# Integrate $\int \frac{\ln(\sin x)}{\sin^2 x}\,\mathrm dx.$

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.

• 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

## 1 Answer

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