Multivariable CalculusSuperchallengeDouble Integrals ShareDouble Integrals · Superchallenge problemEvaluate ∫01 ∫01dx dy1−x2y2\displaystyle\int_0^{1}\!\int_0^{1}\dfrac{dx\,dy}{\sqrt{1 - x^2y^2}}∫01∫011−x2y2dxdy. You may use ∫0π/2ln(sint) dt=−π2ln2\displaystyle\int_0^{\pi/2}\ln(\sin t)\,dt = -\dfrac{\pi}{2}\ln 2∫0π/2ln(sint)dt=−2πln2. Give an exact answer.Check Show answer Walk through it step by step Practice more like this Review this concept