Anti-derivative of a trig equation

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Homework Statement


I just need to find the anti-derivative of this equation:
\int x(sin^2(3x)cos(3x))dx



Homework Equations


sin^2(x)+cos^2(x)=1


The Attempt at a Solution


I'm not really sure where to start here. I tried to do a substitution first but couldn't make it work. So then I tried to put sin^2 in terms of cos^2 but that didn't seem to help me out any more. Anyone have an idea of how to tackle this problem? Thanks in advance!
 
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togame said:

Homework Statement


I just need to find the anti-derivative of this equation:
\int x(sin^2(3x)cos(3x))dx



Homework Equations


sin^2(x)+cos^2(x)=1


The Attempt at a Solution


I'm not really sure where to start here. I tried to do a substitution first but couldn't make it work. So then I tried to put sin^2 in terms of cos^2 but that didn't seem to help me out any more. Anyone have an idea of how to tackle this problem? Thanks in advance!

Integrate by parts, with u = x, dv = \sin^2(3x)\cos(3x)dx.

To integrate \sin^2(3x)\cos(3x)dx, use the substitution y = \sin(3x)
 
Curious3141 said:
Integrate by parts, with u = x, dv = \sin^2(3x)\cos(3x)dx.

To integrate \sin^2(3x)\cos(3x)dx, use the substitution y = \sin(3x)

Thank you very much. I didn't see the ability to do another substitution right off hand.
 
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