$$$\tan^{2}{\left(u \right)}$$$ 的積分

求$$$\int \tan^{2}{\left(u \right)}\, du$$$。

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

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

重寫並拆分分式:

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

逐項積分：

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

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

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

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

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

回顧一下 $$$v=\tan{\left(u \right)}$$$：

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

因此，

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

化簡：

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

加上積分常數：

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

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