Answer
$=\frac{\pi}{3}+\sqrt 3$
Work Step by Step
$A=\frac{1}{2}\int(1+2cos3\theta)^{2}d \theta$
$=\frac{1}{2}\int(1+4cos3\theta+4cos^{2}3\theta)d \theta$
$=\frac{1}{2}[3\theta+\frac{4}{3}sin3\theta+\frac{1}{6}sin6\theta]$
Now, put the limits in, subtracting the small loop from the large one. For the large loop , we are using the interval of the half of the loop, so remember to double it.
$=2.\frac{1}{2}[3\theta+\frac{4}{3}sin3\theta+\frac{1}{6}sin6\theta]_{0}^{2\pi/9}
-\frac{1}{2}[3\theta+\frac{4}{3}sin3\theta+\frac{1}{6}sin6\theta]_{2\pi/9}^{4\pi/9}$
$=\frac{\pi}{3}+\sqrt 3$
