JEE Advanced 2019 · Paper 1, Mathematics question 17

Remove the exponential. Replacing $x$ by $-x$ and adding, using $\frac1{1+e^s}+\frac1{1+e^{-s}}=1$:
$$I=\frac2\pi\int_0^{\pi/4}\frac{dx}{2-\cos2x}=\frac2\pi\int_0^{\pi/4}\frac{\sec^2x\,dx}{1+3\tan^2x}.$$

Integrate.
$$\int_0^{\pi/4}\frac{\sec^2x\,dx}{1+3\tan^2x}=\frac1{\sqrt3}\tan^{-1}(\sqrt3\tan x)\Big|_0^{\pi/4}=\frac\pi{3\sqrt3}.$$

Value. $I=\dfrac2{3\sqrt3}$, so $27I^2=27\cdot\dfrac4{27}=4$.

Answer: 4