∫ 0 π 2 x 3 sin x d x =

Question

∫ 0 π 2 x 3 sin x d x =, 3 π 2 4 - 3 π + 6, 3 π 2 4 + 3 π - 6, 3 π 2 4 + 6, 3 π 2 4 - 6

MARKS App · Accepted Answer

J = ∫ 0 π 2 x 3 sin x d x = ( − x 3 cos x ) 0 π 2 + ∫ 0 π 2 3 x 2 cos x d x = 0 + 3 { ( x 2 sin x ) 0 π 2 − ∫ 0 π 2 2 x sin x d x } = 3 { ( π 2 4 ) − 2 { ( − 2 cos x ) 0 π 2 + ∫ 0 π 2 cos x } } = 3 ( π 2 4 − 2 × 1 ) = 3 π 2 4 − 6

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