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Correct option is B
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+π=13-π6=6-3π63+π, it implies k=23Similar Questions
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