Answer
$$\ln \left( 2 \right)$$
Work Step by Step
$$\eqalign{
& \int_{\pi /6}^{\pi /2} {\cot x} dx \cr
& = \int_{\pi /6}^{\pi /2} {\frac{{\cos x}}{{\sin x}}} dx \cr
& {\text{Let }}u = \sin x,{\text{ }}du = \cos xdx \cr
& {\text{The new limits of integration are:}} \cr
& x = \pi /2 \to u = \sin \left( {\pi /2} \right) = 1 \cr
& x = \pi /6 \to u = \sin \left( {\pi /6} \right) = 1/2 \cr
& {\text{Substituting}} \cr
& \int_{\pi /6}^{\pi /2} {\frac{{\cos x}}{{\sin x}}} dx = \int_{1/2}^1 {\frac{{du}}{u}} \cr
& {\text{Integrating}} \cr
& \int_{1/2}^1 {\frac{{du}}{u}} = \left[ {\ln \left| u \right|} \right]_{1/2}^1 \cr
& = \ln \left( 1 \right) - \ln \left( {\frac{1}{2}} \right) \cr
& = - \ln \left( {\frac{1}{2}} \right) \cr
& = - \ln \left( 1 \right) + \ln \left( 2 \right) \cr
& = \ln \left( 2 \right) \cr} $$
