JEE Main 20203 Sep, Evening shiftMathematicsMediumIntegral Calculus369× since 2002If the value of the integral ∫012x2(1−x2)32dx\int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{\left( {1 - {x^2}} \right)}^{{3 \over 2}}}}}} dx0∫21(1−x2)23x2dx is k6{k \over 6}6k, then k is equal to :AOption A. 23+π2\sqrt 3 + \pi23+πBOption B. 32−π3\sqrt 2 - \pi32−πCOption C. 32+π3\sqrt 2 + \pi32+πDOption D. 23−π2\sqrt 3 - \pi23−πCheckSkip · show solution
