Chapter 11: Parametric Equations and Polar Coordinates - Practice Exercises - Page 688: 49

$2+\dfrac{\pi}{4}$

Here, $A= \int_{0}^{\pi/4} (4) (\dfrac{1}{2}) [1+\cos 2 \theta]^2 d\theta$ This gives: $(2)[\sin (2 \theta) +(\dfrac{1}{2}) \theta +\dfrac{1}{8}(\sin 4 \theta)]_{0}^{(\pi/4)}=2(1+\dfrac{\pi}{8})$ This implies that $A=2(1+\dfrac{\pi}{8})=2+\dfrac{\pi}{4}$