How does $\int_0^{2\pi}\sqrt{(a-\cos\theta)^2+\sin^2\theta}\ d\theta$ end up as an elliptic integral of the second kind?

Question

This answer to Why can't Wolfram Alpha calculate $\int_0^{2\pi}\sqrt{(a-\cos\theta)^2+\sin^2\theta}\ d\theta$? includes the following excerpt:

$$\Re(a (a+2))>-1\land \Re((a-2) a)>-1$$ where appear ellptic integrals of the second kind. In fact, this reduces to $$I=4(a+1)E\left(\frac{4 a}{(a+1)^2}\right)$$

Here $E$ is the Complete Elliptic Integral of the Second Kind:

$$E(a) = \int_0^{\pi/2} \sqrt{1 - a \sin^2 t} \ \ dt$$

The expression inside the square root of my expression $(a-\cos\theta)^2+\sin^2\theta$ reduces to $a^2 - 2a \cos \theta + 1$ but I don't see how it ends up in the proper form for $E$.

Answer 1

$$A=(a-\cos(\theta))^2+\sin^2(\theta)=(a^2+1)-2 a \cos (\theta )$$ $$\cos (\theta )=1-2 \sin ^2\left(\frac{\theta }{2}\right)$$ $$A=(a-1)^2+4a\sin ^2\left(\frac{\theta }{2}\right)=(a-1)^2 \left(1+\frac{4 a }{(a-1)^2}\sin ^2\left(\frac{\theta }{2}\right)\right)$$

Answer 2

Note: I get a different result, but I think the idea is correct.

We have that $\cos(2\alpha)=1-2\sin^2(\alpha)$ so \begin{align} a^2-2a\cos(\theta)+1 &=a^2-2a(1-2\sin^2(\theta/2))+1\\ &=a^2-2a+1+4a\sin^2(\theta/2)\\ &=(a-1)^2+4a\sin^2(\theta/2)\\ &=(a-1)^2 \left(1+\frac{4a}{(a-1)^2}\sin^2(\theta/2)\right) \end{align} So \begin{align} I(a) &=\int_{0}^{2\pi}\sqrt{a^2-2a\cos(\theta)+1}\mathrm{d}\theta \\ &=\int_0^{2\pi} \sqrt{(a-1)^2 \left(1+\frac{4a}{(a-1)^2}\sin^2(\theta/2)\right)}\mathrm{d}\theta \\ &=\sqrt{(a-1)^2} \int_0^{2\pi} \sqrt{ \left(1+\frac{4a}{(a-1)^2}\sin^2(\theta/2)\right)}\mathrm{d}\theta \\ &=2\sqrt{(a-1)^2} \int_0^{\pi} \sqrt{ \left(1+\frac{4a}{(a-1)^2}\sin^2(t)\right)}\mathrm{d}t \\ &=4\sqrt{(a-1)^2} \int_0^{\pi/2} \sqrt{ \left(1+\frac{4a}{(a-1)^2}\sin^2(t)\right)}\mathrm{d}t \\ &=4\sqrt{(a-1)^2}E\left(-\frac{4a}{(a-1)^2}\right) \end{align}