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

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SUMMARY

The integral $\int_{-B}^{B}\frac{\sqrt{B^2 - y^2}}{1-y} dy$ can be transformed using the substitution $y = B \sin \theta$, leading to the expression $B^2 \int_{0}^{\pi} \frac{\sin^2 \theta d\theta}{1-B\cos\theta}$. To further simplify and solve this integral, the substitution $t = \tan(\theta / 2)$ is recommended, along with the identities $\sin(\theta) = \frac{2t}{1+t^2}$ and $\cos(\theta) = \frac{1-t^2}{1+t^2}$. This approach effectively facilitates the integration process.

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Homework Statement



[itex]\int_{-B}^{B}\frac{\sqrt{B^2 - y^2}}{1-y} dy[/itex]

Homework Equations





The Attempt at a Solution



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

[itex]y = B sin \theta,[/itex]
[itex]dy = B cos \theta d\theta,[/itex]

then the integral above becomes:

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

Now my question is, how to integrate this out?
 
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Hi deftist!

The trick is to do the subtitution

[tex]t=\tan(\theta /2)[/tex]

and to apply the formula's

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

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