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Mathematics 14 Online
OpenStudy (wasiqss):

integrate lnsin2x

OpenStudy (anonymous):

do you know the integral for ln x ?

OpenStudy (wasiqss):

yes

OpenStudy (wasiqss):

xlnx-x

OpenStudy (anonymous):

how did you get that value? using u substitution.

OpenStudy (anonymous):

and then use integration by parts

OpenStudy (wasiqss):

man seems like u need my help lol

OpenStudy (anonymous):

u = cos2x

OpenStudy (anonymous):

Oop, continue Romeo! I'm sorry!

OpenStudy (anonymous):

forget it you take over....

OpenStudy (anonymous):

u = sin2x --> u' = 2cos2x (ln sin2x)' = 2cos2x/ sin2x = 2 cot2x

OpenStudy (wasiqss):

thnks dude

OpenStudy (anonymous):

:)

OpenStudy (wasiqss):

zarkon check the link i posted

OpenStudy (zarkon):

from Mathematica V8 \[\int \text{Log}[\text{Sin}[2*x]] \, dx=-x \text{Log}\left[1-e^{4 i x}\right]+x \text{Log}[\text{Sin}[2 x]]+\frac{1}{4} i \left(4 x^2+\text{PolyLog}\left[2,e^{4 i x}\right]\right)\]

OpenStudy (wasiqss):

thanks zarko

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