Definite double integral of a rational trigonometric expression $\int_0^b \int_0^{2\pi}\frac{\rho}{r_0-\rho\cos\alpha} d\alpha d\rho$ $$\int_0^b \int_0^{2\pi}\frac{\rho}{r_0-\rho\cos\alpha} d\alpha d\rho$$
 where  $r_0 \gg b$ 
and are positive real numbers.
The integrand is of the form of a rational trigonometric expression.  So I use the tangent half-angle substitution. 
Let $t=\tan(\frac{\alpha}{2})$.  We get that $\cos\alpha = \frac{1-t^2}{1+t^2}$ and $d\alpha = \frac{2}{1+t^2}dt$.  When I try to change the upper and lower limits of integration in terms of $t$, I clearly get $0$ for both bounds.  Is this substitution not valid here, or am I missing something?
What is the appropriate way to evaluate this double integral?   
 A: Place $0 < b < r_0$, if you want to calculate:
$$
I := \int_0^b \text{d}\rho \int_0^{2\pi}\frac{\rho}{r_0 - \rho\,\cos\alpha} \,\text{d}\alpha
$$
by symmetry is equivalent to calculating:
$$
I 
= \int_0^b \text{d}\rho \int_0^\pi\frac{2\rho}{r_0 - \rho\,\cos\alpha} \,\text{d}\alpha
= \int_0^b \text{d}\rho \lim_{\alpha_0 \to \pi^-}\int_0^{\alpha_0}\frac{2}{\frac{r_0}{\rho} - \cos\alpha} \,\text{d}\alpha
$$
from which:
$$
I = \int_0^b \text{d}\rho \lim_{\alpha_0 \to \pi^-}\left[ \frac{4}{\sqrt{\left(\frac{r_0}{\rho}\right)^2 - 1}}\,\arctan\left(\sqrt{\frac{\frac{r_0}{\rho}+1}{\frac{r_0}{\rho}-1}}\,\tan\left(\frac{\alpha}{2}\right)\right) \right]_{\alpha = 0}^{\alpha = \alpha_0}
$$
i.e.
$$
I 
= \int_0^b \frac{2\pi}{\sqrt{\left(\frac{r_0}{\rho}\right)^2 - 1}}\,\text{d}\rho
= \lim_{b_0 \to 0^+}\int_{b_0}^b \frac{2\pi}{\sqrt{\left(\frac{r_0}{\rho}\right)^2 - 1}}\,\text{d}\rho
$$
from which:
$$
I 
= \lim_{b_0 \to 0^+}\left[-2\pi\,\rho\,\sqrt{\left(\frac{r_0}{\rho}\right)^2 - 1}\,\right]_{\rho=b_0}^{\rho=b}
= 2\pi\left(r_0 - \sqrt{r_0^2 - b^2}\right).
$$
