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How do you find the antiderivative of #sin^2(x)cosx#?

1 Answer

Jul 10, 2016

#int sin^2(x) cos(x) dx#, let #u = sin(x) -> du = cos(x) dx#

Thus

#int u^2 du = 1/3 u^3 + C = (sin^3(x))/3 + C#

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