Over an integral arising from Kepler's problem [also: generally useful integral, NOT DUPLICATE!] This post might appear as a duplicate of the following:
Over an integral arising from Kepler's problem [also: generally useful integral]
So recalling quickly:
$$\Phi(\epsilon) = \frac{1}{2\pi}\int_0^{2\pi}\ \frac{\sin^2\theta\ \text{d}\theta}{(1 + \epsilon\cos\theta)^2}$$
Note: The integral arises in studying the Kepler's problem, which is why $\epsilon\in [0;\ 1]$.
Necessary Note I tried to solve it with complex analysis, as you can see in the above link, but in my book it's solved by series with lots of unknown passages, which I will write down here. 
$$
\begin{align*}
\frac{1}{2\pi}\int_0^{2\pi}\ \frac{\sin^2\theta\ \text{d}\theta}{(1 + \epsilon\cos\theta)^2} & = \sum_{n = 1}^{+\infty}\ (-1)^n\epsilon^n(n+1)\cdot\frac{1}{2\pi}\int_0^{2\pi}\ (1 - \cos\theta)\cos^n\theta\ \text{d}\theta
\\\\
& = \sum_{n = 0}^{+\infty}\ \epsilon^{2n}(2n+1)\left[\frac{1}{2^{2n}} \binom{2n}{n} - \frac{1}{2^{2n+2}}\binom{2n+2}{n+1}\right]
\\\\
& = \sum_{n = 0}^{+\infty}\ \epsilon^{2n}\frac{2n+1}{2^{2n+2}} \frac{(2n)!}{(n!)^2} \left[4 - \frac{(2n+1)(2n+2)}{(n+1)^2}\right]
\\\\
& = \sum_{n = 0}^{+\infty}\ \epsilon^{2n} \frac{(2n+1)!}{2^{2n+1}(n!)^2 (n+1)}
\\\\
& = \frac{1}{\epsilon^2}\ \sum_{n = 0}^{+\infty} \epsilon^{2(n+1)}(-1)^{n+1}\binom{-1/2}{n+1}
\\\\
& = \frac{1}{\epsilon^2}[(1 - \epsilon^2)^{-1/2} - 1]
\end{align*}
$$
Can anybody help with that? It seems really messy..
P.s. I wrote down here exactly the passages of the book.
What I need
I am not familiar with the passage from the first row to the second one (namely: where the integral has been solved, in which we pass to the binomial coefficient). In addition to that, I didn't understand how the very last series has been summed.

There is an elementary path:
$$I = \frac{1}{2\pi}\int_0^{2\pi}\frac{\sin^2\theta}{(1+\varepsilon\cos\theta)^2}\,d\theta =\frac{1}{2\pi\varepsilon}\int_0^{2\pi}\sin\theta\,d\left(\dfrac{1}{1+\varepsilon\cos\theta}\right)$$
By parts:
$$ I = \frac{1}{2\pi\varepsilon}\dfrac{\sin\theta}{1+\varepsilon\cos\theta}\,\biggr|_0^{2\pi} - \dfrac{1}{2\pi\varepsilon}\int_0^{2\pi}\frac{\cos\theta\,d\theta}{1+\varepsilon\cos\theta} = 0 - \frac{1}{2\pi\varepsilon^2}\int_0^{2\pi}\frac{\varepsilon\cos\theta+1-1}{1+\varepsilon\cos\theta}\,d\theta$$$$ = -\frac{1}{2\pi\varepsilon^2}\theta\,\biggr|_0^{2\pi} + \frac{1}{2\pi\varepsilon^2}\int_0^{2\pi}\frac{d\theta}{1+\varepsilon\cos\theta} = -\frac{1}{\varepsilon^2} + \frac{1}{\pi\varepsilon^2}\int_0^{\pi}\frac{d\theta}{1+\varepsilon\cos\theta}.$$
Applying universal trigonometric substitution,
$$\theta = 2\arctan u,\quad d\theta = \dfrac{2}{1+u^2}du,\quad\cos\theta = \dfrac{1-u^2}{1+u^2},$$
we get:
$$I=-\dfrac{1}{\varepsilon^2}+\dfrac{2}{\pi\varepsilon^2}\int_0^\infty\frac{d\theta}{1+\varepsilon+(1-\varepsilon)u^2},$$
$$I=-\dfrac{1}{\varepsilon^2}+\dfrac{2}{\pi\varepsilon^2(1-\varepsilon)}\sqrt{\dfrac{1-\varepsilon}{1+\varepsilon}}\arctan\sqrt{\dfrac{1+\varepsilon}{1-\varepsilon}}u\,\biggr|_0^\infty,$$
$$ \boxed {I = \dfrac{1}{\varepsilon^2}\left(\dfrac{1}{\sqrt{1-\varepsilon^2}}-1\right)}$$
