BlackMamba
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Hello,
I am having trouble getting started with this integral.
[itex]\int sin2x ln(cos2x) dx[/itex]
My first thought goes to using integration by parts to solve this but I am still getting stuck. I also know that [itex]sin2x = 2sinxcosx[/itex] 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 [itex]u = ln(cos2x)[/itex] then I'm having trouble figuring out what [itex]du[/itex] 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 [itex]dv = sin2x dx[/itex] then [itex]v = - \frac{1}{2} cos2x[/itex]Any help or suggestions would be greatly appreciated.
I am having trouble getting started with this integral.
[itex]\int sin2x ln(cos2x) dx[/itex]
My first thought goes to using integration by parts to solve this but I am still getting stuck. I also know that [itex]sin2x = 2sinxcosx[/itex] 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 [itex]u = ln(cos2x)[/itex] then I'm having trouble figuring out what [itex]du[/itex] 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 [itex]dv = sin2x dx[/itex] then [itex]v = - \frac{1}{2} cos2x[/itex]Any help or suggestions would be greatly appreciated.