Solving Intractable Integral: \frac{1}{(1+a cos(\theta - \phi))^2}

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SUMMARY

The integral \int \frac{1}{(1+a cos(\theta - \phi))^2} d\theta presents significant challenges in deriving a solution, particularly when a is a constant. Users in the discussion recommend utilizing software tools such as MATLAB and Mathematica for assistance, noting that Mathematica may yield different or incorrect solutions. A suggested approach involves substituting \theta - \phi = t to simplify the integral, leading to \int\frac{dt}{(1+acos(t))^2}. The discussion highlights the complexity of the solution, which may involve hyperbolic arctangents and partial fraction expansions.

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  • #31
NoobixCube said:
What is the usual course of action if there isn't an analytical inverse?
depends on what you want to use the inverse for?
n0_3sc said:
Ignore the problem and move on.
Its not like this is part of a Masters Thesis Research... :-p

:rolleyes:

what range for theta?
 
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  • #32
say from 0 4pi , or any range that might show the functions periodicy
 
  • #33
Now that I look at your problem - It definitely reminds me of "Elliptical Integrals".
Wiki it and I'm sure you'll find something useful.
 
  • #34
take this as a hint; get mathematica
 

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  • #35
NoobixCube said:
What is the usual course of action if there isn't an analytical inverse?

Apparently its "Get Mathematica".
 
  • #36
If you mean elementary anti derivative, then a) If you have limits of integration, numerical methods, or b) Define it as a new function ! =]
 
  • #37
Thanks for taking the time ice109 :smile:
 
  • #38
You can solve this integral by considering it after the simple substitution:

\theta- \phi=t

as:

\int\frac{dt}{(1+acos(t))^2}

Now use the substitution:

1+acos(t)=\frac{1-a^2}{1-acos(t)}

You will easily arrive at the solution.

Take a look at an older post of me where I explain this substitution a bit more:

https://www.physicsforums.com/showthread.php?t=204639

Hope this helps.
 

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