$$$- \tan^{2}{\left(x \right)}$$$ 的积分

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

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

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

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，

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

改写并拆分该分式:

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

逐项积分：

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

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

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

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

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

回忆一下 $$$u=\tan{\left(x \right)}$$$:

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

因此，

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

化简：

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

加上积分常数：

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

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