$$$\frac{1}{\tan^{2}{\left(x \right)}}$$$ 的積分
您的輸入
求$$$\int \frac{1}{\tan^{2}{\left(x \right)}}\, dx$$$。
解答
令 $$$u=\tan{\left(x \right)}$$$。
則 $$$x=\operatorname{atan}{\left(u \right)}$$$ 與 $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$（步驟見»）。
該積分可改寫為
$${\color{red}{\int{\frac{1}{\tan^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{u^{2} \left(u^{2} + 1\right)} d u}}}$$
進行部分分式分解（步驟可見 »）:
$${\color{red}{\int{\frac{1}{u^{2} \left(u^{2} + 1\right)} d u}}} = {\color{red}{\int{\left(- \frac{1}{u^{2} + 1} + \frac{1}{u^{2}}\right)d u}}}$$
逐項積分：
$${\color{red}{\int{\left(- \frac{1}{u^{2} + 1} + \frac{1}{u^{2}}\right)d u}}} = {\color{red}{\left(\int{\frac{1}{u^{2}} d u} - \int{\frac{1}{u^{2} + 1} d u}\right)}}$$
套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-2$$$：
$$- \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\int{\frac{1}{u^{2}} d u}}}=- \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\int{u^{-2} d u}}}=- \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}=- \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\left(- u^{-1}\right)}}=- \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\left(- \frac{1}{u}\right)}}$$
$$$\frac{1}{u^{2} + 1}$$$ 的積分是 $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$：
$$- {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}} - \frac{1}{u} = - {\color{red}{\operatorname{atan}{\left(u \right)}}} - \frac{1}{u}$$
回顧一下 $$$u=\tan{\left(x \right)}$$$：
$$- \operatorname{atan}{\left({\color{red}{u}} \right)} - {\color{red}{u}}^{-1} = - \operatorname{atan}{\left({\color{red}{\tan{\left(x \right)}}} \right)} - {\color{red}{\tan{\left(x \right)}}}^{-1}$$
因此，
$$\int{\frac{1}{\tan^{2}{\left(x \right)}} d x} = - \operatorname{atan}{\left(\tan{\left(x \right)} \right)} - \frac{1}{\tan{\left(x \right)}}$$
化簡：
$$\int{\frac{1}{\tan^{2}{\left(x \right)}} d x} = - x - \frac{1}{\tan{\left(x \right)}}$$
加上積分常數：
$$\int{\frac{1}{\tan^{2}{\left(x \right)}} d x} = - x - \frac{1}{\tan{\left(x \right)}}+C$$
答案
$$$\int \frac{1}{\tan^{2}{\left(x \right)}}\, dx = \left(- x - \frac{1}{\tan{\left(x \right)}}\right) + C$$$A