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

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Answer 1 · 2016-09-23T20:15:22

#= - 1/8 cos^4 2x + C#

start by noticing that, from the power and chain rules:

#d/dx (cos^4 2x) = 4 cos^3 2x * ( - sin 2x ) * 2#

#= -8 cos^3 2x sin 2x #

and so your integral is actually

#int d/dx ( - 1/8 cos^4 2x) dx#

#= - 1/8 int d/dx ( cos^4 2x) dx#

#= - 1/8 cos^4 2x + C#