$$$\csc{\left(x \right)}$$$ 的積分

求$$$\int \csc{\left(x \right)}\, dx$$$。

將餘割改寫為 $$$\csc\left(x\right)=\frac{1}{\sin\left(x\right)}$$$:

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

使用倍角公式 $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$ 重寫正弦:

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

將分子與分母同時乘以 $$$\sec^2\left(\frac{x}{2} \right)$$$:

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

令 $$$u=\tan{\left(\frac{x}{2} \right)}$$$。

則 $$$du=\left(\tan{\left(\frac{x}{2} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2} dx$$$ (步驟見»)，並可得 $$$\sec^{2}{\left(\frac{x}{2} \right)} dx = 2 du$$$。

該積分變為

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

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

$${\color{red}{\int{\frac{1}{u} d u}}} = {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

回顧一下 $$$u=\tan{\left(\frac{x}{2} \right)}$$$：

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} \right)}}}}\right| \right)}$$

因此，

$$\int{\csc{\left(x \right)} d x} = \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)}$$

加上積分常數：

$$\int{\csc{\left(x \right)} d x} = \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)}+C$$

$$$\int \csc{\left(x \right)}\, dx = \ln\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right|\right) + C$$$A