$$$\frac{\sec{\left(x \right)}}{\tan^{2}{\left(x \right)}}$$$ 的积分

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

将被积函数用正弦和/或余弦表示:

$${\color{red}{\int{\frac{\sec{\left(x \right)}}{\tan^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\cos{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x}}}$$

设$$$u=\sin{\left(x \right)}$$$。

则$$$du=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$ (步骤见»)，并有$$$\cos{\left(x \right)} dx = du$$$。

该积分可以改写为

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

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=-2$$$：

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

回忆一下 $$$u=\sin{\left(x \right)}$$$:

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

因此，

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

加上积分常数：

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

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