Find the area lying outside $r=2\cos\ \theta$ and inside $r=1+\cos\ \theta$.
Using the equation $\int_{\theta}^{2pi}(r^2/2) d\theta$, I would get $\int_{0}^{2\pi}((1+\cos\ \theta)^2/2)-((2\cos \ \theta)/2) d\theta$? Is what I am doing correct with the bounds being $0$ and $2 \pi$?
Your bounds are correct. You are essentially first trying to find where the two curves intersect, so you set the equations $r = 2\cos(\theta)$ and $r = 1 + \cos(\theta)$ equal to each other. Doing that, you obtain:
$2\cos(\theta) = 1 + \cos(\theta) \implies \cos(\theta) = 1$. The values $\theta = 0$ and $\theta = 2\pi$ satisfy this equation, so these will be the correct bounds. We must now think about the formula:
$$\frac{1}{2}\int_\alpha^\beta r^2 d\theta$$. This is the correct formula if you are trying to find the area enclosed by a single polar curve, the lines $\theta_1 = \alpha$ and $\theta_2 = \beta$. However, if you want the area enclosed by two polar curves, we need to instead use the formula
$$\frac{1}{2}\int_\alpha^\beta r_0^2 - r^2 d\theta$$
where $r_0$ is the curve on the "outside" of the area you are looking for while $r^2$ is the curve bounding the inside of the area.
In our case $r = 2\cos(\theta)$ will be the "inside" curve while our $r_0$ will be $1+\cos(\theta)$. You will plug these into the new formula to obtain:
$$\frac{1}{2}\int_0^{2\pi} (1+\cos(\theta))^2 - (2\cos(\theta))^2 d\theta$$
Would you be able to take it from here?