How to Integrate 1/(1+e cos(x))^2?

[Dustinsfl]

I am trying to integrate
$$
\int_0^{\nu}\frac{d\theta}{(1 + e\cos\theta)^2}
$$
where $0<e<1$.
I tried using Residue Theory but that was messy and didn't come up as needed.
The highlights of that were:
$\cos z = \frac{z + \frac{1}{z}}{2}$
Denominator became $z + 2ze^2 + 2z^2e^2 + 2e^2 + 4z^2 + 4e$ and that factor to
$$
\left(z + \frac{1+2e^2+\sqrt{1-4e^2[15e(16+3e)]}}{4e(2+e)}\right)\left(z + \frac{1+2e^2-\sqrt{1-4e^2[15e(16+3e)]}}{4e(2+e)}\right)
$$
It is highly probably there were some errors with that mess. I wasn't sure if I could do:
$$
\cos 2z = \frac{2z + \frac{1}{2z}}{2}
$$
but I did.
Is there another way to integrate this? Any method real or complex is fine.

 

[chisigma]

Re: Integrate 1/(1+e\cos x)^2

With the substitution...

$\displaystyle t= \tan \frac{\theta}{2} \implies \theta= 2\ \tan^{-1} t \implies d \theta = 2\ \frac{d t} {1 + t^{2}} \implies \sin \theta = \frac{2 t} {1 + t^{2}} \implies \cos \theta = \frac{1-t^{2}} {1 + t^{2}}$ (1)

... You arrive to a definite integral containing a rational function...

Kind regards

$\chi$ $\sigma$

 

[Dustinsfl]

Re: Integrate 1/(1+e\cos x)^2

With the substitution...

$\displaystyle t= \tan \frac{\theta}{2} \implies \theta= 2\ \tan^{-1} t \implies d \theta = 2\ \frac{d t} {1 + t^{2}} \implies \sin \theta = \frac{2 t} {1 + t^{2}} \implies \cos \theta = \frac{1-t^{2}} {1 + t^{2}}$ (1)

... You arrive to a definite integral containing a rational function...

Kind regards

$\chi$ $\sigma$

But how do we integrate the rational function
$$
2\int\frac{dt}{(1 + t^2)\left(1+e\frac{1-t^2}{1+t^2}\right)^2}
$$
I couldn't simplify that down to anything useful.

 

[Mathwebster]

Re: Integrate 1/(1+e\cos x)^2

Have you tried looking the integral up in say the CRC Standard Mathematical Tables?

 

[Dustinsfl]

Re: Integrate 1/(1+e\cos x)^2

Have you tried looking the integral up in say the CRC Standard Mathematical Tables?

I would rather learn how to integrate it then look it up.

 

[Mathwebster]

Re: Integrate 1/(1+e\cos x)^2

You know, sometimes looking up the answer give you the clue on how to solve the problem.

 

[Dustinsfl]

Re: Integrate 1/(1+e\cos x)^2

You know, sometimes looking up the answer give you the clue on how to solve the problem.

With that begin said, what is your idea for solving it then? I knew the answer before I ever posted the question.

 

[alyafey22]

Re: Integrate 1/(1+e\cos x)^2

Hey, may I ask what do you mean by e ? I thought it was the constant \(\displaystyle \ln e =1 \) ?

 

[Dustinsfl]

Re: Integrate 1/(1+e\cos x)^2

Hey, may I ask what do you mean by e ? I thought it was the constant \(\displaystyle \ln e =1 \) ?

e is the eccentricity of the orbit. So in this case, we have elliptical orbits.

 

[alyafey22]

Re: Integrate 1/(1+e\cos x)^2

I tried using Residue Theory but that was messy and didn't come up as needed.

I think that a complex method can't be used here , since that applies only to an integration of the form \(\displaystyle \int_{0}^{2\pi}\, f(\cos(\theta), \sin(\theta))\, d\theta \)

another question , what do you mean by v ?

It might be so helpful , if you can illustrate more on what this integration really describes ?

