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How do you find the integral of #sin^3(x) cos^2(x) dx#?

1 Answer

Oct 12, 2015

#I = 1/5cos^5x-1/3cos^3x+C#

Explanation:

#I = int sin^3xcos^2xdx = int sin^2xcos^2xsinxdx#
#I = int (1-cos^2x)cos^2xsinxdx#
#cosx=t => -sinxdx=dt => sinxdx=-dt#

#I = int (1-t^2)t^2(-dt) = int (t^4-t^2)dt = t^5/5-t^3/3+C#

#I = 1/5cos^5x-1/3cos^3x+C#

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