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.
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:
$$\forall a\in(0,1),\qquad \int_{0}^{2\pi}\frac{d\theta}{(1+a\cos\theta)^2} = \color{red}{\frac{2\pi}{ (1-a^2)^{3/2}}}.$$
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 $t=\tan x/2$
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}
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).
For convenience, let’s consider the integral \begin{equation} I(b)=\int_0^{2 \pi} \frac{1}{b+\cos \theta}d\theta =2 \int_0^\pi \frac{1}{b+\cos \theta} d \theta \quad \textrm{ (By symmetry)} \end{equation} By the substitution $\pi-\theta \mapsto \theta$, we take their means and get $$ \begin{aligned} I(b) & =\int_0^\pi\left(\frac{1}{b+\cos \theta}+\frac{1}{b-\cos \theta}\right) d \theta \\ & =2 \int_0^\pi \frac{1}{b^2-\cos ^2 \theta} d \theta \\ & =4 b\int_0^{\frac{\pi}{2}} \frac{\sec ^2 \theta}{b^2 \sec ^2 \theta-1} d \theta \end{aligned} $$ Letting $t=\tan \theta$ yields $$ \begin{aligned} I(b) & =4 b \int_0^{\infty} \frac{d t}{\left(b^2-1\right)+b^2 t^2} \\ & =\frac{4}{\sqrt{b^2-1}}\left[\tan ^{-1} \frac{bt}{\sqrt{b^2-1}}\right]_0^{\infty} \\ & =\frac{2 \pi}{\sqrt{b^2-1}} \end{aligned} $$ Differentiating both sides w.r.t. $b$ gives $$ \int_0^{2 \pi} \frac{1}{(b+\cos \theta)^2} d \theta=\frac{2 \pi b}{\left(b^2-1\right)^{\frac{2}{2}} } $$ Replacing $b$ by $\frac{1}{a} $ yields $$ \boxed{\int_0^{2 \pi} \frac{d \theta}{(1+a \cos \theta)^2}=\frac{2 \pi}{\left(1-a^2\right)^{\frac{3}{2}}}} $$
