$$$\frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}$$$ 的积分
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您的输入
求$$$\int \frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}\, dx$$$。
解答
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,
$${\color{red}{\int{\frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} d x}}} = {\color{red}{\int{\frac{1}{2 u} d u}}}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$${\color{red}{\int{\frac{1}{2 u} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{2}\right)}}$$
$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$\frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$
回忆一下 $$$u=\sin{\left(2 x \right)}$$$:
$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = \frac{\ln{\left(\left|{{\color{red}{\sin{\left(2 x \right)}}}}\right| \right)}}{2}$$
因此,
$$\int{\frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} d x} = \frac{\ln{\left(\left|{\sin{\left(2 x \right)}}\right| \right)}}{2}$$
加上积分常数:
$$\int{\frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} d x} = \frac{\ln{\left(\left|{\sin{\left(2 x \right)}}\right| \right)}}{2}+C$$
答案
$$$\int \frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}\, dx = \frac{\ln\left(\left|{\sin{\left(2 x \right)}}\right|\right)}{2} + C$$$A