Answer
$$\frac{9}{2} \pi$$
Work Step by Step
The area is bounded by $r=2 +\sin4 \theta$ for $\theta=0$ to $\theta=2\pi$.
\begin{aligned}
A & =\int_0^{2 \pi} \frac{1}{2} r^2 d \theta\\
&=\int_0^{2 \pi} \frac{1}{2}(2+\sin 4 \theta)^2 d \theta\\
&=\frac{1}{2} \int_0^{2 \pi}\left(4+4 \sin 4 \theta+\sin ^2 4 \theta\right) d \theta \\
& =\frac{1}{2} \int_0^{2 \pi}\left[4+4 \sin 4 \theta+\frac{1}{2}(1-\cos 8 \theta)\right] d \theta \\
& =\frac{1}{2} \int_0^{2 \pi}\left(\frac{9}{2}+4 \sin 4 \theta-\frac{1}{2} \cos 8 \theta\right) d \theta\\
&=\frac{1}{2}\left[\frac{9}{2} \theta-\cos 4 \theta-\frac{1}{16} \sin 8 \theta\right]_0^{2 \pi} \\
& =\frac{1}{2}[(9 \pi-1)-(-1)]\\
&=\frac{9}{2} \pi
\end{aligned}
