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.
 A: 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:
$$I=-\dfrac{1}{\varepsilon^2}+\dfrac{2}{\pi\varepsilon^2}\int_0^\infty\frac{d\theta}{1+\varepsilon+(1-\varepsilon)u^2},$$
$$I=-\dfrac{1}{\varepsilon^2}+\dfrac{2}{\pi\varepsilon^2(1-\varepsilon)}\sqrt{\dfrac{1-\varepsilon}{1+\varepsilon}}\arctan\sqrt{\dfrac{1+\varepsilon}{1-\varepsilon}}u\,\biggr|_0^\infty,$$
$$ \boxed {I = \dfrac{1}{\varepsilon^2}\left(\dfrac{1}{\sqrt{1-\varepsilon^2}}-1\right)}$$
A: To understand the first line, note that the Taylor series expansion of $\frac 1 {(1 + a \epsilon)^2}$ around $0$ gives
$$\frac 1 {(1 + a \epsilon)^2} = \sum \limits _{n=0} ^\infty (-1)^n (n+1) a^n \epsilon ^n .$$
In this, take $a = \cos \theta$ and multiply the whole equality by $\sin ^2 \theta$ (note that inside the parantheses in the right-hand side there should be $1 - \cos^2 \theta$, produced by replacing $\sin ^2 \theta$, and that summation starts from $n=0$).
Note that, save for some multiplicative factors, the integral in the right-hand side is essentially $\int \limits _0 ^{2 \pi} \cos^n \theta {\rm d} \theta - \int \limits _0 ^{2 \pi} \cos^{n+2} \theta {\rm d} \theta$. In order to understand the passage to the second line, you'll have to evaluate these integrals. Fortunately, I don't have to type that, somebody else did it for me, just take a look at the solutions found there.
The following lines are just elementary algebraic manipulations, you shouldn't have any trouble with them.
Finally, to understand the passage from the last-but-one line to the last one, write first the Taylor expansion of $(1+y)^a$ around $0$ (note that $a$ need not be an integer, that's the whole point in fact!). You'll get that the $n$th derivative ($n \ge 1$) in $0$ is $a (a-1) (a-2) \dots (a-n+1)$ so that
$$(1+y)^a = 1 + \sum \limits _{n=1} ^\infty \frac 1 {n!} a (a-1) (a-2) \dots (a-n+1) y^n .$$
Now do the ultra-smart trick of letting $y = - \epsilon ^2$ and $a = -\frac 1 2$. Plug the resulting expression into the last line of your chain of equalities and after doing plenty of algebraic simplifications you should get the expression on the line preceding it.
The details of these calculations are not difficult, but numerous, and typing them here is not my idea of a nice Sunday evening. In any case, if you get really stuck, leave a comment - but please don't ask for pre-calculus-level help!
