$\begin{aligned}
& \int_0^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+\frac{1}{2} \sin 2 x} d x=\int_0^{\frac{\pi}{4}} \frac{1-\cos 2 x}{2+\sin 2 x} d x \
& \int \frac{1}{2+\sin 2 x}-\int \frac{\cos 2 x}{2+\sin 2 x}
\end{aligned}\left(\mathrm{I}_1\right) \quad-\quad\left(\mathrm{I}_2\right)$
$\begin{aligned}
& \left(\mathrm{I}_1\right)=\int \frac{\mathrm{dx}}{2+\frac{2 \tan \mathrm{x}}{1+\tan ^2 x}} \
& \int_0^{\frac{\pi}{4}} \frac{\sec ^2 \mathrm{xdx}}{2 \tan ^2 \mathrm{x}+2 \tan x+2}
\end{aligned}\begin{aligned}
& \tan x=t \
& \frac{1}{2} \int_0^1 \frac{\mathrm{dt}}{\left(\mathrm{t}+\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{\pi}{6 \sqrt{3}} \
& \mathrm{I}_2=\int_0^{\pi / 4} \frac{\cos 2 \mathrm{x}}{2+\sin 2 \mathrm{x}} \mathrm{dx}=\frac{1}{2}\left(\ln \frac{3}{2}\right) \
& \mathrm{I}_1-\mathrm{I}_2=\frac{1}{\sqrt{3}} \frac{\pi}{6}+\frac{1}{2} \ln \frac{2}{3} \
& \Rightarrow \mathrm{a}=2, \mathrm{~b}=6
\end{aligned}$
Ans. 8