令 $x=\sec \theta$，取 $0\le \theta <\frac{\pi }{2}$。则 $\mathrm{d}x=\sec \theta \tan \theta \mathrm{d}\theta$，且 $\sqrt{x^2-1}=\tan \theta$。

因此原广义积分为

$$\begin{array}{rl}{\int }_0^{\frac{\pi }{2}}\frac{\sec \theta \tan \theta }{\sec \theta \tan \theta }\mathrm{d}\theta & ={\int }_0^{\frac{\pi }{2}}\mathrm{d}\theta \\ & =\frac{\pi }{2}.\end{array}$$