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

Answer 1

Ezekiel Alexander

#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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Answer 2

Charles Arocha

To find the antiderivative of ( \sin^2(x) \cos(x) ), you can use integration by parts. Let ( u = \sin^2(x) ) and ( dv = \cos(x) , dx ). Then, differentiate ( u ) to find ( du ), and integrate ( dv ) to find ( v ). After that, apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Substitute the values of ( u ), ( v ), ( du ), and ( dv ) into the formula and solve for the integral.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.