$\int\limits_0^{\pi /2} {\cfrac{{\sin x}}{{1 + {{\cos }^2}x}}dx}$

: Let$I = \int\limits_0^{\pi /2} {\cfrac{{\sin x}}{{1 + {{\cos }^2}x}}dx}$

Put $\cos x = t$ $\Rightarrow$ $- \sin xdx = dt$

When $x = 0,t = \cos 0 = 1$ and when $x = \cfrac{\pi }{2},t = \cos \cfrac{\pi }{2} = 0$

$\therefore$ $I = \int\limits_1^0 {\cfrac{{ - dt}}{{1 + {t^2}}}} = - \left[ {{{\tan }^{ - 1}}t} \right]_1^0 = \left[ { - {{\tan }^{ - 1}}0 + {{\tan }^{ - 1}}1} \right] = \left[ {0 + \cfrac{\pi }{4}} \right] = \cfrac{\pi }{4}$