 

[Dustinsfl]

Re: Integrate 1/(1+e\cos x)^2

I think that a complex method can't be used here , since that applies only to an integration of the form \(\displaystyle \int_{0}^{2\pi}\, f(\cos(\theta), \sin(\theta))\, d\theta \)

another question , what do you mean by v ?

It might be so helpful , if you can illustrate more on what this integration really describes ?

$\nu$ is just a variable. Like the FTC. $\int_a^xf(t)dt$

 

[chisigma]

Re: Integrate 1/(1+e\cos x)^2

A possible alternative is the substitution...

$\displaystyle a\ \cos \theta= x \implies \cos \theta= \frac{x}{a} \implies d \theta = - \frac{d x}{a\ \sin \theta} = - \frac{d x}{a\ \sqrt{1- (\frac{x}{a})^{2}}}$ (1)

... so that the integral becomes...

$\displaystyle \int \frac{d \theta}{(1+ a\ \cos \theta)^{2}} = - \frac{1}{a} \int \frac{d x} {\sqrt{1 - (\frac{x}{a}) ^{2}}\ (1 + x)^{2}}$ (2)

... and 'Monster Wolfram' says to be...

$\displaystyle - \frac{1}{a} \int \frac{d x} {\sqrt{1 - (\frac{x}{a}) ^{2}}\ (1 + x)^{2}} = \frac{1}{a\ (1-\frac{1}{a^{2}})\ (1+x)}\ \{\sqrt{1-\frac{1}{a^{2}}}\ \sqrt{1-\frac{x^{2}}{a^{2}}} - \frac{1+x}{a^{2}}\ \ln [\sqrt{1-\frac{1}{a^{2}}}\ \sqrt{1-\frac{x^{2}}{a^{2}}} + \frac{x}{a^{2}} + 1] + \frac{1+x}{a^{2}}\ \ln (1+x) \} + c$ (3)
Kind regards

$\chi$ $\sigma$

 

[Mathwebster]

Re: Integrate 1/(1+e\cos x)^2

Seeing the solution in the CRC tables I might try the following.

Consider

$\dfrac{d}{d \theta} \dfrac{\sin \theta}{1 + e \cos \theta} = \dfrac{ \cos \theta + e}{(1 + e \cos \theta)^2}$

so

$\dfrac{d}{d \theta} \dfrac{e \sin \theta}{1 + e \cos \theta} = \dfrac{e \cos \theta + e^2}{(1 + e \cos \theta)^2}= \dfrac{1 + e \cos \theta + e^2-1}{(1 + e \cos \theta)^2} $
$= \dfrac{1}{1 + e \cos \theta} + \dfrac{e^2-1}{(1 + e \cos \theta)^2} $

Thus,

$\displaystyle \int \dfrac{d \theta}{(1 + e \cos \theta)^2} = \dfrac{e}{e^2-1} \dfrac{\sin \theta}{1 + e \cos \theta} - \dfrac{1}{e^2-1} \int \dfrac{d \theta}{1 + e \cos \theta}.$

The Weierstrass substitution would probably work a little better on the new integral here.

 

[DreamWeaver]

Re: Integrate 1/(1+e\cos x)^2

This is NOT a simple integral, despite its apparently innocuous appearance. That's partly why, in Gradshteyn and Ryzhik, you will only find 'complete' examples, where \(\displaystyle v=\pi/2\), say, or \(\displaystyle 2\pi\).

Personally, I think the best way to tackle it would be to regard it as a function of \(\displaystyle e\), the differentiate wrt to e, if lucky find a closed form, and then re-integrate...

I suspect your answer should include elliptic integrals (incomplete ones at that, unless \(\displaystyle v=1\)).

 

[Nagarjuna]

How can I view full answer it's urgent, i have already registered

 

[topsquark]

How can I view full answer it's urgent, i have already registered

We don't often give out full answers, we help you all learn to find it on your own. Since you are reviving a 5 year old problem why don't you tell us what you don't understand about what has been posted so far.

-Dan