The solution for x of the equation int_{sqrt{ | vedclass

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(D) We are given the equation $\int_{\sqrt{2}}^{x} \frac{dt}{t\sqrt{t^2-1}} = \frac{\pi}{2}$.Using the standard integral formula $\int \frac{dt}{t\sqr

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none of these

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(D) We are given the equation $\int_{\sqrt{2}}^{x} \frac{dt}{t\sqrt{t^2-1}} = \frac{\pi}{2}$.Using the standard integral formula $\int \frac{dt}{t\sqrt{t^2-1}} = \sec^{-1} t$,we evaluate the definite integral:$\left[\sec^{-1} t\right]_{\sqrt{2}}^{x} = \frac{\pi}{2}$.Substituting the limits,we get:$\sec^{-1} x - \sec^{-1} \sqrt{2} = \frac{\pi}{2}$.Since $\sec^{-1} \sqrt{2} = \frac{\pi}{4}$,the equation becomes:$\sec^{-1} x - \frac{\pi}{4} = \frac{\pi}{2}$.Solving for $\sec^{-1} x$:$\sec^{-1} x = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.Therefore,$x = \sec\left(\frac{3\pi}{4}\right) = -\sqrt{2}$.Since $-\sqrt{2}$ is not among the given options,the correct choice is $(D)$.

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

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