$$$\frac{\pi}{2 x^{2} \sqrt{x^{2} - 1}}$$$ 的积分

求$$$\int \frac{\pi}{2 x^{2} \sqrt{x^{2} - 1}}\, dx$$$。

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

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

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

因此，

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

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

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

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

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

因此，

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

将被积函数改写为关于双曲正割的形式:

$$\frac{\pi {\color{red}{\int{\frac{1}{\cosh^{2}{\left(u \right)}} d u}}}}{2} = \frac{\pi {\color{red}{\int{\operatorname{sech}^{2}{\left(u \right)} d u}}}}{2}$$

$$$\operatorname{sech}^{2}{\left(u \right)}$$$ 的积分为 $$$\int{\operatorname{sech}^{2}{\left(u \right)} d u} = \tanh{\left(u \right)}$$$:

$$\frac{\pi {\color{red}{\int{\operatorname{sech}^{2}{\left(u \right)} d u}}}}{2} = \frac{\pi {\color{red}{\tanh{\left(u \right)}}}}{2}$$

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

$$\frac{\pi \tanh{\left({\color{red}{u}} \right)}}{2} = \frac{\pi \tanh{\left({\color{red}{\operatorname{acosh}{\left(x \right)}}} \right)}}{2}$$

因此，

$$\int{\frac{\pi}{2 x^{2} \sqrt{x^{2} - 1}} d x} = \frac{\pi \sqrt{x - 1} \sqrt{x + 1}}{2 x}$$

加上积分常数：

$$\int{\frac{\pi}{2 x^{2} \sqrt{x^{2} - 1}} d x} = \frac{\pi \sqrt{x - 1} \sqrt{x + 1}}{2 x}+C$$

$$$\int \frac{\pi}{2 x^{2} \sqrt{x^{2} - 1}}\, dx = \frac{\pi \sqrt{x - 1} \sqrt{x + 1}}{2 x} + C$$$A