Theorem 5.2.1.
定理 1 设 \(u=\varphi(x)\) 可导, 且已知 \(F(u)\) 是 \(f(u)\) 的一个原函数,则有换元公式
\begin{equation}
\displaystyle \int f[\varphi(x)] \varphi^{\prime}(x) \mathrm{d} x=F[\varphi(x)]+C .\tag{5.2.1}
\end{equation}
Section 5.2 换元积分法
应用不定积分的性质和不定积分基本公式,可以计算出一部分比较简单的函数的不定积分, 对计算一些复杂函数的积分问题,还需要学习相应的方法.
Subsection 5.2.1 第一换元法 (凑微分法)
Proof.
证 已知 \(F(u)\) 是 \(f(u)\) 的一个原函数,即
\begin{equation*}
F^{\prime}(u)=f(u), \displaystyle \int f(u) \mathrm{d} u=F(u)+C .
\end{equation*}
\(u\) 是中间变量, \(u=\varphi(x)\) 可导, 根据复合函数微分法, 有
\begin{equation*}
\mathrm{d} F[\varphi(x)]=f[\varphi(x)] \varphi^{\prime}(x) \mathrm{d} x,
\end{equation*}
由不定积分的定义可得 \(\displaystyle \int f[\varphi(x)] \varphi^{\prime}(x) \mathrm{d} x=F[\varphi(x)]+C\text{.}\)
这种方法的基本思想是从被积函数 \(f[\varphi(x)] \varphi^{\prime}(x)\) 中提出因子 \(\varphi^{\prime}(x)\) 和 \(\mathrm{d} x\) 凑成 \(\mathrm{d} \varphi(x)\text{,}\) 从而使
\begin{equation*}
\begin{aligned}
\displaystyle \int f[\varphi(x)] \varphi^{\prime}(x) \mathrm{d} x & =\displaystyle \int f[\varphi(x)] \mathrm{d} \varphi(x) \stackrel{\text { 令 } \varphi(x)=u}{=} \displaystyle \int f(u) \mathrm{d} u \\
& =F(u)+C \stackrel{u=\varphi(x)}{=} F[\varphi(x)]+C .
\end{aligned}
\end{equation*}
即将所求积分转化为可以利用基本积分公式的不定积分. 把公式(5.2.1) 称为不定积分的第一换元法 (也称凑微分法). 应用凑微分法求不定积分 \(\displaystyle \int g(x) \mathrm{d} x\) 的关键是将函数 \(g(x)\) 凑成 \(g(x)=f[\varphi(x)] \varphi^{\prime}(x)\) 的形式, 从而
\begin{equation*}
\displaystyle \int g(x) \mathrm{d} x=\displaystyle \int f[\varphi(x)] \varphi^{\prime}(x) \mathrm{d} x=\displaystyle \int f[\varphi(x)] \mathrm{d} \varphi(x) \stackrel{\text { 令 } u=\varphi(x)}{=} \displaystyle \int f(u) \mathrm{d} u,
\end{equation*}
即将函数 \(g(x)\) 的积分转化为函数 \(f(u)\) 的积分. 如果函数 \(f(u)\) 的原函数 \(F(u)\) 能够求出, 那么也就得到了 \(g(x)\) 的原函数.
Example 5.2.2.
例 1 求不定积分 \(\displaystyle \int \sin (2 x-1) \mathrm{d} x\text{.}\)
Solution.
解 令 \(\varphi(x)=2 x-1, \varphi^{\prime}(x)=2\text{.}\) 被积表达式 \(\sin (2 x-1) \mathrm{d} x=\frac{1}{2} \sin \varphi(x) \mathrm{d} \varphi(x)\text{,}\)把 \(\varphi(x)=2 x-1\) 看作一个变量 \(u\text{,}\)而 \(\sin u\) 有原函数 \(-\cos u\text{,}\)由Theorem 5.2.1 得
\begin{equation*}
\begin{aligned}
\displaystyle \int \sin (2 x-1) \mathrm{d} x & =\frac{1}{2} \displaystyle \int \sin (2 x-1) \mathrm{d}(2 x-1) \stackrel{2 x-1=u}{=} \frac{1}{2} \displaystyle \int \sin u \mathrm{~d} u \\
& =-\frac{1}{2} \cos u+C \stackrel{u=2 x-1}{=}-\frac{1}{2} \cos (2 x-1)+C .
\end{aligned}
\end{equation*}
Example 5.2.3.
