$$$\frac{1}{1 - \cos{\left(2 x \right)}}$$$ 的积分
您的输入
求$$$\int \frac{1}{1 - \cos{\left(2 x \right)}}\, dx$$$。
解答
设$$$u=2 x$$$。
则$$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (步骤见»),并有$$$dx = \frac{du}{2}$$$。
积分变为
$${\color{red}{\int{\frac{1}{1 - \cos{\left(2 x \right)}} d x}}} = {\color{red}{\int{\left(- \frac{1}{2 \left(\cos{\left(u \right)} - 1\right)}\right)d u}}}$$
对 $$$c=- \frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \frac{1}{\cos{\left(u \right)} - 1}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$${\color{red}{\int{\left(- \frac{1}{2 \left(\cos{\left(u \right)} - 1\right)}\right)d u}}} = {\color{red}{\left(- \frac{\int{\frac{1}{\cos{\left(u \right)} - 1} d u}}{2}\right)}}$$
使用二倍角公式 $$$\cos\left( u \right)=1-2\sin^2\left(\frac{ u }{2}\right)$$$ 重写余弦并化简:
$$- \frac{{\color{red}{\int{\frac{1}{\cos{\left(u \right)} - 1} d u}}}}{2} = - \frac{{\color{red}{\int{\left(- \frac{1}{2 \sin^{2}{\left(\frac{u}{2} \right)}}\right)d u}}}}{2}$$
设$$$v=\frac{u}{2}$$$。
则$$$dv=\left(\frac{u}{2}\right)^{\prime }du = \frac{du}{2}$$$ (步骤见»),并有$$$du = 2 dv$$$。
积分变为
$$- \frac{{\color{red}{\int{\left(- \frac{1}{2 \sin^{2}{\left(\frac{u}{2} \right)}}\right)d u}}}}{2} = - \frac{{\color{red}{\int{\left(- \frac{1}{\sin^{2}{\left(v \right)}}\right)d v}}}}{2}$$
对 $$$c=-1$$$ 和 $$$f{\left(v \right)} = \frac{1}{\sin^{2}{\left(v \right)}}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$:
$$- \frac{{\color{red}{\int{\left(- \frac{1}{\sin^{2}{\left(v \right)}}\right)d v}}}}{2} = - \frac{{\color{red}{\left(- \int{\frac{1}{\sin^{2}{\left(v \right)}} d v}\right)}}}{2}$$
将被积函数用余割表示:
$$\frac{{\color{red}{\int{\frac{1}{\sin^{2}{\left(v \right)}} d v}}}}{2} = \frac{{\color{red}{\int{\csc^{2}{\left(v \right)} d v}}}}{2}$$
$$$\csc^{2}{\left(v \right)}$$$ 的积分为 $$$\int{\csc^{2}{\left(v \right)} d v} = - \cot{\left(v \right)}$$$:
$$\frac{{\color{red}{\int{\csc^{2}{\left(v \right)} d v}}}}{2} = \frac{{\color{red}{\left(- \cot{\left(v \right)}\right)}}}{2}$$
回忆一下 $$$v=\frac{u}{2}$$$:
$$- \frac{\cot{\left({\color{red}{v}} \right)}}{2} = - \frac{\cot{\left({\color{red}{\left(\frac{u}{2}\right)}} \right)}}{2}$$
回忆一下 $$$u=2 x$$$:
$$- \frac{\cot{\left(\frac{{\color{red}{u}}}{2} \right)}}{2} = - \frac{\cot{\left(\frac{{\color{red}{\left(2 x\right)}}}{2} \right)}}{2}$$
因此,
$$\int{\frac{1}{1 - \cos{\left(2 x \right)}} d x} = - \frac{\cot{\left(x \right)}}{2}$$
加上积分常数:
$$\int{\frac{1}{1 - \cos{\left(2 x \right)}} d x} = - \frac{\cot{\left(x \right)}}{2}+C$$
答案
$$$\int \frac{1}{1 - \cos{\left(2 x \right)}}\, dx = - \frac{\cot{\left(x \right)}}{2} + C$$$A