$$$\sqrt{\cot{\left(x \right)}} \csc^{2}{\left(x \right)}$$$ 的积分

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

设$$$u=\cot{\left(x \right)}$$$。

则$$$du=\left(\cot{\left(x \right)}\right)^{\prime }dx = - \csc^{2}{\left(x \right)} dx$$$ (步骤见»)，并有$$$\csc^{2}{\left(x \right)} dx = - du$$$。

所以，

$${\color{red}{\int{\sqrt{\cot{\left(x \right)}} \csc^{2}{\left(x \right)} d x}}} = {\color{red}{\int{\left(- \sqrt{u}\right)d u}}}$$

对 $$$c=-1$$$ 和 $$$f{\left(u \right)} = \sqrt{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{\left(- \sqrt{u}\right)d u}}} = {\color{red}{\left(- \int{\sqrt{u} d u}\right)}}$$

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=\frac{1}{2}$$$：

$$- {\color{red}{\int{\sqrt{u} d u}}}=- {\color{red}{\int{u^{\frac{1}{2}} d u}}}=- {\color{red}{\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}=- {\color{red}{\left(\frac{2 u^{\frac{3}{2}}}{3}\right)}}$$

回忆一下 $$$u=\cot{\left(x \right)}$$$:

$$- \frac{2 {\color{red}{u}}^{\frac{3}{2}}}{3} = - \frac{2 {\color{red}{\cot{\left(x \right)}}}^{\frac{3}{2}}}{3}$$

因此，

$$\int{\sqrt{\cot{\left(x \right)}} \csc^{2}{\left(x \right)} d x} = - \frac{2 \cot^{\frac{3}{2}}{\left(x \right)}}{3}$$

加上积分常数：

$$\int{\sqrt{\cot{\left(x \right)}} \csc^{2}{\left(x \right)} d x} = - \frac{2 \cot^{\frac{3}{2}}{\left(x \right)}}{3}+C$$

$$$\int \sqrt{\cot{\left(x \right)}} \csc^{2}{\left(x \right)}\, dx = - \frac{2 \cot^{\frac{3}{2}}{\left(x \right)}}{3} + C$$$A