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

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Answer 1 · 2016-09-09T19:50:36

#int sin(2x)cos^2(2x)dx#

#= -1/2int-2sin(2x)cos^2(2x)dx#

#u = cos(2x)#
#du = -2sin(2x)dx#

Now we have

#-1/2intu^2du#

and as you know it's

#-1/6[u^3] + C#

when you substitute back you have

#-1/6[cos^3(2x)] + C#