$$$8 \tan{\left(x \right)} \sec^{3}{\left(x \right)}$$$ 的积分

求$$$\int 8 \tan{\left(x \right)} \sec^{3}{\left(x \right)}\, dx$$$。

对 $$$c=8$$$ 和 $$$f{\left(x \right)} = \tan{\left(x \right)} \sec^{3}{\left(x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{8 \tan{\left(x \right)} \sec^{3}{\left(x \right)} d x}}} = {\color{red}{\left(8 \int{\tan{\left(x \right)} \sec^{3}{\left(x \right)} d x}\right)}}$$

设$$$u=\sec{\left(x \right)}$$$。

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

所以，

$$8 {\color{red}{\int{\tan{\left(x \right)} \sec^{3}{\left(x \right)} d x}}} = 8 {\color{red}{\int{u^{2} d u}}}$$

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

$$8 {\color{red}{\int{u^{2} d u}}}=8 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=8 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

回忆一下 $$$u=\sec{\left(x \right)}$$$:

$$\frac{8 {\color{red}{u}}^{3}}{3} = \frac{8 {\color{red}{\sec{\left(x \right)}}}^{3}}{3}$$

因此，

$$\int{8 \tan{\left(x \right)} \sec^{3}{\left(x \right)} d x} = \frac{8 \sec^{3}{\left(x \right)}}{3}$$

加上积分常数：

$$\int{8 \tan{\left(x \right)} \sec^{3}{\left(x \right)} d x} = \frac{8 \sec^{3}{\left(x \right)}}{3}+C$$

$$$\int 8 \tan{\left(x \right)} \sec^{3}{\left(x \right)}\, dx = \frac{8 \sec^{3}{\left(x \right)}}{3} + C$$$A