Integrate $\int_{-B}^{B}\frac{\sqrt{B^2 - y^2}}{1-y} dy$

[deftist]

Homework Statement

\int_{-B}^{B}\frac{\sqrt{B^2 - y^2}}{1-y} dy

Homework Equations

The Attempt at a Solution

I tried to get rid of the square root thing, so I started by:

y = B sin \theta,
dy = B cos \theta d\theta,

then the integral above becomes:

B^2 \int_{0}^{\pi} \frac{\sin^2 \theta d\theta}{1-Bcos\theta}d\theta.

Now my question is, how to integrate this out?

 

[micromass]

Hi deftist!

The trick is to do the subtitution

t=\tan(\theta /2)

and to apply the formula's

\sin(\theta)=\frac{2t}{1+t^2},~~\cos(\theta)=\frac{1-t^2}{1+t^2},~~\tan(\theta)=\frac{2t}{1-t^2}