$$$\frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}}$$$ 的积分

求$$$\int \frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}}\, dx$$$。

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

设$$$u=\sin{\left(2 x \right)}$$$。

则$$$du=\left(\sin{\left(2 x \right)}\right)^{\prime }dx = 2 \cos{\left(2 x \right)} dx$$$ (步骤见»)，并有$$$\cos{\left(2 x \right)} dx = \frac{du}{2}$$$。

因此，

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

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{{\color{red}{\int{\frac{1}{u} d u}}}}{4} = \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{4}$$

回忆一下 $$$u=\sin{\left(2 x \right)}$$$:

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{4} = \frac{\ln{\left(\left|{{\color{red}{\sin{\left(2 x \right)}}}}\right| \right)}}{4}$$

因此，

$$\int{\frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}} d x} = \frac{\ln{\left(\left|{\sin{\left(2 x \right)}}\right| \right)}}{4}$$

加上积分常数：

$$\int{\frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}} d x} = \frac{\ln{\left(\left|{\sin{\left(2 x \right)}}\right| \right)}}{4}+C$$

$$$\int \frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}}\, dx = \frac{\ln\left(\left|{\sin{\left(2 x \right)}}\right|\right)}{4} + C$$$A