How to evaluate $\int_0^{2\pi}\frac{1}{(1+a\cos {\theta})^2}\,d\theta$ without contour integration? 
Evaluate $$\int_{0}^{2\pi}\frac{1}{(1+a\cos {\theta})^2}\,d\theta \quad , \, 0\le a<1$$

I know it can be solved using complex analysis, but how do I solve this with real analysis methods?
This integral appears in the general derivation of Kepler's third law. Here $a$ is the eccentricity of the planetary orbit. So $a=0$ was assumed to prove that $T^2\propto r_0^3$ for a circular orbit, where $T$ is the orbital period and $r_0$ is the orbital radius.
 A: It is not difficult to check that for any $b>1$ we have:
$$ J(b) = \int_{0}^{2\pi}\frac{d\theta}{b+\cos\theta} = 4\int_{0}^{\pi/2}\frac{d\varphi}{b+\cos(2\varphi)}\\=4\int_{0}^{+\infty}\frac{dt}{(1+t^2)(b-1+2\cos^2(\arctan t))}=\frac{2\pi}{\sqrt{b^2-1}}$$
hence it follows that:
$$ -J'(b) = \int_{0}^{2\pi}\frac{d\theta}{(b+\cos\theta)^2} = \frac{2b\pi}{(b^2-1)^{3/2}} $$
and by taking $a=\frac{1}{b}$ we get:

$$\forall a\in(0,1),\qquad \int_{0}^{2\pi}\frac{d\theta}{(1+a\cos\theta)^2} = \color{red}{\frac{2\pi}{ (1-a^2)^{3/2}}}.$$

A: We can actually evaluate this integral indefinitely.
This is the most basic thing I could come up with,
Consider $$f(x)=\frac{\sin x}{1+a\cos x}$$
$$\implies f'(x)=\frac{a+\cos x}{(1+a\cos x)^2}$$
Now intregrate both sides,
$$\implies f(x)=\int \frac{(\cos x+a)dx}{(1+a\cos x)^2}$$
$$\implies f(x)=\frac{1}{a}\int \frac{(a\cos x+1)dx}{(1+a\cos x)^2}+\frac{a^2-1}{a}\int\frac{dx}{(1+a\cos x)^2}$$
$$\implies \frac{\sin x}{1+a\cos x}=\frac{1}{a}\int \frac{dx}{1+a\cos x}+\frac{a^2-1}{a}\cdot I$$ 
Now, 'I' is the integral we wanted to evaluate and the only problem left is $$\int \frac{dx}{1+a\cos x}$$ which is easily calculated by putting $\cos x=\frac{1-\tan^2x/2}{1+\tan^2x/2}$.
$$\int \frac{dx}{1+a\cos x}=\frac{2}{1-a}\sqrt{\frac{1-a}{1+a}}\arctan \left(t\sqrt{\frac{1-a}{1+a}}\right)+\mathbb C$$
Where $t=\tan x/2$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\color{#f00}{\int_{0}^{2\pi}
{\dd\theta \over \bracks{1 + a\cos\pars{\theta}}^{\,2}}} & =
\int_{-\pi}^{\pi}{\dd\theta \over \bracks{1 - a\cos\pars{\theta}}^{\,2}} =
\left.-2\,\partiald{}{b}
\int_{0}^{\pi}{\dd\theta \over b - a\cos\pars{\theta}}\right\vert_{\ b\ =\ 1}
\\[3mm] & =
-2\,\partiald{}{b}\bracks{%
\int_{0}^{\pi/2}{\dd\theta \over b - a\cos\pars{\theta}} +
\int_{0}^{\pi/2}{\dd\theta \over b + a\cos\pars{\theta}}}_{\ b\ =\ 1}
\\[3mm] & =
-4\,\partiald{}{b}\bracks{b%
\int_{0}^{\pi/2}{\dd\theta \over b^{2} - a^{2}\cos^{2}\pars{\theta}}}
_{\ b\ =\ 1}
\\[3mm] & =
-4\,\partiald{}{b}\bracks{b%
\int_{0}^{\pi/2}{\sec^{2}\pars{\theta}\dd\theta \over
b^{2}\tan^{2}\pars{\theta} + b^{2} - a^{2}}}_{\ b\ =\ 1}
\\[3mm] & \ \stackrel{\tan\pars{\theta}\ \mapsto\ t}{=}\
-4\,\partiald{}{b}\pars{b%
\int_{0}^{\infty}{\dd t \over b^{2}t^{2} + b^{2} - a^{2}}}_{\ b\ =\ 1}
\\[3mm] & \ \stackrel{bt/\root{b^{2} - a^{2}}\ \mapsto\ t}{=}\
-4\,\partiald{}{b}\pars{{1 \over \root{b^{2} - a^{2}}}%
\int_{0}^{\infty}{\dd t \over t^{2} + 1}}_{\ b\ =\ 1}
\\[3mm] & =
\left.-2\pi\,\partiald{}{b}\pars{{1 \over \root{b^{2} - a^{2}}}}
\right\vert_{\ b\ =\ 1} =
\left.{2\pi b \over \pars{b^{2} - a^{2}}^{3/2}}
\right\vert_{\ b\ =\ 1} =
\color{#f00}{2\pi \over \pars{1 - a^{2}}^{3/2}}
\end{align}
A: Without complex analysis, this is solvable by a Weierstrass substitution
Substitute $\cos \theta = \frac{1-t^2}{1+t^2}$ and $d\theta = \frac{2}{1+t^2}dt$ and proceed from there.
I must clarify, since you asked for a solution in the "simplest" manner - the algebra can get fairly cumbersome with this method, even if the techniques remain elementary (partial fractions, and so forth).
