$$$\tan^{2}{\left(4 x \right)}$$$ 的積分

求$$$\int \tan^{2}{\left(4 x \right)}\, dx$$$。

令 $$$u=4 x$$$。

則 $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{4}$$$。

該積分可改寫為

$${\color{red}{\int{\tan^{2}{\left(4 x \right)} d x}}} = {\color{red}{\int{\frac{\tan^{2}{\left(u \right)}}{4} d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{4}$$$ 與 $$$f{\left(u \right)} = \tan^{2}{\left(u \right)}$$$：

$${\color{red}{\int{\frac{\tan^{2}{\left(u \right)}}{4} d u}}} = {\color{red}{\left(\frac{\int{\tan^{2}{\left(u \right)} d u}}{4}\right)}}$$

令 $$$v=\tan{\left(u \right)}$$$。

則 $$$u=\operatorname{atan}{\left(v \right)}$$$ 與 $$$du=\left(\operatorname{atan}{\left(v \right)}\right)^{\prime }dv = \frac{dv}{v^{2} + 1}$$$（步驟見»）。

該積分變為

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

重寫並拆分分式:

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

逐項積分：

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

配合 $$$c=1$$$，應用常數法則 $$$\int c\, dv = c v$$$：

$$- \frac{\int{\frac{1}{v^{2} + 1} d v}}{4} + \frac{{\color{red}{\int{1 d v}}}}{4} = - \frac{\int{\frac{1}{v^{2} + 1} d v}}{4} + \frac{{\color{red}{v}}}{4}$$

$$$\frac{1}{v^{2} + 1}$$$ 的積分是 $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$：

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

回顧一下 $$$v=\tan{\left(u \right)}$$$：

$$- \frac{\operatorname{atan}{\left({\color{red}{v}} \right)}}{4} + \frac{{\color{red}{v}}}{4} = - \frac{\operatorname{atan}{\left({\color{red}{\tan{\left(u \right)}}} \right)}}{4} + \frac{{\color{red}{\tan{\left(u \right)}}}}{4}$$

回顧一下 $$$u=4 x$$$：

$$\frac{\tan{\left({\color{red}{u}} \right)}}{4} - \frac{\operatorname{atan}{\left(\tan{\left({\color{red}{u}} \right)} \right)}}{4} = \frac{\tan{\left({\color{red}{\left(4 x\right)}} \right)}}{4} - \frac{\operatorname{atan}{\left(\tan{\left({\color{red}{\left(4 x\right)}} \right)} \right)}}{4}$$

因此，

$$\int{\tan^{2}{\left(4 x \right)} d x} = \frac{\tan{\left(4 x \right)}}{4} - \frac{\operatorname{atan}{\left(\tan{\left(4 x \right)} \right)}}{4}$$

化簡：

$$\int{\tan^{2}{\left(4 x \right)} d x} = - x + \frac{\tan{\left(4 x \right)}}{4}$$

加上積分常數：

$$\int{\tan^{2}{\left(4 x \right)} d x} = - x + \frac{\tan{\left(4 x \right)}}{4}+C$$

$$$\int \tan^{2}{\left(4 x \right)}\, dx = \left(- x + \frac{\tan{\left(4 x \right)}}{4}\right) + C$$$A