Answer
$\dfrac{\pi}{8}$
Work Step by Step
Consider $A=\dfrac{1}{2}\int_m^n (r^2) d \theta=\int_{-(\pi/4)}^{(\pi/4)} (\dfrac{1}{2})\cos^2 2 \theta d \theta$
or, $\int_{-(\pi/4)}^{(\pi/4)} (\dfrac{1}{2})\cos^2 2 \theta d \theta=(2)\dfrac{1}{2} \int_{-\pi/4}^{\pi/4} (\dfrac{1+\cos 4 \theta}{2})d \theta$
$\implies A= (\dfrac{1}{2})[ \dfrac{\pi}{4}+\dfrac{ \sin 4 \theta}{\theta}]_{0}^{(\pi/4)} =\dfrac{\pi}{8}$
