$$$\frac{1}{1 - \cos{\left(x \right)}}$$$ 的积分

求$$$\int \frac{1}{1 - \cos{\left(x \right)}}\, dx$$$。

使用二倍角公式 $$$\cos\left(x\right)=1-2\sin^2\left(\frac{x}{2}\right)$$$ 重写余弦并化简:

$${\color{red}{\int{\frac{1}{1 - \cos{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{2 \sin^{2}{\left(\frac{x}{2} \right)}} d x}}}$$

设$$$u=\frac{x}{2}$$$。

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

所以，

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

将被积函数用余割表示:

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

$$$\csc^{2}{\left(u \right)}$$$ 的积分为 $$$\int{\csc^{2}{\left(u \right)} d u} = - \cot{\left(u \right)}$$$:

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

回忆一下 $$$u=\frac{x}{2}$$$:

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

因此，

$$\int{\frac{1}{1 - \cos{\left(x \right)}} d x} = - \cot{\left(\frac{x}{2} \right)}$$

加上积分常数：

$$\int{\frac{1}{1 - \cos{\left(x \right)}} d x} = - \cot{\left(\frac{x}{2} \right)}+C$$

$$$\int \frac{1}{1 - \cos{\left(x \right)}}\, dx = - \cot{\left(\frac{x}{2} \right)} + C$$$A