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integrate lnsin2x
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do you know the integral for ln x ?
yes
xlnx-x
how did you get that value? using u substitution.
and then use integration by parts
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man seems like u need my help lol
u = cos2x
Oop, continue Romeo! I'm sorry!
forget it you take over....
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u = sin2x --> u' = 2cos2x (ln sin2x)' = 2cos2x/ sin2x = 2 cot2x
thnks dude
:)
zarkon check the link i posted
from Mathematica V8 \[\int \text{Log}[\text{Sin}[2*x]] \, dx=-x \text{Log}\left[1-e^{4 i x}\right]+x \text{Log}[\text{Sin}[2 x]]+\frac{1}{4} i \left(4 x^2+\text{PolyLog}\left[2,e^{4 i x}\right]\right)\]
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thanks zarko
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