Let ${x^2}$=t, So, 2xdx=dt
when ${x^2}$=0, t =0 and when ${x^2}$=${\Pi /4}$, t= $\Pi ^{2}/16$
Substituting it in the question, the integral becomes
$\int_{0}^{\Pi /4}$cos(t)/2 dt =$[sin t/2]_{0}^{\Pi^{2}/16}$
=sin(0.617 radians)-0 / 2
=0.5785909/2
=0.2892 (ANS)