$ \int \frac{1}{2+\sin 2 x}-\int \frac{\cos 2 x}{2+\sin 2 x}$
$ \left.\left(\mathrm{I}_1\right)-\mathrm{I}_2\right) $
$ \left(\mathrm{I}_1\right)=\int \frac{\mathrm{dx}}{2+\frac{2 \tan \mathrm{x}}{1+\tan ^2 \mathrm{x}}} $
$ \int_0^{\frac{\pi}{4}} \frac{\sec ^2 \mathrm{xdx}}{2 \tan ^2 \mathrm{x}+2 \tan \mathrm{x}+2}$
$ \frac{1}{2} \int_0^1 \frac{d t}{\left(t+\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{\pi}{6 \sqrt{3}} $
$ I_2=\int_0^{\pi / 4} \frac{\cos 2 x}{2+\sin 2 x} d x=\frac{1}{2}\left(\ln \frac{3}{2}\right) $
$ I_1-I_2=\frac{1}{\sqrt{3}} \frac{\pi}{6}+\frac{1}{2} \ln \frac{2}{3} $
$ \Rightarrow a=2, b=6$
Ans. $8$
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