例 2 求不定积分 \(\displaystyle \int x^{2} \sqrt{x^{3}+2} \mathrm{~d} x\text{.}\)
Solution.
解 令 \(\varphi(x)=x^{3}+2, \varphi^{\prime}(x)=3 x^{2}\text{,}\) 被积函数 \(x^{2} \sqrt{x^{3}+2} \mathrm{~d} x=\frac{1}{3}[\varphi(x)]^{\frac{1}{2}} \mathrm{~d} \varphi(x)\text{,}\)将 \(x^{3}+2\) 看作变量 \(u\text{,}\)而 \(u^{\frac{1}{2}}\) 有原函数 \(\frac{2}{3} u^{\frac{3}{2}}\text{,}\) 由Theorem 5.2.1 得
\begin{equation*}
\begin{aligned}
\displaystyle \int x^{2} \sqrt{x^{3}+2} \mathrm{~d} x & =\frac{1}{3} \displaystyle \int\left(x^{3}+2\right)^{\frac{1}{2}} \mathrm{~d}\left(x^{3}+2\right) \stackrel{x^{3}+2=u}{=} \frac{1}{3} \displaystyle \int u^{\frac{1}{2}} \mathrm{~d} u \\
& =\frac{2}{9} u^{\frac{3}{2}}+C \stackrel{u=x^{3}+2}{=} \frac{2}{9}\left(x^{3}+2\right)^{\frac{3}{2}}+C .
\end{aligned}
\end{equation*}
对第一换元法比较熟练以后,所设换元过程可以省略.
Example 5.2.4.
例 3 求不定积分 \(\displaystyle \int \tan x \mathrm{~d} x\text{.}\)
Solution.
解 \(\displaystyle \int \tan x \mathrm{~d} x=\displaystyle \int \frac{\sin x}{\cos x} \mathrm{~d} x=-\displaystyle \int \frac{1}{\cos x} \mathrm{~d}(\cos x)=-\ln |\cos x|+C\text{.}\) 同理可得
\begin{equation*}
\displaystyle \int \cot x \mathrm{~d} x=\ln |\sin x|+C
\end{equation*}
Example 5.2.5.
例 4 求不定积分 \(\displaystyle \int \frac{\mathrm{d} x}{x(1+\ln x)}\text{.}\)
Solution.
解 \(\displaystyle \int \frac{\mathrm{d} x}{x(1+\ln x)}=\displaystyle \int \frac{1}{1+\ln x} \mathrm{~d}(1+\ln x)=\ln |1+\ln x|+C\text{.}\)
Example 5.2.6.
例 5 求不定积分 \(\displaystyle \int \frac{a^{\frac{1}{x}}}{x^{2}} \mathrm{~d} x \quad(a>0)\text{.}\)
Solution.
解 \(\displaystyle \int \frac{a^{\frac{1}{x}}}{x^{2}} \mathrm{~d} x=-\displaystyle \int a^{\frac{1}{x}} \mathrm{~d}\left(\frac{1}{x}\right)=-\frac{a^{\frac{1}{x}}}{\ln a}+C\text{.}\)
Example 5.2.7.
例 6 求不定积分 \(\displaystyle \int \frac{1}{(\arcsin x)^{3} \sqrt{1-x^{2}}} \mathrm{~d} x\text{.}\)
Solution.
解 \(\displaystyle \int \frac{1}{(\arcsin x)^{3} \sqrt{1-x^{2}}} \mathrm{~d} x=\displaystyle \int(\arcsin x)^{-3} \mathrm{~d}(\arcsin x)\)
\begin{equation*}
\begin{aligned}
& =\frac{1}{-3+1}(\arcsin x)^{-3+1}+C \\
& =-\frac{1}{2(\arcsin x)^{2}}+C .
\end{aligned}
\end{equation*}
Example 5.2.8.
例 7 求不定积分 \(\displaystyle \int \sec x \mathrm{~d} x\text{.}\)
Solution.
解 因为 \(\sec x \mathrm{~d} x=\frac{\sec x(\sec x+\tan x)}{\sec x+\tan x} \mathrm{~d} x=\frac{\sec ^{2} x+\sec x \tan x}{\sec x+\tan x} \mathrm{~d} x\)
\begin{equation*}
=\frac{\mathrm{d}(\sec x+\tan x)}{\sec x+\tan x},
\end{equation*}
所以
\begin{equation*}
\displaystyle \int \sec x \mathrm{~d} x=\displaystyle \int \frac{\mathrm{d}(\sec x+\tan x)}{\sec x+\tan x}=\ln |\sec x+\tan x|+C .
