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