$$$\tan^{3}{\left(97 x \right)} \sec^{3}{\left(97 x \right)}$$$ 的积分

求$$$\int \tan^{3}{\left(97 x \right)} \sec^{3}{\left(97 x \right)}\, dx$$$。

设$$$u=97 x$$$。

则$$$du=\left(97 x\right)^{\prime }dx = 97 dx$$$ (步骤见»)，并有$$$dx = \frac{du}{97}$$$。

所以，

$${\color{red}{\int{\tan^{3}{\left(97 x \right)} \sec^{3}{\left(97 x \right)} d x}}} = {\color{red}{\int{\frac{\tan^{3}{\left(u \right)} \sec^{3}{\left(u \right)}}{97} d u}}}$$

对 $$$c=\frac{1}{97}$$$ 和 $$$f{\left(u \right)} = \tan^{3}{\left(u \right)} \sec^{3}{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{\frac{\tan^{3}{\left(u \right)} \sec^{3}{\left(u \right)}}{97} d u}}} = {\color{red}{\left(\frac{\int{\tan^{3}{\left(u \right)} \sec^{3}{\left(u \right)} d u}}{97}\right)}}$$

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

$$\frac{{\color{red}{\int{\tan^{3}{\left(u \right)} \sec^{3}{\left(u \right)} d u}}}}{97} = \frac{{\color{red}{\int{\left(\sec^{2}{\left(u \right)} - 1\right) \tan{\left(u \right)} \sec^{3}{\left(u \right)} d u}}}}{97}$$

设$$$v=\sec{\left(u \right)}$$$。

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

因此，

$$\frac{{\color{red}{\int{\left(\sec^{2}{\left(u \right)} - 1\right) \tan{\left(u \right)} \sec^{3}{\left(u \right)} d u}}}}{97} = \frac{{\color{red}{\int{v^{2} \left(v^{2} - 1\right) d v}}}}{97}$$

Expand the expression:

$$\frac{{\color{red}{\int{v^{2} \left(v^{2} - 1\right) d v}}}}{97} = \frac{{\color{red}{\int{\left(v^{4} - v^{2}\right)d v}}}}{97}$$

逐项积分：

$$\frac{{\color{red}{\int{\left(v^{4} - v^{2}\right)d v}}}}{97} = \frac{{\color{red}{\left(- \int{v^{2} d v} + \int{v^{4} d v}\right)}}}{97}$$

应用幂法则 $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=4$$$：

$$- \frac{\int{v^{2} d v}}{97} + \frac{{\color{red}{\int{v^{4} d v}}}}{97}=- \frac{\int{v^{2} d v}}{97} + \frac{{\color{red}{\frac{v^{1 + 4}}{1 + 4}}}}{97}=- \frac{\int{v^{2} d v}}{97} + \frac{{\color{red}{\left(\frac{v^{5}}{5}\right)}}}{97}$$

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

$$\frac{v^{5}}{485} - \frac{{\color{red}{\int{v^{2} d v}}}}{97}=\frac{v^{5}}{485} - \frac{{\color{red}{\frac{v^{1 + 2}}{1 + 2}}}}{97}=\frac{v^{5}}{485} - \frac{{\color{red}{\left(\frac{v^{3}}{3}\right)}}}{97}$$

回忆一下 $$$v=\sec{\left(u \right)}$$$:

$$- \frac{{\color{red}{v}}^{3}}{291} + \frac{{\color{red}{v}}^{5}}{485} = - \frac{{\color{red}{\sec{\left(u \right)}}}^{3}}{291} + \frac{{\color{red}{\sec{\left(u \right)}}}^{5}}{485}$$

回忆一下 $$$u=97 x$$$:

$$- \frac{\sec^{3}{\left({\color{red}{u}} \right)}}{291} + \frac{\sec^{5}{\left({\color{red}{u}} \right)}}{485} = - \frac{\sec^{3}{\left({\color{red}{\left(97 x\right)}} \right)}}{291} + \frac{\sec^{5}{\left({\color{red}{\left(97 x\right)}} \right)}}{485}$$

因此，

$$\int{\tan^{3}{\left(97 x \right)} \sec^{3}{\left(97 x \right)} d x} = \frac{\sec^{5}{\left(97 x \right)}}{485} - \frac{\sec^{3}{\left(97 x \right)}}{291}$$

加上积分常数：

$$\int{\tan^{3}{\left(97 x \right)} \sec^{3}{\left(97 x \right)} d x} = \frac{\sec^{5}{\left(97 x \right)}}{485} - \frac{\sec^{3}{\left(97 x \right)}}{291}+C$$

$$$\int \tan^{3}{\left(97 x \right)} \sec^{3}{\left(97 x \right)}\, dx = \left(\frac{\sec^{5}{\left(97 x \right)}}{485} - \frac{\sec^{3}{\left(97 x \right)}}{291}\right) + C$$$A