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$$\int_0^\pi x \ln(\sin (x))dx $$

I tried integrating this by parts but I end up getting integral that doesn't converge, which is this $$ \int_0^\pi \dfrac{x^2\cos (x)}{\sin(x)} \ dx$$ So can anyone help me on this one?

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By making the change of variable $$ u=\pi -x $$ you get that $$ I=\int_0^\pi x \ln(\sin x)\:dx=\int_0^\pi (\pi-u) \ln(\sin u)\:du=\pi\int_0^\pi \ln(\sin u)\:du-I $$ giving $$ I=\frac{\pi}2\int_0^\pi \ln(\sin u)\:du=\pi\int_0^{\pi/2} \ln(\sin u)\:du. $$ Then conclude with the classic evaluation of the latter integral: see many answers here.

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  • $\begingroup$ Just for my learning, how does the last equality hold? :) $\endgroup$ – H. R. Dec 6 '15 at 16:02
  • $\begingroup$ properties of Definite integral , Odd and even integrals $\endgroup$ – CuriousSciDude Dec 6 '15 at 16:04
  • $\begingroup$ The pre-factor of the last integral should be $\pi$ instead of $\pi/4$, no? $\endgroup$ – GDumphart Dec 6 '15 at 16:08
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    $\begingroup$ @H.R. Sorry, I did not notice your comment. "How does the last equality hold?", please have a look at this:math.stackexchange.com/questions/690644/… $\endgroup$ – Olivier Oloa Dec 6 '15 at 17:37
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    $\begingroup$ @OlivierOloa: Thanks. :) $\endgroup$ – H. R. Dec 6 '15 at 17:38

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