Evaluate the following integrals: ^ 1 2 _ 0 1 (1+x^2) 1-x^2 dx - Vidyadip

Question

Evaluate the following integrals:
$\int^\limits{\frac{1}{2}}_{0}\frac{1}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}\text{ dx}$

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Answer

$\text{I}=\int^\limits{\frac{1}{2}}_{0}\frac{1}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}\text{ dx}$
Let $\text{x}=\sin\text{u}$
$\text{dx}=\cos\text{u du}$
$\text{I}=\int^\limits{\frac{\pi}{6}}_{0}\frac{1}{(1+\sin^2\text{u})}\text{ du}$
$\text{I}=\int^\limits{\frac{\pi}{6}}_{0}\frac{\sec^2\text{u}}{(1+2\tan^2\text{u})}\text{ du}$
Let $\tan\text{u}=\text{v}$
$\text{dv}=\sec^2\text{u du}$
$\text{I}=\int^\limits{\frac{1}{\sqrt{3}}}_{0}\frac{1}{(1+2\text{v}^2)}\text{ dv}$
$\text{I}=\frac{1}{\sqrt{2}}\Big[\tan^{-1}\big(\sqrt{2}\text{v}\big)\Big]^{\frac{1}{\sqrt{3}}}_0$
$\text{I}=\frac{1}{\sqrt{2}}\Big[\tan^{-1}\Big(\sqrt{\frac{2}{3}}\Big)\Big]$

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