Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_{\pi /3}^{\pi /2} {\frac{{\sin x}}{{\sqrt {1 - 2\cos x} }}} dx \cr
& {\text{The integrand is not defined for }}x = \frac{\pi }{3},{\text{ then}} \cr
& \int_{\pi /3}^{\pi /2} {\frac{{\sin x}}{{\sqrt {1 - 2\cos x} }}} dx = \mathop {\lim }\limits_{b \to \pi /{3^ + }} \int_b^{\pi /2} {\frac{{\sin x}}{{\sqrt {1 - 2\cos x} }}} dx \cr
& {\text{ }} = \frac{1}{2}\mathop {\lim }\limits_{b \to \pi /{3^ + }} \int_b^{\pi /2} {\frac{{2\sin x}}{{\sqrt {1 - 2\cos x} }}} dx \cr
& {\text{Integrating}} \cr
& {\text{ }} = \frac{1}{2}\mathop {\lim }\limits_{b \to \pi /{3^ + }} \left[ {2\sqrt {1 - 2\cos x} } \right]_b^{\pi /2} \cr
& {\text{ }} = \mathop {\lim }\limits_{b \to \pi /{3^ + }} \left[ {\sqrt {1 - 2\cos x} } \right]_b^{\pi /2} \cr
& {\text{ }} = \mathop {\lim }\limits_{b \to \pi /{3^ + }} \left[ {\sqrt {1 - 2\cos \left( {\frac{\pi }{2}} \right)} - \sqrt {1 - 2\cos \left( b \right)} } \right] \cr
& {\text{Evaluate the limit}} \cr
& {\text{ }} = \left[ {\sqrt 1 - \sqrt {1 - 2\cos \left( {\pi /3} \right)} } \right] \cr
& {\text{ }} = \sqrt 1 - 0 \cr
& {\text{ }} = 1 \cr} $$
