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

Option 1)
$\frac{\sqrt{3}}{2}\;$

Option 2)
$\; 2\sqrt{2}\;$

Option 3)
$\; 2\;$

Option 4)
$\; \pi$

Answers (1)

best_answer

As learnt in cocept

Integration of Rational and irrational function -

Integration in the form of :

(i) $\int \frac{dx}{x^{2}+a^{2}}$ (ii) $\int \frac{dx}{x^{2}-a^{2}}$ (iii) $\int \frac{dx}{a^{2}-x^{2}}$

(iv) $\int \frac{dx}{\sqrt{x^{2}+a^{2}}}$ (v) $\frac{dx}{\sqrt{x^{2}-a^{2}}}$ (vi) $\frac{dx}{\sqrt{a^{2}-x^{2}}}$

-

$A=\int_{\sqrt{2}}^{x}\frac{dt}{t\sqrt{t^2 -1 }} =[sec^{-1}t]^x_{\sqrt{2}}$

As per the formula of integration

Given $[sec^{-1}t]^x_{\sqrt{2}} = \frac{\pi }{2}$

$sec^{-1}x-{sec^{-1}\sqrt{2}} = \frac{\pi }{2}$

$sec^{-1}x = \frac{\pi }{2}+ \frac{\pi }{4} = \frac{3\pi}{4}$

$x = sec\frac{3\pi }{4} =\:-\sqrt{2}$

Option 1)

$\frac{\sqrt{3}}{2}\;$

Option 2)

$\; 2\sqrt{2}\;$

Option 3)

$\; 2\;$

Option 4)

$\; \pi$

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prateek

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