I'm having trouble evaluating the integral
$$\int \frac{\ln(\sin x)}{\sin^2 x}\,\mathrm dx.$$
I tried $u$-substitution and integration by parts but they didn't work.
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2$\begingroup$ integration by parts works, pick f = ln term, so that f' can be found. Then g' is csc²x of which g is a known anti derivative $\endgroup$– imranfatSep 27, 2013 at 15:54
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$\begingroup$ You r confusing me can u explain me more please? $\endgroup$– Morgan StoneSep 27, 2013 at 16:03
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$\begingroup$ Follow the technique given in the comment by imranfat. $\endgroup$– Mhenni BenghorbalSep 27, 2013 at 16:03
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$\begingroup$ I don't understand what he meant..... $\endgroup$– Morgan StoneSep 27, 2013 at 16:07
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1 Answer
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Here is a start. using integration by parts,
$$ \int u dv = u v - \int v du .$$
Let
$$ u=\ln(\sin(x)) \implies u'=\frac{\cos(x)}{\sin(x)}=\cot(x),\quad v=\int \frac{dx}{\sin^2 x}=-\cot(x). $$
Can you finish it now?
answered Sep 27, 2013 at 16:07
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$\begingroup$ Ok I think I got it, thank you very much! $\endgroup$ Sep 27, 2013 at 16:16
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$\begingroup$ @MorganStone: you are very welcome. $\endgroup$ Sep 27, 2013 at 18:22