Answer
$$\frac{\pi }{4}$$
Work Step by Step
$$\eqalign{
& \int_{\pi /2}^\pi {\frac{{\sin x}}{{1 + {{\cos }^2}x}}} dx \cr
& {\text{Let }}u = \cos x,{\text{ }}du = - \sin xdx,{\text{ }}dx = - \frac{1}{{\sin x}}du \cr
& {\text{The new limits of integration are}} \cr
& x = \pi ,{\text{ }}u = \cos \pi = - 1 \cr
& x = \pi /2,{\text{ }}u = \cos \frac{\pi }{2} = 0 \cr
& {\text{Substituting}} \cr
& \int_{\pi /2}^\pi {\frac{{\sin x}}{{1 + {{\cos }^2}x}}} dx = \int_0^{ - 1} {\frac{{\sin x}}{{1 + {u^2}}}} \left( { - \frac{1}{{\sin x}}} \right)du \cr
& = - \int_0^{ - 1} {\frac{1}{{1 + {u^2}}}} du \cr
& {\text{Integrating using basic integration rules}} \cr
& = - \left[ {{{\tan }^{ - 1}}u} \right]_0^{ - 1} \cr
& = - \left[ {{{\tan }^{ - 1}}\left( { - 1} \right) - {{\tan }^{ - 1}}\left( 0 \right)} \right] \cr
& = - \left[ { - \frac{\pi }{4} - 0} \right] \cr
& = \frac{\pi }{4} \cr} $$
