$$$\tan^{5}{\left(x \right)} \sec^{4}{\left(x \right)}$$$ 的积分

求$$$\int \tan^{5}{\left(x \right)} \sec^{4}{\left(x \right)}\, dx$$$。

抽出一个正切，并将其余部分用正割表示，使用公式 $$$\tan^2\left(x \right)=\sec^2\left(x \right)-1$$$:

$${\color{red}{\int{\tan^{5}{\left(x \right)} \sec^{4}{\left(x \right)} d x}}} = {\color{red}{\int{\left(\sec^{2}{\left(x \right)} - 1\right)^{2} \tan{\left(x \right)} \sec^{4}{\left(x \right)} d x}}}$$

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，

$${\color{red}{\int{\left(\sec^{2}{\left(x \right)} - 1\right)^{2} \tan{\left(x \right)} \sec^{4}{\left(x \right)} d x}}} = {\color{red}{\int{u^{3} \left(u^{2} - 1\right)^{2} d u}}}$$

Expand the expression:

$${\color{red}{\int{u^{3} \left(u^{2} - 1\right)^{2} d u}}} = {\color{red}{\int{\left(u^{7} - 2 u^{5} + u^{3}\right)d u}}}$$

逐项积分：

$${\color{red}{\int{\left(u^{7} - 2 u^{5} + u^{3}\right)d u}}} = {\color{red}{\left(\int{u^{3} d u} - \int{2 u^{5} d u} + \int{u^{7} d u}\right)}}$$

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=3$$$：

$$- \int{2 u^{5} d u} + \int{u^{7} d u} + {\color{red}{\int{u^{3} d u}}}=- \int{2 u^{5} d u} + \int{u^{7} d u} + {\color{red}{\frac{u^{1 + 3}}{1 + 3}}}=- \int{2 u^{5} d u} + \int{u^{7} d u} + {\color{red}{\left(\frac{u^{4}}{4}\right)}}$$

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=7$$$：

$$\frac{u^{4}}{4} - \int{2 u^{5} d u} + {\color{red}{\int{u^{7} d u}}}=\frac{u^{4}}{4} - \int{2 u^{5} d u} + {\color{red}{\frac{u^{1 + 7}}{1 + 7}}}=\frac{u^{4}}{4} - \int{2 u^{5} d u} + {\color{red}{\left(\frac{u^{8}}{8}\right)}}$$

对 $$$c=2$$$ 和 $$$f{\left(u \right)} = u^{5}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$\frac{u^{8}}{8} + \frac{u^{4}}{4} - {\color{red}{\int{2 u^{5} d u}}} = \frac{u^{8}}{8} + \frac{u^{4}}{4} - {\color{red}{\left(2 \int{u^{5} d u}\right)}}$$

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=5$$$：

$$\frac{u^{8}}{8} + \frac{u^{4}}{4} - 2 {\color{red}{\int{u^{5} d u}}}=\frac{u^{8}}{8} + \frac{u^{4}}{4} - 2 {\color{red}{\frac{u^{1 + 5}}{1 + 5}}}=\frac{u^{8}}{8} + \frac{u^{4}}{4} - 2 {\color{red}{\left(\frac{u^{6}}{6}\right)}}$$

回忆一下 $$$u=\sec{\left(x \right)}$$$:

$$\frac{{\color{red}{u}}^{4}}{4} - \frac{{\color{red}{u}}^{6}}{3} + \frac{{\color{red}{u}}^{8}}{8} = \frac{{\color{red}{\sec{\left(x \right)}}}^{4}}{4} - \frac{{\color{red}{\sec{\left(x \right)}}}^{6}}{3} + \frac{{\color{red}{\sec{\left(x \right)}}}^{8}}{8}$$

因此，

$$\int{\tan^{5}{\left(x \right)} \sec^{4}{\left(x \right)} d x} = \frac{\sec^{8}{\left(x \right)}}{8} - \frac{\sec^{6}{\left(x \right)}}{3} + \frac{\sec^{4}{\left(x \right)}}{4}$$

加上积分常数：

$$\int{\tan^{5}{\left(x \right)} \sec^{4}{\left(x \right)} d x} = \frac{\sec^{8}{\left(x \right)}}{8} - \frac{\sec^{6}{\left(x \right)}}{3} + \frac{\sec^{4}{\left(x \right)}}{4}+C$$

$$$\int \tan^{5}{\left(x \right)} \sec^{4}{\left(x \right)}\, dx = \left(\frac{\sec^{8}{\left(x \right)}}{8} - \frac{\sec^{6}{\left(x \right)}}{3} + \frac{\sec^{4}{\left(x \right)}}{4}\right) + C$$$A