Find the area inside the curve $r^2=2\cos(5\theta)$ and outside the unit circle. I found the area of one full rose-petal($A_1$) and the area enclosed by the petal and the unit circle($A_2$), subtracted these from one another to get the area enclosed by the curve outside of the circle, and multiplied the answer by the amount of rose pedals:
$f(\theta)=\pm\sqrt(2\cos(5\theta))$
$g(\theta)=1$
$f(\theta)= g(\theta) \Leftrightarrow \theta = {\pm\frac{\pi}{15}}$
$\cos(\theta)=0 \Leftrightarrow \theta = {\pm\frac{\pi}{10}}$
$$A_{1}=\frac{1}{2}\int_{-\frac{\pi}{15}}^{\frac{\pi}{15}}2\cos(5\theta)d\theta$$
$$A_{2}=\frac{1}{2}\int_{-\frac{\pi}{10}}^{-\frac{\pi}{15}}2\cos(5\theta)d\theta+\frac{1}{2}\int_{-\frac{\pi}{15}}^{\frac{\pi}{15}}1^2d\theta+\frac{1}{2}\int_{\frac{\pi}{15}}^{\frac{\pi}{10}}2\cos(5\theta)d\theta$$
$A = 6*(A_1-A_2)=\frac{-2}{5\pi} - \frac{1}{5}(-12\sqrt3 + 12)$, not the correct answer?


The intersection points of one of the petals of the  curve $r^2 = \sqrt{2 \cos(5\theta)}$ and unit circle $r = 1$ are indeed $r = 1, \theta = \pm \frac{\pi}{15}$.
But as you need area outside the unit circle, the integral should be
$A = 6 \cdot  \displaystyle \frac 12 \int_{-\pi/15}^{\pi/15} \left(2 \cos(5\theta) - 1\right) ~ d\theta$
$ = \dfrac{6 \sqrt3 - 2 \pi}{5}$
Mistake in your work is in the calculation of $A_1$. It should be,
$ \displaystyle \frac{1}{2}\int_{-\frac{\pi}{10}}^{\frac{\pi}{10}}2 \cos(5\theta)~d\theta$
