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

Answer 1

Cameron Alexander

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

#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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Answer 2

Charles Arocha

To integrate sin^3(x) cos^2(x) dx, you can use trigonometric identities to simplify the integral. Specifically, you can use the identity sin^2(x) = 1 - cos^2(x) to rewrite sin^3(x) as sin(x) * (1 - cos^2(x)), which can then be expanded using the distributive property. After expanding, you will have a combination of sine and cosine terms that can be integrated separately.

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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.