Mathematics - Definite Integration Question with Solution | TestHub

$\begin{array}{rl}& {\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}x}{1+\frac{1}{2}\sin 2x}dx={\int }_0^{\frac{\pi }{4}}\frac{1-\cos 2x}{2+\sin 2x}dx\\ & \int \frac{1}{2+\sin 2x}-\int \frac{\cos 2x}{2+\sin 2x}\end{array}$ $\left({\mathrm{I}}_1\right)-\left({\mathrm{I}}_2\right)$
$\begin{array}{rl}& \left({\mathrm{I}}_1\right)=\int \frac{\mathrm{dx}}{2+\frac{2\tan \mathrm{x}}{1+{\tan }^{2}x}}\\ & {\int }_0^{\frac{\pi }{4}}\frac{{\sec }^2\mathrm{xdx}}{2{\tan }^2\mathrm{x}+2\tan x+2}\end{array}$ $\begin{array}{rl}& \tan x=t\\ & \frac{1}{2}{\int }_0^1\frac{\mathrm{dt}}{{\left(\mathrm{t}+\frac{1}{2}\right)}^2+\frac{3}{4}}=\frac{\pi }{6\sqrt{3}}\\ & {\mathrm{I}}_2={\int }_0^{\pi /4}\frac{\cos 2\mathrm{x}}{2+\sin 2\mathrm{x}}\mathrm{dx}=\frac{1}{2}\left(\ln \frac{3}{2}\right)\\ & {\mathrm{I}}_1-{\mathrm{I}}_2=\frac{1}{\sqrt{3}}\frac{\pi }{6}+\frac{1}{2}\ln \frac{2}{3}\\ & \Rightarrow \mathrm{a}=2,\mathrm{b}=6\end{array}$
Ans. 8