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

Solution :
$$\eqalign{ & \int_{\sqrt 2 }^x {\frac{{dt}}{{t\sqrt {{t^2} - 1} }} = \frac{\pi }{2}} \cr & \therefore \left[ {{{\sec }^{ - 1}}t} \right]_{\sqrt 2 }^x = \frac{\pi }{2}\,\,\,\,\,\,\left[ {\because \int {\frac{{dx}}{{x\sqrt {{x^2} - 1} }} = {{\sec }^{ - 1}}x} } \right] \cr & \Rightarrow {\sec ^{ - 1}}x - {\sec ^{ - 1}}\sqrt 2 = \frac{\pi }{2} \cr & \Rightarrow {\sec ^{ - 1}}x - \frac{\pi }{4} = \frac{\pi }{2} \cr & \Rightarrow {\sec ^{ - 1}}x = \frac{\pi }{2} + \frac{\pi }{4} \cr & \Rightarrow {\sec ^{ - 1}}x = \frac{{3\pi }}{4} \cr & \Rightarrow x = \sec \frac{{3\pi }}{4} \cr & \Rightarrow x = - \sqrt 2 \cr} $$