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

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The integral discussed, ∫(1/(1+a cos(θ - φ))^2) dθ, poses significant challenges, with participants noting its complexity and the potential for lengthy, intricate solutions involving hyperbolic functions. Suggestions include using software like MATLAB and Mathematica for assistance, though discrepancies in their outputs have been noted. A substitution method is proposed to simplify the integral, transforming it into a more manageable form. The conversation also touches on the difficulty of finding an analytical inverse for the resulting function, with numerical methods recommended as alternatives. Overall, the thread emphasizes the intricacies of solving this integral and the utility of computational tools in tackling such problems.
  • #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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