Answer
The area that the curve encloses
$$ r=2+ \cos 4 \theta \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
A =\int_{0}^{2\pi} \frac{1}{2} r^{2} d \theta =\frac{9}{2} \pi
$$
Work Step by Step
The area that the curve encloses
$$ r=2+ \cos 4 \theta \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
\begin{aligned} A &=\int_{0}^{2\pi} \frac{1}{2} r^{2} d \theta \\
&=\int_{0}^{2 \pi} \frac{1}{2}(2+\sin 4 \theta)^{2} d \theta \\
&=\frac{1}{2} \int_{0}^{2 \pi}\left(4+4 \sin 4 \theta+\sin ^{2} 4 \theta\right) d \theta \\
&=\frac{1}{2} \int_{0}^{2 \pi}\left[4+4 \sin 4 \theta+\frac{1}{2}(1-\cos 8 \theta)\right] d \theta
\\
&=\frac{1}{2} \int_{0}^{2 \pi}\left(\frac{9}{2}+4 \sin 4 \theta-\frac{1}{2} \cos 8 \theta\right) d \theta
\\
&=\frac{1}{2}\left[\frac{9}{2} \theta-\cos 4 \theta-\frac{1}{16} \sin 8 \theta\right]_{0}^{2 \pi} \\ &=\frac{1}{2}[(9 \pi-1)-(-1)]=\frac{9}{2} \pi \\
&=\frac{1}{2}(9\pi)\\
&=\frac{9}{2} \pi
\end{aligned}
$$
