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