$$$\operatorname{acsc}{\left(x \right)}$$$ 的积分

求$$$\int \operatorname{acsc}{\left(x \right)}\, dx$$$。

对于积分$$$\int{\operatorname{acsc}{\left(x \right)} d x}$$$，使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

设 $$$\operatorname{u}=\operatorname{acsc}{\left(x \right)}$$$ 和 $$$\operatorname{dv}=dx$$$。

则 $$$\operatorname{du}=\left(\operatorname{acsc}{\left(x \right)}\right)^{\prime }dx=- \frac{\left|{x}\right|}{x^{2} \sqrt{x^{2} - 1}} dx$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{1 d x}=x$$$ (步骤见 »)。

因此，

$${\color{red}{\int{\operatorname{acsc}{\left(x \right)} d x}}}={\color{red}{\left(\operatorname{acsc}{\left(x \right)} \cdot x-\int{x \cdot \left(- \frac{\left|{x}\right|}{x^{2} \sqrt{x^{2} - 1}}\right) d x}\right)}}={\color{red}{\left(x \operatorname{acsc}{\left(x \right)} - \int{\left(- \frac{\left|{x}\right|}{x \sqrt{x^{2} - 1}}\right)d x}\right)}}$$

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

$$x \operatorname{acsc}{\left(x \right)} - {\color{red}{\int{\left(- \frac{\left|{x}\right|}{x \sqrt{x^{2} - 1}}\right)d x}}} = x \operatorname{acsc}{\left(x \right)} - {\color{red}{\left(- \int{\frac{1}{\sqrt{x^{2} - 1}} d x}\right)}}$$

设$$$x=\cosh{\left(u \right)}$$$。

则$$$dx=\left(\cosh{\left(u \right)}\right)^{\prime }du = \sinh{\left(u \right)} du$$$（步骤见»）。

此外，可得$$$u=\operatorname{acosh}{\left(x \right)}$$$。

被积函数变为

$$$\frac{1}{\sqrt{x^{2} - 1}} = \frac{1}{\sqrt{\cosh^{2}{\left( u \right)} - 1}}$$$

利用恒等式 $$$\cosh^{2}{\left( u \right)} - 1 = \sinh^{2}{\left( u \right)}$$$：

$$$\frac{1}{\sqrt{\cosh^{2}{\left( u \right)} - 1}}=\frac{1}{\sqrt{\sinh^{2}{\left( u \right)}}}$$$

假设$$$\sinh{\left( u \right)} \ge 0$$$，我们得到如下结果：

$$$\frac{1}{\sqrt{\sinh^{2}{\left( u \right)}}} = \frac{1}{\sinh{\left( u \right)}}$$$

积分变为

$$x \operatorname{acsc}{\left(x \right)} + {\color{red}{\int{\frac{1}{\sqrt{x^{2} - 1}} d x}}} = x \operatorname{acsc}{\left(x \right)} + {\color{red}{\int{1 d u}}}$$

应用常数法则 $$$\int c\, du = c u$$$，使用 $$$c=1$$$：

$$x \operatorname{acsc}{\left(x \right)} + {\color{red}{\int{1 d u}}} = x \operatorname{acsc}{\left(x \right)} + {\color{red}{u}}$$

回忆一下 $$$u=\operatorname{acosh}{\left(x \right)}$$$:

$$x \operatorname{acsc}{\left(x \right)} + {\color{red}{u}} = x \operatorname{acsc}{\left(x \right)} + {\color{red}{\operatorname{acosh}{\left(x \right)}}}$$

因此，

$$\int{\operatorname{acsc}{\left(x \right)} d x} = x \operatorname{acsc}{\left(x \right)} + \operatorname{acosh}{\left(x \right)}$$

加上积分常数：

$$\int{\operatorname{acsc}{\left(x \right)} d x} = x \operatorname{acsc}{\left(x \right)} + \operatorname{acosh}{\left(x \right)}+C$$

$$$\int \operatorname{acsc}{\left(x \right)}\, dx = \left(x \operatorname{acsc}{\left(x \right)} + \operatorname{acosh}{\left(x \right)}\right) + C$$$A