I'm trying to find the area of the region both inside the circle $r= \sinθ$ and outside the circle $r=\sqrt3 \cosθ$ (both equations are in polar coordinates). Here is what it looks like: 
The two graphs intersect at the origin and the polar point $(r, \theta) = \left(\frac{\pi}3, \frac{\sqrt3}{2}\right)$.
I thought the obvious answer would be to use the formula $A = \frac{1}{2}\int_{\theta_{1}}^{\theta_{2}} [R^2 - r^2] d\theta$ and integrate from $0$ to $\pi/3$ as seen below:
$$A = \frac{1}{2}\int_0^{\pi/3} \left[(\sin\theta)^2 - (\sqrt3\cos\theta)^2\right] d\theta,$$
however, this is wrong. What they have instead is:
$$A = \frac{1}{2}\int_{\pi/3}^\pi (\sin\theta)^2 d\theta \quad - \quad \frac{1}{2}\int_{\pi/3}^{\pi/2}(\sqrt3\cos\theta)^2 d\theta.$$
Can someone please explain how they got the bounds in their answer? And if there is a simpler way, for example, a slight alteration to the way I initially tried to do it, please let me know.
Thank you!
You seem to have a misunderstanding about how the curves in the figure are drawn, and what part of the enclosed area comprises the region of interest.
Refer to the following animation, which sweeps out the curves for $0 \le \theta \le \pi$.
You can see that when $0 \le \theta < \pi/3$, the portion of the curves drawn are the outer arc of the orange circle (where "outer" means that this arc is outside the blue circle) and the inner arc of the blue circle (where "inner" means this arc is inside the orange circle).
Only when $\pi/3 \le \theta \le \pi$ do the curves swept by the black line represent the region you wish to integrate: specifically, as you stated, inside the circle $r = \sin \theta$ and outside the circle $r = \sqrt{3} \cos \theta$. However, an important subtlety occurs when $\theta = \pi/2$: This is where $r < 0$ for the orange circle. This is why the integral is split up at this point, since if you apply your formula blindly, you would find yourself unnecessarily subtracting the orange circle's contribution from the blue circle from $\pi/2 < \theta < \pi$.
To recap, the area inside the blue circle and outside the orange comprises the wedge-shaped region in the first quadrant $$\frac{1}{2}\int_{\theta = \pi/3}^{\pi/2} ((\sin \theta)^2 - (\sqrt{3} \cos \theta)^2) \, d\theta,$$ since the blue curve's distance from the origin is greater than the orange curve's for angles in that interval; and the semicircular region $$\frac{1}{2} \int_{\theta = \pi/2}^\pi (\sin \theta)^2 \, d\theta.$$ What the provided solution has simply done is taken the blue curve's total contribution and subtracted the orange--it is simply an algebraic rearrangement.