How do you integrate #int sin^5(2x)cos2xdx#?

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Answer 1 · 2016-12-08T07:52:47

#u#-substitution!!

I'm so happy I just learned about these.

Just looking at this beautifully set up equation, we can see that we want to substitute #sin(2x)# as #u# so as to have a #cos(2x)dx# in the equation.

Thus:

Let #sin(2x)=u#

#du=2cos(2x)dx#
#1/2du=cos(2x)dx#

So:

#intsin^5(2x)cos(2x)dx#
#=1/2intu^5du#

#=1/2*u^6/6+c=u^6/12+c#

Plug #u# back in:

#=sin^6(2x)/12+c#