If the value of the integral ∫ 0 1 2 x 2 1 - x 2 3 2 d x is k 6 , then k is equal to:

If the value of the integral ∫012x21-x232dx is k6, then k is equal to:
  1. 23+π
  2. 23-π
  3. 32+π
  4. 32-π

Solution

k6=∫012x21-x232dx

Let x=sinθ⇒dx=cosθdθ

⇒k6=∫0π6sin2 θ1-sin2 θ32cosθdθ⇒k6=∫0π6sin2 θcos3 θcosθdθ

⇒k6=∫0π6tan2θdθ=∫0π6sec2θ-1dθ⇒k6=tanθ-θ0π6=13-π6=23-π6

⇒k=23-π

Asked in: JEE Main 2020 (03 Sep Shift 2)

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