rational functions that can be integrated using previously learned techniques.

• 部分分式分解是一种将有理函数拆分成若干简单有理函数之和的技术，这些简单有理函数可用已学的方法积分。

• When applying partial fraction decomposition, we must make sure that the degree of the numerator is less than the degree of the denominator. If not, we need to perform long division before attempting partial fraction decomposition.

• 进行部分分式分解时，必须保证分子的次数小于分母的次数。否则，在尝试部分分式分解之前需要先做多项式长除法。

• The form the decomposition takes depends on the type of factors in the denominator. The types of factors include nonrepeated linear factors, repeated linear factors, nonrepeated irreducible quadratic factors, and repeated irreducible quadratic factors.

• 分解所取的形式取决于分母中因式的类型。因式类型包括互异一次因式、重复的一次因式、互异不可约二次因式和重复的不可约二次因式。

3.5 Other Strategies for Integration 3.5 其他积分策略

• An integration table may be used to evaluate indefinite integrals.

• 积分表可用于计算不定积分。

• A CAS (or computer algebra system) may be used to evaluate indefinite integrals.

• CAS（计算机代数系统）可用于计算不定积分。

• It may require some effort to reconcile equivalent solutions obtained using different methods.

• 协调用不同方法得到的等价解可能需要一些努力。

3.6 Numerical Integration 3.6 数值积分

• We can use numerical integration to estimate the values of definite integrals when a closed form of the integral is difficult to find or when an approximate value only of the definite integral is needed.

• 当积分的闭式解难以求得，或只需要定积分的近似值时，可用数值积分估计定积分的值。

• The most commonly used techniques for numerical integration are the midpoint rule, trapezoidal rule, and Simpson’s rule.

• 数值积分最常用的方法是中点法则、梯形法则和辛普森法则。

• The midpoint rule approximates the definite integral using rectangular regions whereas the trapezoidal rule approximates the definite integral using trapezoidal approximations.

• 中点法则用矩形区域来逼近定积分，而梯形法则用梯形近似来逼近定积分。

• Simpson’s rule approximates the definite integral by first approximating the original function using piecewise quadratic functions.

• 辛普森法则先使用分段二次函数逼近原函数，再据此逼近定积分。

3.7 Improper Integrals 3.7 反常积分

• Integrals of functions over infinite intervals are defined in terms of limits.

• 函数在无穷区间上的积分用极限来定义。

• Integrals of functions over an interval for which the function has a discontinuity at an endpoint may be defined in terms of limits.

• 函数在某端点处不连续的区间上的积分，也可用极限来定义。

• The convergence or divergence of an improper integral may be determined by comparing it with the value of an improper integral for which the convergence or divergence is known.

• 反常积分的收敛或发散性，可通过与已知收敛或发散性的反常积分的值进行比较来确定。

Review Exercises 复习题

For the following exercises, determine whether the statement is true or false. Justify your answer with a proof or a counterexample.

408\.

$\int{e^{x}\text{sin}(x)dx}$ cannot be integrated by parts.

409.

$\int{\frac{1}{x^{4} + 1}dx}$ cannot be integrated using partial fractions.

410\.

In numerical integration, increasing the number of points decreases the error.

411.

Integration by parts can always yield the integral.

For the following exercises, evaluate the integral using the specified method.

412\.

$\int{x^{2}\text{sin}(4x)dx}$ using integration by parts

413.

$\int{\frac{1}{x^{2}\sqrt{x^{2} + 16}}dx}$ using trigonometric substitution

414\.

$\int{\sqrt{x}\mspace{2mu}\text{ln}(x)dx}$ using integration by parts

415.

$\int{\frac{3x}{x^{3} + 2x^{2} - 5x - 6}dx}$ using partial fractions

416\.

$\int{\frac{x^{5}}{\left( {4x^{2} + 4} \right)^{5\text{/}2}}dx}$ using trigonometric substitution

417.

$\int{\frac{\sqrt{4 - \text{sin}^{2}(x)}}{\text{sin}^{2}(x)}\text{cos}(x)dx}$ using a table of integrals or a CAS

For the following exercises, integrate using whatever method you choose.

418\.

$\int{\text{sin}^{2}(x)\text{cos}^{2}(x)dx}$

419.

$\int{x^{3}\sqrt{x^{2} + 2}\mspace{2mu} dx}$

420\.

$\int{\frac{3x^{2} + 1}{x^{4} - 2x^{3} - x^{2} + 2x}dx}$

421.

$\int{\frac{1}{x^{4} + 4}dx}$

422\.

$\int{\frac{\sqrt{3 + 16x^{4}}}{x^{3}}dx}$

For the following exercises, approximate the integrals using the midpoint rule, trapezoidal rule, and Simpson’s rule using four subintervals, rounding to three decimals.

423.

\[T\] ${\int_{1}^{2}\sqrt{x^{5} + 2}}dx$

424\.

\[T\] ${\int_{0}^{\sqrt{\pi}}e^{\text{−}\text{sin}{(x^{2})}}}dx$

425.

\[T\] ${\int_{1}^{4}\frac{\text{ln}\left( {1\text{/}x} \right)}{x}}dx$

For the following exercises, evaluate the integrals, if possible.

426\.

${\int_{1}^{\infty}\frac{1}{x^{n}}}dx,$ for what values of $n$ does this integral converge or diverge?

427.

${\int_{1}^{\infty}\frac{e^{\text{−}x}}{x}}dx$

For the following exercises, consider the gamma function given by $\text{Γ}(a) = {\int_{0}^{\infty}{e^{\text{−}y}y^{a - 1}dy}}.$

428\.

Show that $\text{Γ}(a) = (a - 1)\text{Γ}(a - 1).$

429\.

Extend to show that $\text{Γ}(a) = (a - 1)\text{!},$ assuming $a$ is a positive integer.

The fastest car in the world, the Bugati Veyron, can reach a top speed of 408 km/h. The graph represents its velocity.

430\.

\[T\] Use the graph to estimate the velocity every 20 sec and fit to a graph of the form $v(t) = a\mspace{2mu}\text{exp}^{bx}\text{sin}(cx) + d.$ (*Hint:* Consider the time units.)

431.

\[T\] Using your function from the previous problem, find exactly how far the Bugati Veyron traveled in the 1 min 40 sec included in the graph.