\end{equation*}
同理可得
\begin{equation*}
\displaystyle \int \csc x \mathrm{~d} x=\ln |\csc x-\cot x|+C .
\end{equation*}
Example 5.2.9.
例 8 求不定积分 \(\displaystyle \int \sin ^{3} x \mathrm{~d} x\text{.}\)
Solution.
解 \(\displaystyle \int \sin ^{3} x \mathrm{~d} x=\displaystyle \int \sin ^{2} x \sin x \mathrm{~d} x=-\displaystyle \int\left(1-\cos ^{2} x\right) \mathrm{d}(\cos x)\)
\begin{equation*}
\begin{aligned}
& =-\displaystyle \int \mathrm{d}(\cos x)+\displaystyle \int \cos ^{2} x \mathrm{~d}(\cos x) \\
& =-\cos x+\frac{1}{3} \cos ^{3} x+C .
\end{aligned}
\end{equation*}
Example 5.2.10.
例 9 求不定积分 \(\displaystyle \int \sec ^{6} x \mathrm{~d} x\text{.}\)
Solution.
解 \(\displaystyle \int \sec ^{6} x \mathrm{~d} x=\displaystyle \int\left(\sec ^{2} x\right)^{2} \sec ^{2} x \mathrm{~d} x=\displaystyle \int\left(1+\tan ^{2} x\right)^{2} \mathrm{~d}(\tan x)\)
\begin{equation*}
\begin{aligned}
& =\displaystyle \int\left(1+2 \tan ^{2} x+\tan ^{4} x\right) \mathrm{d}(\tan x) \\
& =\tan x+\frac{2}{3} \tan ^{3} x+\frac{1}{5} \tan ^{5} x+C .
\end{aligned}
\end{equation*}
Example 5.2.11.
例 10 求不定积分 \(\displaystyle \int \frac{\cos ^{3} x}{\sqrt{\sin x}} \mathrm{~d} x\text{.}\)
Solution.
解 \(\displaystyle \int \frac{\cos ^{3} x}{\sqrt{\sin x}} \mathrm{~d} x=\displaystyle \int \frac{\cos ^{2} x \cos x}{\sqrt{\sin x}} \mathrm{~d} x=\displaystyle \int\left(1-\sin ^{2} x\right) \sin ^{-\frac{1}{2}} x \mathrm{~d}(\sin x)\)
\begin{equation*}
=\displaystyle \int \sin ^{-\frac{1}{2}} x \mathrm{~d}(\sin x)-\displaystyle \int \sin ^{\frac{3}{2}} x \mathrm{~d}(\sin x)=2 \sin ^{\frac{1}{2}} x-\frac{2}{5} \sin ^{\frac{5}{2}} x+C .
\end{equation*}
Example 5.2.12.
例 11 求不定积分 \(\displaystyle \int \frac{\mathrm{e}^{3 \sqrt{x}}}{\sqrt{x}} \mathrm{~d} x\text{.}\)
Solution.
解 \(\displaystyle \int \frac{\mathrm{e}^{3 \sqrt{x}}}{\sqrt{x}} \mathrm{~d} x=\frac{2}{3} \displaystyle \int \mathrm{e}^{3 \sqrt{x}} \mathrm{~d}(3 \sqrt{x})=\frac{2}{3} \mathrm{e}^{3 \sqrt{x}}+C\text{.}\)
Example 5.2.13.
例 12 求不定积分 \(\displaystyle \int \frac{1}{\sqrt{a^{2}-x^{2}}} \mathrm{~d} x \quad(a>0)\text{.}\)
Solution.
解 \(\displaystyle \int \frac{1}{\sqrt{a^{2}-x^{2}}} \mathrm{~d} x=\displaystyle \int \frac{1}{a \sqrt{1-\left(\frac{x}{a}\right)^{2}}} \mathrm{~d} x\)
\begin{equation*}
\begin{aligned}
& =\displaystyle \int \frac{1}{\sqrt{1-\left(\frac{x}{a}\right)^{2}}} \mathrm{~d}\left(\frac{x}{a}\right) \\
& =\arcsin \frac{x}{a}+C .
\end{aligned}
\end{equation*}
Example 5.2.14.
例 13 求不定积分 \(\displaystyle \int \frac{1}{a^{2}+x^{2}} \mathrm{~d} x\text{.}\)
Solution.
