$$$\frac{1}{\sin^{2}{\left(2 x \right)}}$$$ 的积分
您的输入
求$$$\int \frac{1}{\sin^{2}{\left(2 x \right)}}\, dx$$$。
解答
设$$$u=2 x$$$。
则$$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (步骤见»),并有$$$dx = \frac{du}{2}$$$。
该积分可以改写为
$${\color{red}{\int{\frac{1}{\sin^{2}{\left(2 x \right)}} d x}}} = {\color{red}{\int{\frac{1}{2 \sin^{2}{\left(u \right)}} d u}}}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \frac{1}{\sin^{2}{\left(u \right)}}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$${\color{red}{\int{\frac{1}{2 \sin^{2}{\left(u \right)}} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{\sin^{2}{\left(u \right)}} d u}}{2}\right)}}$$
将被积函数用余割表示:
$$\frac{{\color{red}{\int{\frac{1}{\sin^{2}{\left(u \right)}} d u}}}}{2} = \frac{{\color{red}{\int{\csc^{2}{\left(u \right)} d u}}}}{2}$$
$$$\csc^{2}{\left(u \right)}$$$ 的积分为 $$$\int{\csc^{2}{\left(u \right)} d u} = - \cot{\left(u \right)}$$$:
$$\frac{{\color{red}{\int{\csc^{2}{\left(u \right)} d u}}}}{2} = \frac{{\color{red}{\left(- \cot{\left(u \right)}\right)}}}{2}$$
回忆一下 $$$u=2 x$$$:
$$- \frac{\cot{\left({\color{red}{u}} \right)}}{2} = - \frac{\cot{\left({\color{red}{\left(2 x\right)}} \right)}}{2}$$
因此,
$$\int{\frac{1}{\sin^{2}{\left(2 x \right)}} d x} = - \frac{\cot{\left(2 x \right)}}{2}$$
加上积分常数:
$$\int{\frac{1}{\sin^{2}{\left(2 x \right)}} d x} = - \frac{\cot{\left(2 x \right)}}{2}+C$$
答案
$$$\int \frac{1}{\sin^{2}{\left(2 x \right)}}\, dx = - \frac{\cot{\left(2 x \right)}}{2} + C$$$A