Expert Answers

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The integral `int sin x*cos^2x dx` has to be determined

`int sin x*cos^2x dx`

let `cos x = y` , `-dy = sin x*dx`

=> `-int y^2 dy`

=> `-y^3/3`

=> `(-cos^3 x)/3`

The integral `int sin x*cos^2x dx = (-cos^3 x)/3`

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