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0-pi-2-sin-2-xcos-3-xdx-

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Question Number 21721 by Isse last updated on 02/Oct/17
∫_0 ^(π/2) sin^2 xcos^3 xdx
∫0π/2sin2xcos3xdx
Answered by sma3l2996 last updated on 02/Oct/17
=∫_0 ^(π/2) sin^2 x(1−sin^2 x)cosxdx=∫_0 ^(π/2) (cosxsin^2 x−cosxsin^4 x)dx =[(1/3)sin^3 x−(1/5)sin^5 x]_0 ^(π/2) =(1/3)−(1/5)=(2/(15))
=∫0π/2sin2x(1−sin2x)cosxdx=∫0π/2(cosxsin2x−cosxsin4x)dx=[13sin3x−15sin5x]0π/2=13−15=215

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