# Evaluate integral trouble

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!

-
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

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?

-
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