MathematicsMedium210×since 2002Q3378If 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 :A23+π2\sqrt 3 + \pi23+πB32−π3\sqrt 2 - \pi32−πC32+π3\sqrt 2 + \pi32+πD23−π2\sqrt 3 - \pi23−πCheck answerSkip