典型例题汇总：不定积分（凑微分、分部积分、一般有理式积分，三角函数有理式积分等）

题目 03

$$I=\frac{1}{\sin x\cos 2x}\mathrm{d}x=?$$

解析 03

Tips:

一般情况下，遇到二倍角的三角函数，首先尝试转换为一倍角的三角函数：

$\sin 2x=2\sin x\cos x$

$\cos 2x=2{\cos }^{2}x–1=1–2{\sin }^{2}x$

$2\cos 2x=2{\cos }^{2}x–1+1–2{\sin }^{2}x$ $\Rightarrow$ $\cos 2x={\cos }^{2}x–{\sin }^{2}x$

先尝试都转换为 $\sin x$

$$I=\int \frac{\mathrm{d}(\sin x)}{\sin x\cdot \cos x(1-2{\sin }^{2}x)}$$

但由于很难都凑成 $\sin x$, 因此，上面的步骤几乎走不通。

于是，尝试都转换为 $\cos x$:

$$I=\int \frac{-\mathrm{d}(\cos x)}{{\sin }^{2}x(2{\cos }^{2}x-1)}=\int \frac{-\mathrm{d}(\cos x)}{(1-{\cos }^{2}x)(2{\cos }^{2}x-1)}$$

令：

$$u=\cos x$$

则：

$$I=\int \frac{-\mathrm{d}u}{(1-u^2)(2u^2-1)}\Rightarrow$$

设：

$$\frac{Au+B}{1-u^2}+\frac{Cu+D}{2u^2-1}$$

则有：

$$\frac{(Au+B)(2u^2-1)+(Cu+D)(1-u^2)}{(1-u^2)(2u^2-1)}\Rightarrow$$

$$2A{u}^3-Au+2B{u}^2-B+Cu-C{u}^3+D-D{u}^2=1\Rightarrow$$

$$\{\begin{array}{l}2A–C=0\\ –A+C=0\\ 2B–D=0\\ –B+D=1\end{array}\Rightarrow \{\begin{array}{l}A=0\\ C=0\\ B=1\\ D=2\end{array}$$

于是：

$$I=-\int (\frac{1}{1-u^2}+\frac{2}{2u^2-1})\mathrm{d}u\Rightarrow$$

$$I=+\int \frac{1}{u^2-1}\mathrm{d}u-2\int \frac{1}{2u^2-1}\mathrm{d}u\Rightarrow$$

$$I=\int \frac{1}{(u+1)(u-1)}\mathrm{d}u-2\int \frac{1}{(\sqrt{2}u+1)(\sqrt{2}u-1)}\mathrm{d}u\Rightarrow$$

又：

$$\frac{1}{u-1}-\frac{1}{u+1}=\frac{u+1-u+1}{(u-1)(u+1)}=$$

$$\frac{2}{(u-1)(u+1)}$$

$$\frac{1}{\sqrt{2}u-1}-\frac{1}{\sqrt{2}u+1}=\frac{\sqrt{2}u+1-\sqrt{2}u+1}{(\sqrt{2}u-1)(\sqrt{2}u+1)}=$$

$$\frac{2}{(\sqrt{2}u-1)(\sqrt{2}u+1)}$$

于是：

$$I=$$

$$\frac{1}{2}\int (\frac{1}{u-1}-\frac{1}{u+1})\mathrm{d}u-2\times \frac{1}{2}\int (\frac{1}{\sqrt{2}u-1}-\frac{1}{\sqrt{2}u+1})\mathrm{d}u\Rightarrow$$

$$I=\frac{1}{2}\ln \left|\frac{u-1}{u+1}\right|-[\int \frac{1}{\sqrt{2}u-1}\mathrm{d}u-\int \frac{1}{\sqrt{2}u+1}\mathrm{d}u]$$

$$I=$$

$$\frac{1}{2}\ln \left|\frac{u-1}{u+1}\right|+[\frac{1}{\sqrt{2}}\ln |\sqrt{2}u-1|-\frac{1}{\sqrt{2}}\ln |\sqrt{2}u+1|]+C\Rightarrow$$

$$I=\frac{1}{2}\ln \left|\frac{u-1}{u+1}\right|-\frac{1}{\sqrt{2}}\ln \left|\frac{\sqrt{2}u-1}{\sqrt{2}u+1}\right|+C\Rightarrow$$

$$I=\frac{1}{2}\ln \left|\frac{\cos x-1}{\cos x+1}\right|-\frac{1}{\sqrt{2}}\ln \left|\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}\right|+C$$