It follows from leibniz rule that if $\frac{\partial f}{\partial \theta_0}(\theta,\theta_0)$ exists then
$$\frac{d}{d\theta_0}\bigg(\int_0^{\theta_0}f(\theta,\theta_0)d\theta\bigg)=\int _0^{\theta_0}\frac{\partial f}{\partial\theta_0}(\theta,\theta_0)d\theta+f(\theta_0,\theta_0)$$
I am trying to figure out the if there is a way to differentiate the improper integral $g(\theta_0)=\int_0^{\theta_0}\frac{1-a\cos\theta}{\sqrt{\cos\theta-\cos\theta_0}}d\theta$. That is, I want to find an expression for
$$\frac{\partial g}{\partial\theta_0}(\theta_0)=\frac{\partial}{\partial\theta_0}\int_0^{\theta_0}\frac{1-a\cos\theta}{\sqrt{\cos\theta-\cos\theta_0}}d\theta$$
I obviously run into trouble if I simply attempt to evaluate the integrand at $\theta=\theta_0$
$$\frac{\partial g}{\partial\theta_0}(\theta_0) = \int_0^{\theta_0}\frac{\partial}{\partial\theta_0}\frac{1-a\cos\theta}{\sqrt{\cos\theta-\cos\theta_0}}d\theta +\frac{1-a\cos\theta_0}{\sqrt{\cos\theta_0-\cos\theta_0}}$$
This seems to imply that the derivative of $g(\theta_0)$ undefined everywhere, but if I evaluate the function numerically it appears to be smooth 
I feel like there should be a way to evaluate the derivative of $g$. I tried the replacing the right endpoint with a limit and applied the definition of the derivative but had no success. Am I missing something?
I'm assuming $0 < \theta_0 < \pi/2$ (so the integrand is real).
Break up the integral into two parts, $J_1$ where $0 < \theta < \theta_0/2$ and $J_2$ where $\theta_0/2 < \theta < \theta_0$.
Differentiating $J_1$ is straightforward:
$$ \dfrac{d J_1}{d\theta_0} = \dfrac{\sin(\theta_0)}{2} \int_0^{\theta_0/2} \dfrac{a \cos(\theta) - 1}{(\cos(\theta)-\cos(\theta_0))^{3/2}}\; d\theta + \dfrac{1 - a \cos(\theta_0/2)}{2 \sqrt{\cos(\theta_0/2) - \cos(\theta_0)}}$$
For $J_2$ we first do a change of variables $s = \sqrt{\cos(\theta) - \cos(\theta_0)}$. I get
$$J_2 = -2 \int_0^{\sqrt{\cos(\theta_0/2)-\cos(\theta_0)}} \dfrac{1 - a \cos(\theta_0) - a s^2}{\sqrt{1 - \cos(\theta_0)^2 - 2 \cos(\theta_0) s^2 - s^4}}\; ds$$
$$ \eqalign{\dfrac{dJ_2}{d\theta_0} &= 2 \sin(\theta_0) \int_0^{\sqrt{\cos(\theta_0/2)-\cos(\theta_0)}} \dfrac{s^2 + \cos(\theta_0) - a}{\left(1 - \cos(\theta_0)^2 - 2 \cos(\theta_0) s^2 - s^4\right)^{3/2}}\; ds\cr &+ \dfrac{1 + 2 a - (a+4) \cos(\theta_0/2) + 2 a \cos(\theta_0)}{2 \sqrt{\cos(\theta_0/2) - \cos(\theta_0)}}} $$