How do you find the antiderivative of $Cos(2x)Sin(x)dx$ ?

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Answer 1 · 2017-08-30T01:57:12

$intcos2xsinxdx = cosx-2/3cos^3x + "constant"$

$cos2x -= 2cos^2x-1$

$therefore intcos2xsinxdx = int(2cos^2x -1)sinxdx = 2intcos^2xsinxdx -intsinxdx$

Let $u = cosx$ and $du = -sinxdx$

Then $cosx +2intcos^2xsinxdx = cosx-2intu^2du = cosx-2/3u^3=cosx-2/3cos^3x +"constant"$

How do you find the antiderivative of Cos(2x)Sin(x)dxCos(2x)Sin(x)dx?