The solution for x of the equation ∫2x 1tt2−1dt=π2, is

A
$\frac{\sqrt{3}}{2}$
B
$\sqrt{2}$
C
$2$
D
$- \sqrt{2}$

Solution:

We have,

$\begin{array}{l} \int_{\sqrt{2}}^{r} \frac{1}{t \sqrt{t^{2} - 1}} d t = \frac{π}{2} \\ \Rightarrow {[\sec^{- 1}]}_{\sqrt{2}}^{x} = \frac{π}{2} \\ \Rightarrow \sec^{- 1} x - \sec^{- 1} \sqrt{2} = \frac{π}{2} \\ \Rightarrow \sec^{- 1} x - \frac{π}{4} = \frac{π}{2} \\ \Rightarrow \sec^{- 1} x = (\frac{π}{2} + \frac{π}{4}) \Rightarrow x = \sec (\frac{π}{2} + \frac{π}{4}) = - cosec \frac{π}{4} = - \sqrt{2} \end{array}$

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

Solution:

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