$$$\csc{\left(x \right)}$$$ 的积分
您的输入
求$$$\int \csc{\left(x \right)}\, dx$$$。
解答
将余割改写为$$$\csc\left(x\right)=\frac{1}{\sin\left(x\right)}$$$:
$${\color{red}{\int{\csc{\left(x \right)} d x}}} = {\color{red}{\int{\frac{1}{\sin{\left(x \right)}} d x}}}$$
使用二倍角公式 $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$ 改写正弦:
$${\color{red}{\int{\frac{1}{\sin{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} \right)} \cos{\left(\frac{x}{2} \right)}} d x}}}$$
将分子和分母同时乘以 $$$\sec^2\left(\frac{x}{2} \right)$$$:
$${\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} \right)} \cos{\left(\frac{x}{2} \right)}} d x}}} = {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2 \tan{\left(\frac{x}{2} \right)}} d x}}}$$
设$$$u=\tan{\left(\frac{x}{2} \right)}$$$。
则$$$du=\left(\tan{\left(\frac{x}{2} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2} dx$$$ (步骤见»),并有$$$\sec^{2}{\left(\frac{x}{2} \right)} dx = 2 du$$$。
该积分可以改写为
$${\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2 \tan{\left(\frac{x}{2} \right)}} d x}}} = {\color{red}{\int{\frac{1}{u} d u}}}$$
$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$${\color{red}{\int{\frac{1}{u} d u}}} = {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
回忆一下 $$$u=\tan{\left(\frac{x}{2} \right)}$$$:
$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} \right)}}}}\right| \right)}$$
因此,
$$\int{\csc{\left(x \right)} d x} = \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)}$$
加上积分常数:
$$\int{\csc{\left(x \right)} d x} = \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)}+C$$
答案
$$$\int \csc{\left(x \right)}\, dx = \ln\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right|\right) + C$$$A