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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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$\int_{\sqrt{2}}^x \frac{d t}{t \sqrt{t^2-1}}=\frac{\pi}{2}$ is
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None of these
$\int_{\sqrt{2}}^x \frac{d t}{t \sqrt{t^2-1}}=\frac{\pi}{2}$
$\left[\sec ^{-1} t\right]_{\sqrt{2}}^x=\frac{\pi}{2}$
$\sec ^{-1} x-\frac{\pi}{4}=\frac{\pi}{2}$
$\sec ^{-1} x=\frac{3 \pi}{4}$
$x=-\sqrt{2}$.
$\left[\sec ^{-1} t\right]_{\sqrt{2}}^x=\frac{\pi}{2}$
$\sec ^{-1} x-\frac{\pi}{4}=\frac{\pi}{2}$
$\sec ^{-1} x=\frac{3 \pi}{4}$
$x=-\sqrt{2}$.
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