$$$\frac{\tan^{2}{\left(x \right)}}{2}$$$ 的積分

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

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

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

令 $$$u=\tan{\left(x \right)}$$$。

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

因此，

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

重寫並拆分分式:

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

逐項積分：

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

配合 $$$c=1$$$，應用常數法則 $$$\int c\, du = c u$$$：

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

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

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

回顧一下 $$$u=\tan{\left(x \right)}$$$：

$$- \frac{\operatorname{atan}{\left({\color{red}{u}} \right)}}{2} + \frac{{\color{red}{u}}}{2} = - \frac{\operatorname{atan}{\left({\color{red}{\tan{\left(x \right)}}} \right)}}{2} + \frac{{\color{red}{\tan{\left(x \right)}}}}{2}$$

因此，

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

化簡：

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

加上積分常數：

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

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