How do you find the antiderivative of #int cos^3xsin^2xdx#?

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Answer 1 · 2017-01-18T03:57:26

#1/3sin^3x - 1/5sin^5x + C#

Factor:

#intcosx(cos^2x)sin^2xdx#

Rewrite using the pythagorean identity #sin^2theta + cos^2theta = 1#:

#intcosx(1 - sin^2x)sin^2xdx#

Expand:

#intcosx(sin^2x - sin^4x)dx#

Let #u = sinx#. Then #du = cosxdx = dx = (du)/cosx#.

#intcosx(u^2 - u^4) * (du)/cosx#

#intu^2 - u^4du#

This can be integrated as #intx^ndx = x^(n + 1)/(n +1) + C#, where #n != -1#

#1/3u^3 - 1/5u^5 + C#

Reverse the substitution:

#1/3sin^3x - 1/5sin^5x + C#

Hopefully this helps!