Answer
$$ - \frac{1}{8}\pi $$
Work Step by Step
$$\eqalign{
& \int_{\pi /3}^{\pi /2} {\sin \theta \sqrt {1 - 4{{\cos }^2}\theta } } d\theta ;{\text{ }}u = 2\cos \theta \cr
& {\text{Let }}u = 2\cos \theta ,{\text{ }}du = - 2\sin \theta d\theta ,{\text{ }}\sin \theta d\theta = - \frac{1}{2}du \cr
& {\text{The new limits of integration are:}} \cr
& x = \pi /2{\text{ }} \Rightarrow u = 2\cos \left( {\pi /2} \right) = 0 \cr
& x = \pi /3{\text{ }} \Rightarrow u = 2\cos \left( {\pi /2} \right) = 1 \cr
& {\text{Apply the substitution}} \cr
& \int_{\pi /3}^{\pi /2} {\sin \theta \sqrt {1 - 4{{\cos }^2}\theta } } d\theta = \int_0^1 {\sqrt {1 - {u^2}} \left( { - \frac{1}{2}} \right)} du \cr
& {\text{ }} = - \frac{1}{2}\int_0^1 {\sqrt {1 - {u^2}} } du \cr
& {\text{The integral of the form }}\int_0^1 {\sqrt {1 - {u^2}} du} ,{\text{ represents a part of a circle}} \cr
& {\text{centered at the origin with radius }}a{\text{ in the first quadrant}},{\text{ then }} \cr
& {\text{ }} = - \frac{1}{2}\left( {\frac{1}{4}\pi {{\left( 1 \right)}^2}} \right) \cr
& {\text{ }} = - \frac{1}{8}\pi \cr} $$
