How do I find the area shared by the circles $r = 2\cos(\theta)$ and $r = 1$?

Question

I figured out the intersection points:

$r=2\cos(\theta)$, $r=1$

$2\cos(\theta) = 1$

$\cos(\theta) = \frac{1}{2}$

$\arccos(1/2) = π/3$ (I), $5π/3$ (IV)

Answer 1

The last integral is just $$2\int_{\pi/3}^{\pi/2}4\cos^2\theta\; \mathrm d\,\theta$$ All you have to do then is to linearise $\cos^2\theta$ with the duplication formula: from $$\cos 2\theta=2\cos^2\theta-1=1-2\sin^2 \theta,$$ you deduce the linearisation formulae:\begin{align*}\cos^2\theta&=\frac{1+\cos 2\theta}2,\\\sin^2\theta&=\frac{1-\cos 2\theta}2. \end{align*}