$$$\cot^{2}{\left(x \right)}$$$ 的積分

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

令 $$$u=\cot{\left(x \right)}$$$。

則 $$$du=\left(\cot{\left(x \right)}\right)^{\prime }dx = - \csc^{2}{\left(x \right)} dx$$$ (步驟見»)，並可得 $$$\csc^{2}{\left(x \right)} dx = - du$$$。

因此，

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

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

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

重寫並拆分分式:

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

逐項積分：

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

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

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

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

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

回顧一下 $$$u=\cot{\left(x \right)}$$$：

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

因此，

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

加上積分常數：

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

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