Question

Evaluate the integral: $∫_0^{\frac \pi2} \frac {sinx}{1+cos^2x}dx$

Answer 1

$∫_0^{\frac \pi2} \frac {sin\ x}{1+cos^2x}dx$

$Let \ cosx=t ⇒ -sinx \ dx=dt$

$When\ x=0,t=1 \ and\ when \ x=\frac \pi2,t=0$

⇒∫$∫_0^{\frac \pi2} \frac {sin\ x}{1+cos^2x}dx$ = $-∫_1^0\frac {dt}{1+t^2}$

                               =$-[tan^{-1}t]_1^0$

                               =$-[tan^{-1}0-tan^{-1}1]$

                               =$-[-\frac \pi4]$

                               =$\frac \pi4$