$$$\tan^{3}{\left(7 x \right)}$$$ 的积分

求$$$\int \tan^{3}{\left(7 x \right)}\, dx$$$。

设$$$u=7 x$$$。

则$$$du=\left(7 x\right)^{\prime }dx = 7 dx$$$ (步骤见»)，并有$$$dx = \frac{du}{7}$$$。

积分变为

$${\color{red}{\int{\tan^{3}{\left(7 x \right)} d x}}} = {\color{red}{\int{\frac{\tan^{3}{\left(u \right)}}{7} d u}}}$$

对 $$$c=\frac{1}{7}$$$ 和 $$$f{\left(u \right)} = \tan^{3}{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{\frac{\tan^{3}{\left(u \right)}}{7} d u}}} = {\color{red}{\left(\frac{\int{\tan^{3}{\left(u \right)} d u}}{7}\right)}}$$

设$$$v=\tan{\left(u \right)}$$$。

则 $$$u=\operatorname{atan}{\left(v \right)}$$$ 且 $$$du=\left(\operatorname{atan}{\left(v \right)}\right)^{\prime }dv = \frac{dv}{v^{2} + 1}$$$（步骤见»）。

该积分可以改写为

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

由于分子次数不小于分母次数，进行多项式长除法（步骤见»）:

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

逐项积分：

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

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

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

设$$$w=v^{2} + 1$$$。

则$$$dw=\left(v^{2} + 1\right)^{\prime }dv = 2 v dv$$$ (步骤见»)，并有$$$v dv = \frac{dw}{2}$$$。

该积分可以改写为

$$\frac{v^{2}}{14} - \frac{{\color{red}{\int{\frac{v}{v^{2} + 1} d v}}}}{7} = \frac{v^{2}}{14} - \frac{{\color{red}{\int{\frac{1}{2 w} d w}}}}{7}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(w \right)} = \frac{1}{w}$$$ 应用常数倍法则 $$$\int c f{\left(w \right)}\, dw = c \int f{\left(w \right)}\, dw$$$：

$$\frac{v^{2}}{14} - \frac{{\color{red}{\int{\frac{1}{2 w} d w}}}}{7} = \frac{v^{2}}{14} - \frac{{\color{red}{\left(\frac{\int{\frac{1}{w} d w}}{2}\right)}}}{7}$$

$$$\frac{1}{w}$$$ 的积分为 $$$\int{\frac{1}{w} d w} = \ln{\left(\left|{w}\right| \right)}$$$:

$$\frac{v^{2}}{14} - \frac{{\color{red}{\int{\frac{1}{w} d w}}}}{14} = \frac{v^{2}}{14} - \frac{{\color{red}{\ln{\left(\left|{w}\right| \right)}}}}{14}$$

回忆一下 $$$w=v^{2} + 1$$$:

$$\frac{v^{2}}{14} - \frac{\ln{\left(\left|{{\color{red}{w}}}\right| \right)}}{14} = \frac{v^{2}}{14} - \frac{\ln{\left(\left|{{\color{red}{\left(v^{2} + 1\right)}}}\right| \right)}}{14}$$

回忆一下 $$$v=\tan{\left(u \right)}$$$:

$$- \frac{\ln{\left(1 + {\color{red}{v}}^{2} \right)}}{14} + \frac{{\color{red}{v}}^{2}}{14} = - \frac{\ln{\left(1 + {\color{red}{\tan{\left(u \right)}}}^{2} \right)}}{14} + \frac{{\color{red}{\tan{\left(u \right)}}}^{2}}{14}$$

回忆一下 $$$u=7 x$$$:

$$- \frac{\ln{\left(1 + \tan^{2}{\left({\color{red}{u}} \right)} \right)}}{14} + \frac{\tan^{2}{\left({\color{red}{u}} \right)}}{14} = - \frac{\ln{\left(1 + \tan^{2}{\left({\color{red}{\left(7 x\right)}} \right)} \right)}}{14} + \frac{\tan^{2}{\left({\color{red}{\left(7 x\right)}} \right)}}{14}$$

因此，

$$\int{\tan^{3}{\left(7 x \right)} d x} = - \frac{\ln{\left(\tan^{2}{\left(7 x \right)} + 1 \right)}}{14} + \frac{\tan^{2}{\left(7 x \right)}}{14}$$

加上积分常数：

$$\int{\tan^{3}{\left(7 x \right)} d x} = - \frac{\ln{\left(\tan^{2}{\left(7 x \right)} + 1 \right)}}{14} + \frac{\tan^{2}{\left(7 x \right)}}{14}+C$$

$$$\int \tan^{3}{\left(7 x \right)}\, dx = \left(- \frac{\ln\left(\tan^{2}{\left(7 x \right)} + 1\right)}{14} + \frac{\tan^{2}{\left(7 x \right)}}{14}\right) + C$$$A