Answer
\[\frac{π}{4}\]
Work Step by Step
Let \[I=\int_{0}^{\frac{π}{2}}\frac{\cos x}{1+\sin^2 x}\,dx\]
Put \[t=\sin x\;\;\;...(1)\]
\[\Rightarrow dt=\cos x\,dx\]
At $x=0\Rightarrow t=0$
and at $x=\frac{π}{2}\Rightarrow t=1$
\[I=\int_{0}^{1}\frac{dt}{1+t^2}\]
\[\Rightarrow I=\left[\tan^{-1} (t)\right]_{0}^{1}\]
\[\Rightarrow I=\tan^{-1} (1)-\tan^{-1} (0)\]
\[\Rightarrow I=\frac{π}{4}\]
Hence , \[\int_{0}^{\frac{π}{2}}\frac{\cos x}{1+\sin^2 x}\,dx=\frac{π}{4}\]
