Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm having trouble evaluating this integral. I tried $u$-substitution and integration by parts but they didn't work. Any ideas guys?

Evaluate $$\int \frac{\ln(\sin x)}{\sin^2 x} dx$$

Thanks you for the help!

share|cite|improve this question
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 – imranfat Sep 27 '13 at 15:54
You r confusing me can u explain me more please? – Morgan Stone Sep 27 '13 at 16:03
Follow the technique given in the comment by imranfat. – Mhenni Benghorbal Sep 27 '13 at 16:03
I don't understand what he meant..... – Morgan Stone Sep 27 '13 at 16:07
up vote 3 down vote accepted

Here is a start. using integration by parts,

$$ \int u dv = u v - \int v du .$$


$$ u=\ln(\sin(x)) \implies u'=\frac{\cos(x)}{\sin(x)}=\cot(x),\quad v=\int \frac{dx}{\sin^2 x}=-\cot(x). $$

Can you finish it now?

share|cite|improve this answer
Ok I think I got it, thank you very much! – Morgan Stone Sep 27 '13 at 16:16
@MorganStone: you are very welcome. – Mhenni Benghorbal Sep 27 '13 at 18:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.