Hello, DonnieDarko!
$\displaystyle \int^{\frac{2\pi}{3}}_{\frac{\pi}{6}} \cos x\ln(\sin x)\,dx$
$\displaystyle \text{So, I use substitution: }\:t \,=\, \sin x,\;dt\,=\,\cos x\,dx$
$\displaystyle \text{When I put it in, i got: }\:\int \ln t\,dt$
$\displaystyle \text{Now I know I have to use partial integration, but how do I set limits?}$
Plug the $\displaystyle x$-limits into your substitution.
You have: .$\displaystyle t \:=\:\sin x$
$\displaystyle \text{When }x = \tfrac{\pi}{6}\!:\;\;t \:=\:\sin\tfrac{\pi}{6} \:=\:\tfrac{1}{2}$
$\displaystyle \text{When }x = \tfrac{2\pi}{3}\!:\;\;t \:=\:\sin\tfrac{2\pi}{3} \:=\:\tfrac{\sqrt{3}}{2}$
$\displaystyle \displaystyle\text{The integral becomes: }\int^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}}\!\! \ln t\,dt$
I assume that by "partial integration", you mean what I would call "integration by parts":
$\displaystyle \int_a^b udv= \left[uv\right]_a^b- \int_a^b vdu$
Here, once you have made the substitution, t= sin(x), and have [itex]\int ln(t)dt[/itex], let u= ln(t), dv= dt. Then du= (1/t)dt and v= t so we have
$\displaystyle \left[t ln(t)\right]_{1/2}^{\sqrt{3}/2}- \int_{1/2}^{\sqrt{3}/2} dt= \left[tln(t)- t\right]_{1/2}^{\sqrt{3}/2}$
That's how Plato got his answer.