Question

double integral 0 to 1 integral 0 to 1 dxdy /[√ (1 - x ^2) √( 1 - x ^2)]​

Answer 1

Given to evaluate,

$\displaystyle\longrightarrow I=\int\limits_{x=0}^1\int\limits_{y=0}^1\dfrac{dx\ dy}{\sqrt{(1-x^2)(1-y^2)}}$

$\displaystyle\longrightarrow I=\int\limits_{x=0}^1\int\limits_{y=0}^1\dfrac{dx}{\sqrt{1-x^2}}\cdot\dfrac{dy}{\sqrt{1-y^2}}$

It's equivalent to,

$\displaystyle\longrightarrow I=\int\limits_0^1\dfrac{dx}{\sqrt{1-x^2}}\cdot\int\limits_0^1\dfrac{dy}{\sqrt{1-y^2}}$

$\displaystyle\longrightarrow I=\Big[\sin^{-1}x\Big]_0^1\Big[\sin^{-1}y\Big]_0^1$

$\displaystyle\longrightarrow I=\Big(\sin^{-1}1-\sin^{-1}0\Big)\Big(\sin^{-1}1-\sin^{-1}0\Big)$

$\displaystyle\longrightarrow I=\left(\dfrac{\pi}{2}-0\right)\left(\dfrac{\pi}{2}-0\right)$

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{\pi^2}{4}}}$