The solution for $x$ of the equation ${\int }_{\sqrt{2}}^x \frac{1}{t\sqrt{t^2-1}}dt=\frac{\pi }{2}$, is 

Solution:

We have, 

$\begin{array}{l}{\int }_{\sqrt{2}}^r \frac{1}{t\sqrt{t^2-1}}dt=\frac{\pi }{2}\\ \Rightarrow {\left[{\sec }^{-1}\right]}_{\sqrt{2}}^x=\frac{\pi }{2}\\ \Rightarrow {\sec }^{-1}x-{\sec }^{-1}\sqrt{2}=\frac{\pi }{2}\\ \Rightarrow {\sec }^{-1}x-\frac{\pi }{4}=\frac{\pi }{2}\\ \Rightarrow {\sec }^{-1}x=\left(\frac{\pi }{2}+\frac{\pi }{4}\right)\Rightarrow x=\sec \left(\frac{\pi }{2}+\frac{\pi }{4}\right)=-\mathrm{cosec}\frac{\pi }{4}=-\sqrt{2}\end{array}$