Question

$intlimits^frac{pi}{2}_0 {frac{cosx}{1+sin^{2}x} , dx$

Answer 1

$intlimits^{frac{pi}{2}}_0 {frac{cos x}{1+sin^2 x}} , dx =left[begin{array}{c}{sin x=t}&{dt=d(sin x)=(sin x)prime dx=cos x dx}&{0leq xleqfrac{pi}{2}rightarrow 0leq tleq 1}end{array}right] =\ \ =intlimits^1_0 {frac{dt}{1+t^2}} , =arctg t|limits^1 _0=arctg1-arctg0=frac{pi}{4}-0= frac{pi}{4}$