Answer
$$\frac{{\sqrt 3 }}{3}$$
Work Step by Step
$$\eqalign{
& \int_{\pi /6}^{\pi /3} {{{\csc }^2}2\theta } d\theta \cr
& u = 2\theta ,{\text{ then }}du = 2d\theta dx,{\text{ }} \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = \pi /9,{\text{ }}u = 2\left( {\pi /3} \right) = 2\pi /3 \cr
& {\text{if }}x = \pi /6,{\text{ }}u = 2\left( {\pi /6} \right) = \pi /3 \cr
& {\text{so}} \cr
& \int_{\pi /6}^{\pi /3} {{{\csc }^2}2\theta } d\theta = \int_{\pi /3}^{2\pi /3} {{{\csc }^2}u} \left( {\frac{1}{2}du} \right) \cr
& = \frac{1}{2}\int_{\pi /3}^{2\pi /3} {{{\csc }^2}u} du \cr
& {\text{find the antiderivative }} \cr
& = \frac{1}{2}\left( { - \cot u} \right)_{\pi /3}^{2\pi /3} \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = \frac{1}{2}\left( { - \cot \left( {\frac{{2\pi }}{3}} \right) + \cot \left( {\frac{\pi }{3}} \right)} \right) \cr
& = \frac{1}{2}\left( {\frac{{\sqrt 3 }}{3} + \frac{{\sqrt 3 }}{3}} \right) \cr
& = \frac{{\sqrt 3 }}{3} \cr} $$
