Deriving Rutherford's Formula: Integrating to Find the Angle of Deflection

[jameson2]

Homework Statement

Homework Equations

This is in my mechanics book. It gives the final formula, only saying "carrying out the elementary integration..."
Derive Rutherford's Formula:
$$\phi_0=cos^{-1}\frac{a}{\sqrt{1+a^2}}$$
where
$$a=\frac{\alpha}{mv_\infty^2 \rho}$$

From the equation
$$\phi_0 = \int_{r_{min}}^\infty \frac{\frac{\rho}{r^2}dr}{\sqrt{1-\frac{\rho^2}{r^2}-\frac{2U}{mv_\infty^2}}}$$

Using $$U=\frac{\alpha}{r}$$

The Attempt at a Solution

This question comes up regularly on exams, and the hint that is given is that

$$\int_{x_+}^\infty \frac{1}{x\sqrt{(x-x_+)(x-x_-}} = arccos \frac{a}{\sqrt{1+a^2}}$$

where $$x_+=a+\sqrt{1+a^2}$$ and $$x_-=a-\sqrt{1+a^2}$$

I don't see how this helps, since the integral is from r min, which corresponds to $$x_-$$ I think. Also, I have $$\rho$$ instead of 1 in the expressions for $$x_-$$ and $$x_+$$

 

[kurt.physics]

. So, if I substitute U=\frac{\alpha}{r} into the integral, I get: \phi_0 = \int_{r_{min}}^\infty \frac{\frac{\rho}{r^2}dr}{\sqrt{1-\frac{\rho^2}{r^2}-\frac{2\alpha}{mrv_\infty^2}}} Now, how can I rewrite this so that it is in the form given in the hint?