$$$\frac{\tan^{2}{\left(x \right)}}{2}$$$ 的积分

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

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

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

设$$$u=\tan{\left(x \right)}$$$。

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

所以，

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

改写并拆分该分式:

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

逐项积分：

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

应用常数法则 $$$\int c\, du = c u$$$，使用 $$$c=1$$$：

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

$$$\frac{1}{u^{2} + 1}$$$ 的积分为 $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

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

回忆一下 $$$u=\tan{\left(x \right)}$$$:

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

因此，

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

化简：

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

加上积分常数：

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

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