Answer
$=\frac{\pi+3\sqrt 3}{4}$
Work Step by Step
$A=\int(\frac{1}{2}+cos\theta)^{2}d \theta$
$=\int(\frac{1}{4}+cos\theta+cos^{2}\theta)d \theta$
$=\frac{3}{4}\theta+sin\theta+\frac{1}{4}sin2\theta$
Now, put the limits in, subtracting the small loop from the big one
$=[\frac{3}{4}\theta+sin\theta+\frac{1}{4}sin2\theta]_{0}^{2\pi/3}-[\frac{3}{4}\theta+sin\theta+\frac{1}{4}sin2\theta]_{2\pi/3}^{\pi}$
$=\frac{\pi+3\sqrt 3}{4}$
