In Roger Newton's book Scattering theory of waves and particles, the orbit equation of a scattered particle is given by
$$ \vartheta=\int _{r_{min}}^{\infty}dr\ \frac{1}{r^{2}}\left[\frac{1}{b^{2}}\left(1-\frac{V}{E}\right)-\frac{1}{r^{2}}\right]^{-1/2}$$
where $ r_{min} $ is the largest value of $r$ at which the radical vanishes. The deflection angle is
$$\Theta =\pi -2\vartheta$$
When $V=\alpha /r$, which is the Coulomb potential with $\alpha =Z_{1}Z_{2}e^{2}$, the cross section obtained is Rutherford formula
$$\frac{d\sigma}{d\Omega}=\frac{b}{\sin(\theta)|d\Theta/db|}=\left(\frac{Z_{1}Z_{2}e^{2}}{4E\sin^{2}(\theta/2)}\right)^{2}.$$
In order to calculate this formula, one must evaluate the integral above. How can I do that? The result, given in the book, has to be
$$ \vartheta =\cot ^{-1}\left(\frac{\alpha}{2Eb}\right) .$$
Evaluating the integral is pretty simple. Let $u = \frac{1}{r}$, then we see that \begin{align} \int^\infty_{r_{\text{min}}} \frac{1}{r^2}\left[\frac{1}{b^2}\left(1-\frac{\alpha}{Er} \right)-\frac{1}{r^2} \right]^{-1/2}=& \int^{1/r_{\text{min}}}_0\left[\frac{1}{b^2}\left(1-\frac{\alpha}{E}u\right)-u^2\right]^{-1/2}. \end{align} Note that \begin{align} \frac{1}{b^2}-\frac{\alpha}{Eb^2}u-u^2=\frac{1}{b^2}+\frac{\alpha^2}{4E^2b^4}-\left(u+\frac{\alpha}{2Eb^2}\right)^2 \end{align} which means \begin{align} \frac{1}{r_{\text{min}}}= \frac{-\alpha+\sqrt{4E^2b^2+\alpha^2}}{2Eb^2}. \end{align}
Any way, by direct integration, we see that \begin{align} \int^{1/r_{\text{min}}}_0\left[\frac{1}{b^2}\left(1-\frac{\alpha}{E}u\right)-u^2\right]^{-1/2}=\tan^{-1}\left(\frac{u+\frac{\alpha}{2Eb^2}}{\sqrt{\frac{1}{b^2}\left(1-\frac{\alpha}{E}u\right)-u^2}} \right)\Bigg|^{1/r_\text{min}}_0=\frac{\pi}{2}-\tan^{-1}\left(\frac{\alpha}{2Eb} \right). \end{align}
Finally, using the fact that \begin{align} \cot\vartheta=\tan\left(\frac{\pi}{2}-\vartheta\right) = \frac{\alpha}{2Eb} \ \ \implies \ \ \ \vartheta = \cot^{-1}\left(\frac{\alpha}{2Eb} \right). \end{align}