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$$$\cot{\left(\pi x \right)}$$$ 的积分
您的输入
求$$$\int \cot{\left(\pi x \right)}\, dx$$$。
解答
设$$$u=\pi x$$$。
则$$$du=\left(\pi x\right)^{\prime }dx = \pi dx$$$ (步骤见»)，并有$$$dx = \frac{du}{\pi}$$$。
因此，
$${\color{red}{\int{\cot{\left(\pi x \right)} d x}}} = {\color{red}{\int{\frac{\cot{\left(u \right)}}{\pi} d u}}}$$
对 $$$c=\frac{1}{\pi}$$$ 和 $$$f{\left(u \right)} = \cot{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：
$${\color{red}{\int{\frac{\cot{\left(u \right)}}{\pi} d u}}} = {\color{red}{\frac{\int{\cot{\left(u \right)} d u}}{\pi}}}$$
将余切改写为 $$$\cot\left( u \right)=\frac{\cos\left( u \right)}{\sin\left( u \right)}$$$:
$$\frac{{\color{red}{\int{\cot{\left(u \right)} d u}}}}{\pi} = \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{\sin{\left(u \right)}} d u}}}}{\pi}$$
设$$$v=\sin{\left(u \right)}$$$。
则$$$dv=\left(\sin{\left(u \right)}\right)^{\prime }du = \cos{\left(u \right)} du$$$ (步骤见»)，并有$$$\cos{\left(u \right)} du = dv$$$。
因此，
$$\frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{\sin{\left(u \right)}} d u}}}}{\pi} = \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{\pi}$$
$$$\frac{1}{v}$$$ 的积分为 $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:
$$\frac{{\color{red}{\int{\frac{1}{v} d v}}}}{\pi} = \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{\pi}$$
回忆一下 $$$v=\sin{\left(u \right)}$$$:
$$\frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{\pi} = \frac{\ln{\left(\left|{{\color{red}{\sin{\left(u \right)}}}}\right| \right)}}{\pi}$$
回忆一下 $$$u=\pi x$$$:
$$\frac{\ln{\left(\left|{\sin{\left({\color{red}{u}} \right)}}\right| \right)}}{\pi} = \frac{\ln{\left(\left|{\sin{\left({\color{red}{\pi x}} \right)}}\right| \right)}}{\pi}$$
因此，
$$\int{\cot{\left(\pi x \right)} d x} = \frac{\ln{\left(\left|{\sin{\left(\pi x \right)}}\right| \right)}}{\pi}$$
加上积分常数：
$$\int{\cot{\left(\pi x \right)} d x} = \frac{\ln{\left(\left|{\sin{\left(\pi x \right)}}\right| \right)}}{\pi}+C$$
答案
$$$\int \cot{\left(\pi x \right)}\, dx = \frac{\ln\left(\left|{\sin{\left(\pi x \right)}}\right|\right)}{\pi} + C$$$A