解 \(\displaystyle \int \frac{1}{a^{2}+x^{2}} \mathrm{~d} x=\displaystyle \int \frac{1}{a^{2}\left[1+\left(\frac{x}{a}\right)^{2}\right]} \mathrm{d} x\)
\begin{equation*}
\begin{aligned}
& =\frac{1}{a} \displaystyle \int \frac{1}{1+\left(\frac{x}{a}\right)^{2}} \mathrm{~d}\left(\frac{x}{a}\right) \\
& =\frac{1}{a} \arctan \frac{x}{a}+C .
\end{aligned}
\end{equation*}
Example 5.2.15.
例 14 求不定积分 \(\displaystyle \int \frac{1}{a^{2}-x^{2}} \mathrm{~d} x\text{.}\)
Solution.
解 因为 \(\frac{1}{a^{2}-x^{2}}=\frac{1}{2 a}\left(\frac{1}{a+x}+\frac{1}{a-x}\right)\text{,}\) 所以
\begin{equation*}
\begin{aligned}
\displaystyle \int \frac{\mathrm{d} x}{a^{2}-x^{2}} & =\frac{1}{2 a}\left(\displaystyle \int \frac{1}{a+x} \mathrm{~d} x+\displaystyle \int \frac{1}{a-x} \mathrm{~d} x\right) \\
& =\frac{1}{2 a}\left[\displaystyle \int \frac{1}{a+x} \mathrm{~d}(a+x)-\displaystyle \int \frac{1}{a-x} \mathrm{~d}(a-x)\right] \\
& =\frac{1}{2 a} \ln \left|\frac{a+x}{a-x}\right|+C .
\end{aligned}
\end{equation*}
Example 5.2.16.
例 15 求不定积分 \(\displaystyle \int \frac{x}{2+x^{4}} \mathrm{~d} x\text{.}\)
Solution.
解 \(\displaystyle \int \frac{x}{2+x^{4}} \mathrm{~d} x=\frac{1}{2} \displaystyle \int \frac{1}{(\sqrt{2})^{2}+\left(x^{2}\right)^{2}} \mathrm{~d} x^{2}=\frac{1}{2 \sqrt{2}} \arctan \frac{x^{2}}{\sqrt{2}}+C\text{.}\)
Example 5.2.17.
例 16 求不定积分 \(\displaystyle \int \frac{\ln \tan \frac{x}{2}}{\sin x} \mathrm{~d} x\text{.}\)
Solution.
\begin{equation*}
\text { 解 } \begin{aligned}
\displaystyle \int \frac{\ln \tan \frac{x}{2}}{\sin x} \mathrm{~d} x & =\displaystyle \int \frac{\ln \tan \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \mathrm{~d} x=\displaystyle \int \frac{\ln \tan \frac{x}{2}}{\tan \frac{x}{2} \cos ^{2} \frac{x}{2}} \mathrm{~d}\left(\frac{x}{2}\right) \\
& =\displaystyle \int \frac{\ln \tan \frac{x}{2}}{\tan \frac{x}{2}} \mathrm{~d}\left(\tan \frac{x}{2}\right)=\displaystyle \int \ln \tan \frac{x}{2} \mathrm{~d}\left(\ln \tan \frac{x}{2}\right) \\
& =\frac{1}{2} \ln ^{2}\left(\tan \frac{x}{2}\right)+C .
\end{aligned}
\end{equation*}
Example 5.2.4、Example 5.2.8 及Example 5.2.13 \(\sim \) Example 5.2.15的结论可作为新的积分公式, 即 (16)\(\displaystyle \int \tan x \mathrm{~d} x=-\ln |\cos x|+C\text{;}\)
(17)\(\displaystyle \int \cot x \mathrm{~d} x=\ln |\sin x|+C\text{;}\) (18)\(\displaystyle \int \sec x \mathrm{~d} x=\ln |\sec x+\tan x|+C\text{;}\)
(19)\(\displaystyle \int \csc x \mathrm{~d} x=\ln |\csc x-\cot x|+C\text{;}\)
(20)\(\displaystyle \int \frac{1}{\sqrt{a^{2}-x^{2}}} \mathrm{~d} x=\arcsin \frac{x}{a}+C\text{;}\) (21)\(\displaystyle \int \frac{1}{a^{2}+x^{2}} \mathrm{~d} x=\frac{1}{a} \arctan \frac{x}{a}+C\text{;}\) (22)\(\displaystyle \int \frac{\mathrm{d} x}{a^{2}-x^{2}}=\frac{1}{2 a} \ln \left|\frac{a+x}{a-x}\right|+C\text{.}\)
熟练运用凑微分法的关键是要熟记基本积分公式. 在掌握求不定积分基本方法之后, 可直接将被积函数凑成 \(\displaystyle \int f[\varphi(x)] \mathrm{d} \varphi(x)\) 的形式来计算不定积分, 并把函数 \(\varphi(x)\) 看作一个整体变量,然后应用基本积分公式,因此称为“凑微分法”.
