$$$\frac{\sec{\left(x \right)}}{\sin{\left(x \right)}}$$$ 的積分

求$$$\int \frac{\sec{\left(x \right)}}{\sin{\left(x \right)}}\, dx$$$。

簡化被積函數:

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

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

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

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

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

令 $$$u=\tan{\left(x \right)}$$$。

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

因此，

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

$$$\frac{1}{u}$$$ 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 $$$u=\tan{\left(x \right)}$$$：

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

因此，

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

加上積分常數：

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

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