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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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    $\begingroup$ 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 $\endgroup$ – imranfat Sep 27 '13 at 15:54
  • $\begingroup$ You r confusing me can u explain me more please? $\endgroup$ – Morgan Stone Sep 27 '13 at 16:03
  • $\begingroup$ Follow the technique given in the comment by imranfat. $\endgroup$ – Mhenni Benghorbal Sep 27 '13 at 16:03
  • $\begingroup$ I don't understand what he meant..... $\endgroup$ – Morgan Stone Sep 27 '13 at 16:07
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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?

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  • $\begingroup$ Ok I think I got it, thank you very much! $\endgroup$ – Morgan Stone Sep 27 '13 at 16:16
  • $\begingroup$ @MorganStone: you are very welcome. $\endgroup$ – Mhenni Benghorbal Sep 27 '13 at 18:22

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