Subsection 5.2.2 第二换元法
在不定积分的计算中, 常常会出现与前面介绍的情况刚好相反的问题, 即不定积分 \(\displaystyle \int f(x) \mathrm{d} x\) 并不复杂, 却较难直接算出. 但若通过适当的变量代换 \(x=\varphi(t)\text{,}\)将积分 \(\displaystyle \int f(x) \mathrm{d} x\) 化为易于计算的积分 \(\displaystyle \int f[\varphi(t)] \varphi^{\prime}(t) \mathrm{d} t\) 的形式, 这就是另一种通过变量代换计算不定积分的方法,称为第二换元法.
Theorem 5.2.18.
定理 2 设 \(x=\varphi(t)\) 是单调、可导的函数, 并且 \(\varphi^{\prime}(t) \neq 0\text{,}\) 又设 \(f[\varphi(t)] \varphi^{\prime}(t)\)具有原函数,则有换元公式
\begin{equation*}
\displaystyle \int f(x) \mathrm{d} x \stackrel{x=\varphi(t)}{=}\left\{\displaystyle \int f[\varphi(t)] \varphi^{\prime}(t) \mathrm{d} t\right\}_{t=\varphi^{-1}(x)},
\end{equation*}
其中 \(t=\varphi^{-1}(x)\) 为 \(x=\varphi(t)\) 的反函数.
Proof.
证 已知 \(f[\varphi(t)] \varphi^{\prime}(t)\) 具有原函数, 设其原函数为 \(F(t)\text{,}\) 只须证明
\begin{equation*}
\left\{F\left[\varphi^{-1}(x)\right]+C\right\}^{\prime}=f(x) .
\end{equation*}
由复合函数的求导法则得
\begin{equation*}
\left\{F\left[\varphi^{-1}(x)\right]+C\right\}^{\prime}=F^{\prime}(t) \frac{\mathrm{d} t}{\mathrm{~d} x}=f[\varphi(t)] \varphi^{\prime}(t) \frac{1}{\varphi^{\prime}(t)}=f[\varphi(t)]=f(x) .
\end{equation*}
第二换元法是将变量 \(x\) 替换成适当的函数 \(x=\varphi(t)\text{,}\)将积分化成下面形式:
\begin{equation*}
\displaystyle \int f(x) \mathrm{d} x=\displaystyle \int f[\varphi(t)] \varphi^{\prime}(t) \mathrm{d} t
\end{equation*}
求出上式右端的积分后, 再用 \(x=\varphi(t)\) 的反函数 \(t=\varphi^{-1}(x)\) 代回.
这个换元公式的成立是需要一定条件的, 由于积分 \(\displaystyle \int g[\varphi(t)] \varphi^{\prime}(t) \mathrm{d} t\) 求出后必须用 \(x=\varphi(t)\) 的反函数 \(t=\varphi^{-1}(x)\) 代回去, 为了保证该反函数存在而且是单值可导的,所以在定理的条件中要求函数 \(x=\varphi(t)\) 在 \(t\) 的某一区间 (这区间和所考虑的 \(x\)的积分区间相对应) 上是单调、可导的, 并且 \(\varphi^{\prime}(t) \neq 0\text{,}\) 即要保证反函数 \(t=\varphi^{-1}(x)\)存在.
Example 5.2.19.
例 17 求不定积分 \(\displaystyle \int \frac{\mathrm{d} x}{1+\sqrt{1+x}}\text{.}\)
Solution.
解 为去掉被积函数分母中的根式, 设 \(\sqrt{1+x}=t\text{,}\) 则 \(x=t^{2}-1, \mathrm{~d} x=2 t \mathrm{~d} t\text{.}\)
\begin{equation*}
\begin{aligned}
& \displaystyle \int \frac{\mathrm{d} x}{1+\sqrt{1+x}}=\displaystyle \int \frac{2 t}{1+t} \mathrm{~d} t=2 \displaystyle \int \frac{t+1-1}{1+t} \mathrm{~d} t=2\left(\displaystyle \int \mathrm{d} t-\displaystyle \int \frac{1}{1+t} \mathrm{~d} t\right) \\
& =2(t-\ln |1+t|)+C \\
& \stackrel{t=\sqrt{1+x}}{=} 2(\sqrt{1+x}-\ln |1+\sqrt{1+x}|)+C .
