1. We need to solve $\int_{\sqrt 2}^{x} \frac{dt}{t\sqrt{t^2-1}}=\frac{\pi}{2}.$

2. Recall the standard integral: $\int \frac{dt}{t\sqrt{t^2-1}}=\sec^{-1} t + C$ for $t>1$.

Since the lower limit is $\sqrt{2}>1$, this formula is valid on the interval of interest.

3. Apply the Fundamental Theorem of Calculus: $\int_{\sqrt 2}^{x} \frac{dt}{t\sqrt{t^2-1}} = \sec^{-1} x - \sec^{-1}(\sqrt 2).$

So the equation becomes $\sec^{-1} x - \sec^{-1}(\sqrt 2)=\frac{\pi}{2}.$

4. Now evaluate $\sec^{-1}(\sqrt 2)=\frac{\pi}{4}$ because $\sec\left(\frac{\pi}{4}\right)=\sqrt 2.$

Hence, $\sec^{-1} x - \frac{\pi}{4}=\frac{\pi}{2}.$ So, $\sec^{-1} x = \frac{3\pi}{4}.$

5. But for the principal value of $\sec^{-1}x$, the range is usually taken as $[0,\pi]\setminus \left\{\frac{\pi}{2}\right\}.$ Then $x=\sec\left(\frac{3\pi}{4}\right)=\frac{1}{\cos(3\pi/4)}=\frac{1}{-\frac{\sqrt2}{2}}=-\sqrt2.$

However, this is impossible here because the integrand $\frac{1}{t\sqrt{t^2-1}}$ is real only for $|t|\ge 1$, and also the antiderivative used from $\sqrt2$ to $x$ along the real interval requires staying in the real domain. Moving from $\sqrt2$ to $-\sqrt2$ would cross the interval $(-1,1)$ where $\sqrt{t^2-1}$ is not real, so this cannot be a valid real definite integral.

6. A better way is to use the trigonometric substitution directly. Let $t=\sec\theta, \qquad dt=\sec\theta\tan\theta\,d\theta, \qquad \sqrt{t^2-1}=\tan\theta.$ Then $\frac{dt}{t\sqrt{t^2-1}} = \frac{\sec\theta\tan\theta\,d\theta}{\sec\theta\tan\theta}=d\theta.$

Thus, $\int_{\sqrt2}^{x} \frac{dt}{t\sqrt{t^2-1}} = \theta_x-\theta_{\sqrt2}.$

For $t=\sqrt2$, $\sec\theta_{\sqrt2}=\sqrt2 \implies \theta_{\sqrt2}=\frac{\pi}{4}.$

So,