Answer
The area that the curve encloses
$$ r=\sqrt{1+\cos ^{2} 5 \theta} \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
\begin{aligned} A &=\int_{0}^{2 \pi} \frac{1}{2} r^{2} d \theta \\
&=\frac{3}{2} \pi \end{aligned}
$$
Work Step by Step
The area that the curve encloses
$$ r=\sqrt{1+\cos ^{2} 5 \theta} \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
\begin{aligned} A &=\int_{0}^{2 \pi} \frac{1}{2} r^{2} d \theta \\
&=\int_{0}^{2 \pi} \frac{1}{2}(\sqrt{1+\cos ^{2} 5 \theta})^{2} d \theta \\ &=\frac{1}{2} \int_{0}^{2 \pi}\left(1+\cos ^{2} 5 \theta\right) d \theta \\
&=\frac{1}{2} \int_{0}^{2 \pi}\left[1+\frac{1}{2}(1+\cos 10 \theta)\right] d \theta \\
&=\frac{1}{2}\left[\frac{3}{2} \theta+\frac{1}{20} \sin 10 \theta\right]_{0}^{2 \pi} \\
&=\frac{1}{2}(3 \pi) \\
&=\frac{3}{2} \pi \end{aligned}
$$
