Answer
The area that the curve encloses
$$ r=1+5 \sin 6 \theta \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
\begin{aligned} A &=\int_{0}^{2 \pi} \frac{1}{2} r^{2} d \theta \\
&=\frac{27}{2} \pi \end{aligned}
$$
Work Step by Step
The area that the curve encloses
$$ r=1+5 \sin 6 \theta \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
\begin{aligned} A &=\int_{0}^{2 \pi} \frac{1}{2} r^{2} d \theta \\
&=\int_{0}^{2 \pi} \frac{1}{2}(1+5 \sin 6 \theta)^{2} d \theta \\
&=\frac{1}{2} \int_{0}^{2 \pi}\left(1+10 \sin 6 \theta+25 \sin ^{2} 6 \theta\right) d \theta
\\ &=\frac{1}{2} \int_{0}^{2 \pi}\left[1+10 \sin 6 \theta+25 \cdot \frac{1}{2}(1-\cos 12 \theta)\right] d \theta \\
&=\frac{1}{2} \int_{0}^{2 \pi}\left[\frac{27}{2}+10 \sin 6 \theta-\frac{25}{2} \cos 12 \theta\right] d \theta \\
&=\frac{1}{2}\left[\frac{27}{2} \theta-\frac{5}{3} \cos 6 \theta-\frac{25}{24} \sin 12 \theta\right]_{0}^{2 \pi} \\
&=\frac{1}{2}\left[\left(27 \pi-\frac{5}{3}\right)-\left(-\frac{5}{3}\right)\right] \\
&=\frac{27}{2} \pi \end{aligned}
$$
