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How do you find the antiderivative of ${sin}^{2} (x) cos x$ $sin^2(x)cosx$ ?

1 Answer

Jul 10, 2016

$\int {sin}^{2} (x) cos (x) d x$ $int sin^2(x) cos(x) dx$ , let $u = sin (x) \to d u = cos (x) d x$ $u = sin(x) -> du = cos(x) dx$

Thus

$\int u^{2} d u = \frac{1}{3} u^{3} + C = \frac{{sin}^{3} (x)}{3} + C$ $int u^2 du = 1/3 u^3 + C = (sin^3(x))/3 + C$

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See all questions in Integrals of Trigonometric Functions

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