$$$\cot^{2}{\left(\theta \right)}$$$ 的積分
您的輸入
求$$$\int \cot^{2}{\left(\theta \right)}\, d\theta$$$。
解答
令 $$$u=\cot{\left(\theta \right)}$$$。
則 $$$du=\left(\cot{\left(\theta \right)}\right)^{\prime }d\theta = - \csc^{2}{\left(\theta \right)} d\theta$$$ (步驟見»)，並可得 $$$\csc^{2}{\left(\theta \right)} d\theta = - du$$$。
該積分可改寫為
$${\color{red}{\int{\cot^{2}{\left(\theta \right)} d \theta}}} = {\color{red}{\int{\left(- \frac{u^{2}}{u^{2} + 1}\right)d u}}}$$
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 $$$u=\cot{\left(\theta \right)}$$$：
$$\operatorname{atan}{\left({\color{red}{u}} \right)} - {\color{red}{u}} = \operatorname{atan}{\left({\color{red}{\cot{\left(\theta \right)}}} \right)} - {\color{red}{\cot{\left(\theta \right)}}}$$
因此，
$$\int{\cot^{2}{\left(\theta \right)} d \theta} = - \cot{\left(\theta \right)} + \operatorname{atan}{\left(\cot{\left(\theta \right)} \right)}$$
加上積分常數：
$$\int{\cot^{2}{\left(\theta \right)} d \theta} = - \cot{\left(\theta \right)} + \operatorname{atan}{\left(\cot{\left(\theta \right)} \right)}+C$$
答案
$$$\int \cot^{2}{\left(\theta \right)}\, d\theta = \left(- \cot{\left(\theta \right)} + \operatorname{atan}{\left(\cot{\left(\theta \right)} \right)}\right) + C$$$A