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

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$\int_{\sqrt{2}}^x \frac{d t}{t \sqrt{t^2-1}}=\frac{\pi}{2}$ $\left[\sec ^{-1} t\right]_{\sqrt{2}}^x=\frac{\pi}{2}$ $\sec ^{-1} x-\frac{\pi}{4}=\frac{\pi}{2}$ $\sec ^{-1} x=\frac{3 \pi}{4}$ $x=-\sqrt{2}$.

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

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