If the value of the integral ∫01/2x21-x23/2 dx is k6 , then k is equal to:

Consider the integral I=∫012x21-x232dx Substitute x=sinθ,dx=cosθdθ and the limits are 0,π6 Hence, the given integral becomes I=∫0π6sin2θcos3θcosθdθ=∫0π6tan2θ  dθ=∫0π6sec2θ−1  dθ=tanθ−θ0π6=13−π6 Therefore, k6=13-π6=6-3π63, it implies k=23-π