Can you think of a good approximation to this integral? So, I have


$$\int_{0}^{\arcsin\left(\frac{r}{g}\right)}\left(g\cos\theta-\sqrt{r^{2}-g^{2}\sin^{2}\theta}\right)^{2}d\theta$$


but the integral, which one can evaluate with mathematica, has a singularity at the upper limit to the integral. Can anyone think of a good approximation? You can use g<<1.
 A: Sorry, but I do not see any singularity so long as $r \le g$, which I believe is the case.  Expanding the square in the integrand, you get three very doable integrals:
$$g^2 \int_0^{\arcsin{r/g}} d\theta \, \cos^2{\theta} = \frac12 g^2 \arcsin{\frac{r}{g}} + \frac12 r \sqrt{g^2-r^2}$$
$$\int_0^{\arcsin{r/g}} d\theta \, (r^2-g^2 \sin^2{\theta}) = r^2 \arcsin{\frac{r}{g}} - \frac12 g^2 \arcsin{\frac{r}{g}} + \frac12 r \sqrt{g^2-r^2}$$
$$-2 g \int_0^{\arcsin{r/g}} d\theta \, \cos{\theta} \sqrt{r^2-g^2 \sin^2{\theta}} = -\frac{\pi}{2} r^2$$
Adding, I get the integral is, without approximation,
$$ r^2 \arcsin{\frac{r}{g}} + r \sqrt{g^2-r^2} - \frac{\pi}{2} r^2$$
I did all of these out by hand, by the way.  If you need to see the steps, I will be happy to oblige.
A: I would start by using Taylor approximation in the square root. We get:
$$\sqrt{r^2-g^2\sin^2\theta} = r+g^2\frac{\sin^2\theta}{2r}+O(g^4)$$
This way the integral becomes (omitting the $O(g^4)$ terms):
$$\int_0^{\arcsin\left(\frac{r}{g}\right)}\left[g\cos\theta-r-g^2\frac{\sin^2\theta}{2r}\right]^2d\theta$$
And this you should be able to integrate. The only problem left is to study the nature of the relation $\frac{r}{g}$ and to make sure that the approximation I made above is sound.
A: See also the appendix in http://arxiv.org/abs/0706.3682 for a slightly more general result.
