JEE Main 2024 — 4 April, Shift 1 — Question 30

$\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \mathrm{x}}{1+\frac{1}{2} \sin 2 \mathrm{x}} d x=\int_{0}^{\frac{\pi}{4}} \frac{1-\cos 2 x}{2+\sin 2 x} d x$

$\int \frac{1}{2+\sin 2 x}-\int \frac{\cos 2 x}{2+\sin 2 x}$

$\left(\mathrm{I}_{1}\right)$ $\left(I_{2}\right)$ $\left(I_{1}\right)=\int \frac{d x}{2+\frac{2 \tan x}{1+\tan ^{2} x}}$ $\int_{0}^{\frac{\pi}{4}} \frac{\sec ^{2} x d x}{2 \tan ^{2} x+2 \tan x+2}$

$\tan \mathrm{x}=\mathrm{t}$

$\frac{1}{2} \int_{0}^{1} \frac{d t}{\left(t+\frac{1}{2}\right)^{2}+\frac{3}{4}}=\frac{\pi}{6 \sqrt{3}}$

$\mathrm{I}_{2}=\int_{0}^{\pi / 4} \frac{\cos 2 \mathrm{x}}{2+\sin 2 \mathrm{x}} \mathrm{dx}=\frac{1}{2}\left(\ln \frac{3}{2}\right)$

$\mathrm{I}_{1}-\mathrm{I}_{2}=\frac{1}{\sqrt{3}} \frac{\pi}{6}+\frac{1}{2} \ln \frac{2}{3}$

$\Rightarrow \mathrm{a}=2, \mathrm{~b}=6$

Ans. $8$