Answer
$$\frac{3}{2} \pi$$
Work Step by Step
The area is bounded by $r=\sqrt{1+\cos ^2 5 \theta}$ 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}\left(\sqrt{1+\cos ^2 5 \theta}\right)^2 d \theta \\
& =\frac{1}{2} \int_0^{2 \pi}\left(1+\cos ^2 5 \theta\right) d \theta\\
&=\frac{1}{2} \int_0^{2 \pi}\left[1+\frac{1}{2}(1+\cos 10 \theta)\right] d \theta \\
& =\frac{1}{2}\left[\frac{3}{2} \theta+\frac{1}{20} \sin 10 \theta\right]_0^{2 \pi}\\
&=\frac{1}{2}(3 \pi)\\
&=\frac{3}{2} \pi
\end{aligned}