\end{aligned}
\end{equation*}
应用第二换元法时必须注意积分后一定要将变量代换回来.
Example 5.2.20.
例 18 求不定积分 \(\displaystyle \int \sqrt{a^{2}-x^{2}} \mathrm{~d} x \quad(a>0)\text{.}\)
Solution.
解 见图 \(5-2,|x| \leqslant a\text{,}\) 设 \(x=a \sin t\text{,}\) 当 \(-\frac{\pi}{2} \leqslant t \leqslant \frac{\pi}{2}\) 时, \(x=a \sin t\) 存在单值反函数 \(t=\arcsin \frac{x}{a}\text{,}\) 且 \(|\cos t|=\cos t, \mathrm{~d} x=\) \(a \cos t \mathrm{~d} t\text{,}\) 所以
\begin{equation*}
\begin{aligned}
\displaystyle \int \sqrt{a^{2}-x^{2}} \mathrm{~d} x & =\displaystyle \int \sqrt{a^{2}-a^{2} \sin ^{2} t} \cdot a \cos t \mathrm{~d} t=a^{2} \displaystyle \int|\cos t| \cos t \mathrm{~d} t \\
& =a^{2} \displaystyle \int \cos ^{2} t \mathrm{~d} t=\frac{a^{2}}{2} \displaystyle \int(1+\cos 2 t) \mathrm{d} t=\frac{a^{2}}{2}\left(\displaystyle \int \mathrm{d} t+\displaystyle \int \cos 2 t \mathrm{~d} t\right) \\
& =\frac{a^{2}}{2}\left(t+\frac{1}{2} \sin 2 t\right)+C=\frac{a^{2}}{2} t+\frac{a^{2}}{4} \sin 2 t+C .
\end{aligned}
\end{equation*}
由于 \(t=\arcsin \frac{x}{a}\text{,}\) 则 \(\sin 2 t=2 \sin t \cos t=2 \frac{x}{a^{2}} \sqrt{a^{2}-x^{2}}\text{,}\)于是
\begin{equation*}
\displaystyle \int \sqrt{a^{2}-x^{2}} \mathrm{~d} x=\frac{a^{2}}{2} \arcsin \frac{x}{a}+\frac{x}{2} \sqrt{a^{2}-x^{2}}+C .
\end{equation*}
Example 5.2.21.
例 19 求不定积分 \(\displaystyle \int \frac{1}{\sqrt{x^{2}+a^{2}}} \mathrm{~d} x \quad(a>0)\text{.}\)
Solution.
解 见图 5-3, 设 \(x=a \tan t\text{,}\) 则 \(\mathrm{d} x=a \sec ^{2} t \mathrm{~d} t\text{,}\) 当 \(-\frac{\pi}{2}<t<\frac{\pi}{2}\) 时, \(x=a \tan t\) 存在单值反函数, 且 \(|\sec t|=\sec t\text{.}\) 于是
\begin{equation*}
\displaystyle \int \frac{\mathrm{d} x}{\sqrt{x^{2}+a^{2}}}=\displaystyle \int \frac{a \sec ^{2} t}{a \sqrt{1+\tan ^{2} t}} \mathrm{~d} t=\displaystyle \int \sec t \mathrm{~d} t=\ln |\sec t+\tan t|+C .
\end{equation*}
又因为 \(\tan t=\frac{x}{a}\text{,}\) 则 \(\sec t=\frac{\sqrt{x^{2}+a^{2}}}{a}\text{.}\)
所以
\begin{equation*}
\displaystyle \int \frac{\mathrm{d} x}{\sqrt{x^{2}+a^{2}}}=\ln \left|\frac{\sqrt{a^{2}+x^{2}}}{a}+\frac{x}{a}\right|+C_{1}=\ln \left|\sqrt{a^{2}+x^{2}}+x\right|+C,
\end{equation*}
其中 \(C=C_{1}-\ln a\) 仍为任意常数.
Example 5.2.22.
例 20 求不定积分 \(\displaystyle \int \frac{\mathrm{d} x}{\sqrt{x^{2}-a^{2}}}\text{.}\)
Solution.
