Having trouble finding the Integral

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BlackMamba
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Hello,

I am having trouble getting started with this integral.

\int sin2x ln(cos2x) dx

My first thought goes to using integration by parts to solve this but I am still getting stuck. I also know that sin2x = 2sinxcosx but again I'm not seeing how that would help me so I'm back to integration by parts.

My issue is if I set u = ln(cos2x) then I'm having trouble figuring out what du is. I know it's just finding the derivative but I can't for the life of me remember how to do that. Then if I set dv = sin2x dx then v = - \frac{1}{2} cos2xAny help or suggestions would be greatly appreciated.
 
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Set u=cos(2x).
 
Oh my gosh...I was making that WAY harder then it needed to be. Thank you very much arildno.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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