$$$\csc^{2}{\left(x \right)} + 1$$$ 的积分

求$$$\int \left(\csc^{2}{\left(x \right)} + 1\right)\, dx$$$。

逐项积分：

$${\color{red}{\int{\left(\csc^{2}{\left(x \right)} + 1\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{\csc^{2}{\left(x \right)} d x}\right)}}$$

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=1$$$：

$$\int{\csc^{2}{\left(x \right)} d x} + {\color{red}{\int{1 d x}}} = \int{\csc^{2}{\left(x \right)} d x} + {\color{red}{x}}$$

$$$\csc^{2}{\left(x \right)}$$$ 的积分为 $$$\int{\csc^{2}{\left(x \right)} d x} = - \cot{\left(x \right)}$$$:

$$x + {\color{red}{\int{\csc^{2}{\left(x \right)} d x}}} = x + {\color{red}{\left(- \cot{\left(x \right)}\right)}}$$

因此，

$$\int{\left(\csc^{2}{\left(x \right)} + 1\right)d x} = x - \cot{\left(x \right)}$$

加上积分常数：

$$\int{\left(\csc^{2}{\left(x \right)} + 1\right)d x} = x - \cot{\left(x \right)}+C$$

$$$\int \left(\csc^{2}{\left(x \right)} + 1\right)\, dx = \left(x - \cot{\left(x \right)}\right) + C$$$A