Having trouble finding the Integral

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SUMMARY

The integral \(\int \sin(2x) \ln(\cos(2x)) \, dx\) can be effectively approached using integration by parts. The user initially struggled with determining the derivative of \(u = \ln(\cos(2x))\) but found clarity by simplifying the problem to setting \(u = \cos(2x)\). This adjustment significantly streamlined the integration process, demonstrating the importance of correctly identifying substitution variables in integration techniques.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by parts.
  • Familiarity with trigonometric identities, particularly \(\sin(2x)\) and \(\cos(2x)\).
  • Knowledge of logarithmic differentiation and its application in calculus.
  • Basic skills in calculus, including differentiation and integration of standard functions.
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  • Practice integration by parts with various functions to build confidence.
  • Study trigonometric identities and their applications in integration.
  • Learn about logarithmic differentiation techniques for complex integrals.
  • Explore advanced integration techniques, such as substitution and partial fractions.
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Students and educators in calculus, mathematicians tackling complex integrals, and anyone seeking to improve their integration skills in trigonometric and logarithmic functions.

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.
 

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