## Return to Answer

2 fixed upper endpoint of last integral

Let me set

$$k^2=\frac{4ar}{(r+a)^2} <1.$$

If we replace $\theta$ with $2\theta$ we reduce this to an integral;

$$2 \underbrace{\int_0^\pi \sqrt{1-k^2\cos^2\theta} d\theta}_{=I}.$$

Now set

$$x=\cos\theta$$

so that $$dx=-\sqrt{1-x^2} d\theta$$

and

$$I=\int_{-1}^1\frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}} dx= 2\underbrace{\int_0^2 2\underbrace{\int_0^1 \frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}} dx}_{=:E_2(k)}.$$

The integral $E_2(k)$ is called Jacobi's complete elliptic integral of the second kind. There is no simple formula for it but you can have a look at the beautiful book by H. McKean and V. Moll, Elliptic Curves. Function Theory, Geometry Arithmetic Cambridge University Press,1997.

1

Let me set

$$k^2=\frac{4ar}{(r+a)^2} <1.$$

If we replace $\theta$ with $2\theta$ we reduce this to an integral;

$$2 \underbrace{\int_0^\pi \sqrt{1-k^2\cos^2\theta} d\theta}_{=I}.$$

Now set

$$x=\cos\theta$$

so that $$dx=-\sqrt{1-x^2} d\theta$$

and

$$I=\int_{-1}^1\frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}} dx= 2\underbrace{\int_0^2 \frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}} dx}_{=:E_2(k)}.$$

The integral $E_2(k)$ is called Jacobi's complete elliptic integral of the second kind. There is no simple formula for it but you can have a look at the beautiful book by H. McKean and V. Moll, Elliptic Curves. Function Theory, Geometry Arithmetic Cambridge University Press,1997.
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