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

$-\sqrt{2}$

We have,

$\int\limits_{\sqrt{2}}^{x}\frac{1}{t\sqrt{t^2-1}}dt=\frac{π}{2}$

$⇒\left[\sec^{-1}\right]_{\sqrt{2}}^{x}=\frac{π}{2}$

$⇒\sec^{-1}x-\sec^{-1}\sqrt{2}=\frac{π}{2}$

$⇒\sec^{-1}x-\frac{π}{4}=\frac{π}{2}$

$⇒\sec^{-1}x=(\frac{π}{2}+\frac{π}{4})⇒x=\sec(\frac{π}{2}+\frac{π}{4})=-cosec\frac{π}{4}=-\sqrt{2}$