$$$\frac{1}{\left(1 - \tan{\left(x \right)}\right)^{2} \cos^{2}{\left(x \right)}}$$$ 的積分

求$$$\int \frac{1}{\left(1 - \tan{\left(x \right)}\right)^{2} \cos^{2}{\left(x \right)}}\, dx$$$。

令 $$$u=1 - \tan{\left(x \right)}$$$。

則 $$$du=\left(1 - \tan{\left(x \right)}\right)^{\prime }dx = - \sec^{2}{\left(x \right)} dx$$$ (步驟見»)，並可得 $$$\sec^{2}{\left(x \right)} dx = - du$$$。

因此，

$${\color{red}{\int{\frac{1}{\left(1 - \tan{\left(x \right)}\right)^{2} \cos^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\left(- \frac{1}{u^{2}}\right)d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=-1$$$ 與 $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$：

$${\color{red}{\int{\left(- \frac{1}{u^{2}}\right)d u}}} = {\color{red}{\left(- \int{\frac{1}{u^{2}} d u}\right)}}$$

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-2$$$：

$$- {\color{red}{\int{\frac{1}{u^{2}} d u}}}=- {\color{red}{\int{u^{-2} d u}}}=- {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}=- {\color{red}{\left(- u^{-1}\right)}}=- {\color{red}{\left(- \frac{1}{u}\right)}}$$

回顧一下 $$$u=1 - \tan{\left(x \right)}$$$：

$${\color{red}{u}}^{-1} = {\color{red}{\left(1 - \tan{\left(x \right)}\right)}}^{-1}$$

因此，

$$\int{\frac{1}{\left(1 - \tan{\left(x \right)}\right)^{2} \cos^{2}{\left(x \right)}} d x} = \frac{1}{1 - \tan{\left(x \right)}}$$

化簡：

$$\int{\frac{1}{\left(1 - \tan{\left(x \right)}\right)^{2} \cos^{2}{\left(x \right)}} d x} = - \frac{1}{\tan{\left(x \right)} - 1}$$

加上積分常數：

$$\int{\frac{1}{\left(1 - \tan{\left(x \right)}\right)^{2} \cos^{2}{\left(x \right)}} d x} = - \frac{1}{\tan{\left(x \right)} - 1}+C$$

$$$\int \frac{1}{\left(1 - \tan{\left(x \right)}\right)^{2} \cos^{2}{\left(x \right)}}\, dx = - \frac{1}{\tan{\left(x \right)} - 1} + C$$$A