解 设 \(x=a \sec t\text{,}\) 则 \(\mathrm{d} x=a \sec t \tan t \mathrm{~d} t\text{,}\) 见图 5-4.
当 \(0<t<\frac{\pi}{2}\) 或 \(\frac{\pi}{2}<t<\pi\) 时, \(x=a \sec t\) 存在反函数, 这里仅讨论 \(0<t<\frac{\pi}{2}\) 的情况, 同理可讨论 \(\frac{\pi}{2}<t<\pi\) 的情况.
当 \(0<t<\frac{\pi}{2}\) 时, \(|\tan t|=\tan t\) 则
\begin{equation*}
\displaystyle \int \frac{\mathrm{d} x}{\sqrt{x^{2}-a^{2}}}=\displaystyle \int \frac{a \sec t \tan t \mathrm{~d} t}{a \tan t}=\displaystyle \int \sec t \mathrm{~d} t=\ln |\sec t+\tan t|+C .
\end{equation*}
因为 \(\sec t=\frac{x}{a}\text{,}\) 则 \(\tan t=\frac{\sqrt{x^{2}-a^{2}}}{a}\text{.}\) 于是
\begin{equation*}
\displaystyle \int \frac{\mathrm{d} x}{\sqrt{x^{2}-a^{2}}}=\ln \left|\frac{x}{a}+\frac{\sqrt{x^{2}-a^{2}}}{a}\right|+C_{1}=\ln \left|x+\sqrt{x^{2}-a^{2}}\right|+C
\end{equation*}
(23)\(\displaystyle \int \sqrt{a^{2}-x^{2}} \mathrm{~d} x=\frac{a^{2}}{2} \arcsin \frac{x}{a}+\frac{x}{2} \sqrt{a^{2}-x^{2}}+C\text{;}\)
(24)\(\displaystyle \int \frac{\mathrm{d} x}{\sqrt{x^{2}+a^{2}}}=\ln \left|\sqrt{a^{2}+x^{2}}+x\right|+C\text{;}\)
(25)\(\displaystyle \int \frac{\mathrm{d} x}{\sqrt{x^{2}-a^{2}}}=\ln \left|x+\sqrt{x^{2}-a^{2}}\right|+C\text{.}\)
第二换元法主要用于两类不定积分 : 一类是被积函数中根式无法用第一换元积分法求解, 这时就用第二换元积分法消去根式, 然后求解, 如Example 5.2.19 . 二类是被积函数中含 \(\sqrt{a^{2}-x^{2}}, \sqrt{a^{2}+x^{2}}, \sqrt{x^{2}-a^{2}}\) 的因式,而不能用第一换元积分法求解,如Example 5.2.20\(\sim\)Example 5.2.22 的换元法.
Example 5.2.23.
例 21 求不定积分 \(\displaystyle \int \frac{\mathrm{d} x}{(1+x) \sqrt{1-x^{2}}}\text{.}\)
Solution.
解 设 \(x=\sin t, \mathrm{~d} x=\cos t \mathrm{~d} t\text{,}\) 所以
\begin{equation*}
\begin{aligned}
\displaystyle \int \frac{\mathrm{d} x}{(1+x) \sqrt{1-x^{2}}} & =\displaystyle \int \frac{\cos t \mathrm{~d} t}{(1+\sin t) \sqrt{1-\sin ^{2} t}}=\displaystyle \int \frac{1}{1+\sin t} \mathrm{~d} t=\displaystyle \int \frac{1-\sin t}{\cos ^{2} t} \mathrm{~d} t \\
& =\displaystyle \int \frac{1}{\cos ^{2} t} \mathrm{~d} t+\displaystyle \int \frac{1}{\cos ^{2} t} \mathrm{~d}(\cos t)=\tan t-\frac{1}{\cos t}+C \\
& =\frac{x}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-x^{2}}}+C .
\end{aligned}
\end{equation*}
Example 5.2.24.
例 22 求不定积分 \(\displaystyle \int \frac{1}{\sqrt{\mathrm{e}^{x}+1}} \mathrm{~d} x\text{.}\)
Solution.
