$$$\frac{\sec^{2}{\left(x \right)}}{9 \tan^{2}{\left(x \right)}}$$$ 的积分

求$$$\int \frac{\sec^{2}{\left(x \right)}}{9 \tan^{2}{\left(x \right)}}\, dx$$$。

对 $$$c=\frac{1}{9}$$$ 和 $$$f{\left(x \right)} = \frac{\sec^{2}{\left(x \right)}}{\tan^{2}{\left(x \right)}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\frac{\sec^{2}{\left(x \right)}}{9 \tan^{2}{\left(x \right)}} d x}}} = {\color{red}{\left(\frac{\int{\frac{\sec^{2}{\left(x \right)}}{\tan^{2}{\left(x \right)}} d x}}{9}\right)}}$$

设$$$u=\tan{\left(x \right)}$$$。

则$$$du=\left(\tan{\left(x \right)}\right)^{\prime }dx = \sec^{2}{\left(x \right)} dx$$$ (步骤见»)，并有$$$\sec^{2}{\left(x \right)} dx = du$$$。

积分变为

$$\frac{{\color{red}{\int{\frac{\sec^{2}{\left(x \right)}}{\tan^{2}{\left(x \right)}} d x}}}}{9} = \frac{{\color{red}{\int{\frac{1}{u^{2}} d u}}}}{9}$$

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

$$\frac{{\color{red}{\int{\frac{1}{u^{2}} d u}}}}{9}=\frac{{\color{red}{\int{u^{-2} d u}}}}{9}=\frac{{\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}}{9}=\frac{{\color{red}{\left(- u^{-1}\right)}}}{9}=\frac{{\color{red}{\left(- \frac{1}{u}\right)}}}{9}$$

回忆一下 $$$u=\tan{\left(x \right)}$$$:

$$- \frac{{\color{red}{u}}^{-1}}{9} = - \frac{{\color{red}{\tan{\left(x \right)}}}^{-1}}{9}$$

因此，

$$\int{\frac{\sec^{2}{\left(x \right)}}{9 \tan^{2}{\left(x \right)}} d x} = - \frac{1}{9 \tan{\left(x \right)}}$$

加上积分常数：

$$\int{\frac{\sec^{2}{\left(x \right)}}{9 \tan^{2}{\left(x \right)}} d x} = - \frac{1}{9 \tan{\left(x \right)}}+C$$

$$$\int \frac{\sec^{2}{\left(x \right)}}{9 \tan^{2}{\left(x \right)}}\, dx = - \frac{1}{9 \tan{\left(x \right)}} + C$$$A