题目 03
$$
I=\frac{1}{\sin x \cos 2 x} \mathrm{~ d} x=?
$$
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：
$\sin 2 x = 2 \sin x \cos x$
$\cos 2x = 2 \cos^{2} x – 1 = 1 – 2 \sin^{2} x$
$2\cos 2x = 2 \cos^{2} x – 1 + 1 – 2 \sin^{2} x$ $\Rightarrow$ $\cos 2x = \cos ^{2} x – \sin ^{2} x$
先尝试都转换为 $\sin x$
$$
I=\int \frac{\mathrm{~ d} (\sin x)}{\sin x \cdot \cos x\left(1-2 \sin ^{2} x\right)}
$$
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 $\cos x$:
$$
I=\int \frac{-\mathrm{~ d} (\cos x)}{\sin ^{2} x\left(2 \cos ^{2} x-1\right)}=\int \frac{-\mathrm{~ d} (\cos x)}{\left(1-\cos ^{2} x\right)\left(2 \cos ^{2} x-1\right)}
$$
令：
$$
u=\cos x
$$
则：
$$
I=\int \frac{-\mathrm{~ d} u}{\left(1-u^{2}\right)\left(2 u^{2}-1\right)} \Rightarrow
$$
设：
$$
\frac{A u+B}{1-u^{2}}+\frac{C u+D}{2 u^{2}-1}
$$
则有：
$$
\frac{(A u+B)\left(2 u^{2}-1\right)+(C u+D)\left(1-u^{2}\right)}{\left(1-u^{2}\right)\left(2 u^{2}-1\right)} \Rightarrow
$$
$$
2 A u^{3}-A u+2 B u^{2}-B+ C u-C u^{3}+D-D u^{2}=1 \Rightarrow
$$
$$
\left\{\begin{array} { l }
{ 2 A – C = 0 } \\ { – A + C = 0 } \\ { 2 B – D = 0 } \\ { – B + D = 1 }
\end{array} \quad \Rightarrow \left\{\begin{array}{l} A=0 \\ C=0 \\ B=1 \\ D=2
\end{array} \right.\right.
$$
于是：
$$
I=-\int \left(\frac{1}{1-u^{2}}+\frac{2}{2 u^{2}-1}\right) \mathrm{~ d} u \Rightarrow
$$
$$
I=+\int \frac{1}{u^{2}-1} \mathrm{~ d} u-2 \int \frac{1}{2 u^{2}-1} \mathrm{~ d} u \Rightarrow
$$
$$
I=\int \frac{1}{(u+1) (u-1)} \mathrm{~ d} u-2 \int \frac{1}{(\sqrt{2} u+1)(\sqrt{2} u-1)} \mathrm{~ d} u \Rightarrow
$$
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：
$$
I=
$$
$$
\frac{1}{2} \int\left(\frac{1}{u-1}-\frac{1}{u+1}\right) \mathrm{~ d} u-2 \times \frac{1}{2} \int\left(\frac{1}{\sqrt{2} u-1}-\frac{1}{\sqrt{2} u+1}\right) \mathrm{~ d} u \Rightarrow
$$
$$
I=\frac{1}{2} \ln \left|\frac{u-1}{u+1}\right|-\left[\int \frac{1}{\sqrt{2} u-1} \mathrm{~ d} u-\int \frac{1}{\sqrt{2} u+1} \mathrm{~ d} u\right]
$$
$$
I=
$$
$$
\frac{1}{2} \ln \left|\frac{u-1}{u+1}\right|+\left[\frac{1}{\sqrt{2}} \ln |\sqrt{2} u-1|-\right. \left.\frac{1}{\sqrt{2}} \ln |\sqrt{2} u+1|\right]+ C \Rightarrow
$$
$$
I=\frac{1}{2} \ln \left|\frac{u-1}{u+1}\right|-\frac{1}{\sqrt{2}} \ln \left|\frac{\sqrt{2} u-1}{\sqrt{2} u+1}\right|+ C \Rightarrow
$$
$$
I=\frac{1}{2} \ln \left|\frac{\cos x-1}{\cos x+1}\right|-\frac{1}{\sqrt{2}} \ln \left|\frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}\right|+ C
$$