解 令 \(\sqrt{\mathrm{e}^{x}+1}=t\text{,}\) 则 \(\mathrm{e}^{x}=t^{2}-1, x=\ln \left(t^{2}-1\right), \mathrm{d} x=\frac{2 t}{t^{2}-1} \mathrm{~d} t\text{.}\) 于是
\begin{equation*}
\begin{aligned}
\displaystyle \int \frac{1}{\sqrt{\mathrm{e}^{x}+1}} \mathrm{~d} x & =\displaystyle \int \frac{1}{t} \cdot \frac{2 t}{t^{2}-1} \mathrm{~d} t=\displaystyle \int \frac{2}{(t+1)(t-1)} \mathrm{d} t=\displaystyle \int\left(\frac{1}{t-1}-\frac{1}{t+1}\right) \mathrm{d} t \\
& =\ln \left|\frac{t-1}{t+1}\right|+C=\ln \left|\frac{\sqrt{\mathrm{e}^{x}+1}-1}{\sqrt{\mathrm{e}^{x}+1}+1}\right|+C .
\end{aligned}
\end{equation*}
Example 5.2.25.
例 23 求不定积分 \(\displaystyle \int \frac{\mathrm{d} x}{x^{2} \sqrt{1+x^{2}}}\text{.}\)
Solution.
解 令 \(x=\tan t\text{,}\) 则 \(\mathrm{d} x=\sec ^{2} t \mathrm{~d} t\text{.}\) 于是
\begin{equation*}
\begin{aligned}
\displaystyle \int \frac{\mathrm{d} x}{x^{2} \sqrt{1+x^{2}}} & =\displaystyle \int \frac{\sec ^{2} t}{\tan ^{2} t \sec t} \mathrm{~d} t=\displaystyle \int \frac{\sec t}{\tan ^{2} t} \mathrm{~d} t=\displaystyle \int \frac{\mathrm{d} t}{\frac{\sin ^{2} t}{\cos ^{2} t} \cdot \cos t}=\displaystyle \int \frac{\mathrm{d} t}{\sin t \cdot \frac{\sin t}{\cos t}} \\
& =\displaystyle \int \csc t \cot t \mathrm{~d} t=-\csc t+C=-\frac{\sqrt{1+x^{2}}}{x}+C .
\end{aligned}
\end{equation*}
注以上几例说明第二换元法的一个主要目标就是将被积函数中根式有理化.
Subsection 5.2.3 习题
-
求下列不定积分:
- \(\displaystyle \int \frac{1}{2 x+1} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int x \sqrt{1-x^{2}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{1}{1+9 x^{2}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{1}{x} \sqrt{\ln x} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \cos x \sin ^{2} x \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{(\arcsin x)^{-2}}{\sqrt{1-x^{2}}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int x^{2} \mathrm{e}^{-x^{3}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{x}{3-2 x^{2}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{2 x-3}{x^{2}-3 x+8} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1-x^{4}}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{\sin x}{2+\cos ^{2} x} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{1}{\cos ^{2} x(1+\tan x)} \mathrm{d} x\text{;}\)
- \(\displaystyle \int \frac{\mathrm{e}^{x}}{4+\mathrm{e}^{2 x}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{1}{\sqrt{x} \sin ^{2} \sqrt{x}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{\mathrm{d} x}{\sqrt{x}(1+2 x)}\text{;}\)
- \(\displaystyle \int \frac{1}{\mathrm{e}^{x}+\mathrm{e}^{-x}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{1+x}{\sqrt{1-x^{2}}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{1}{3+2 x-x^{2}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{\mathrm{d} x}{x \ln x \ln (\ln x)}\text{;}\)
- \(\displaystyle \int x \cos x^{2} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{\cos ^{2} x}{\sin 2 x} \mathrm{~d} x\text{.}\)
-
求下列不定积分:
- \(\displaystyle \int\left(1-x^{2}\right)^{-\frac{3}{2}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{x^{2}}{\sqrt{a^{2}-x^{2}}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{\mathrm{d} x}{x^{2} \sqrt{x^{2}-9}}\text{;}\)
- \(\displaystyle \int \frac{\sin 2 x}{\sqrt{2-\cos ^{4} x}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{\mathrm{d} x}{x \sqrt{x^{2}-a^{2}}}\text{;}\)
- \(\displaystyle \int \frac{\mathrm{d} x}{\sqrt{\left(x^{2}+1\right)^{3}}}\text{;}\)
- \(\displaystyle \int \frac{x^{2}}{\sqrt{x-x^{2}}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{x^{2}+x+1}{(x+1)^{50}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \tan ^{4} x \mathrm{~d} x\text{;}\)
- \(\displaystyle \int x^{3} \sqrt{1+x^{2}} \mathrm{~d} x\text{;}\)
- \(\displaystyle \int \frac{x}{1+\sqrt{1+x^{2}}} \mathrm{~d} x\text{.}\)
