How to integrate 1/(1-A*cosx)^3?

[Arash.]

Hello every body.
I'm searching for a solution to integrate the function below but i couldn't find anything suitable yet.I don't know even how to start it !

can anybody help me finding the integral?
f(x)=1/(1-A*cosx)^3

Thanks you ...

 

[MarcusAgrippa]

Simplify the denominator with a substitution. Try the obvious substitution.

 

[Emmanuel_Euler]

here is the answer, but i have no idea how to solve this integral.
http://www.wolframalpha.com/input/?i=1/(1-A*cosx)^3&dataset=&equal=Submit

 

[MarcusAgrippa]

Ok. Judging from Emmanuel_Euler's Mathematica solution, evaluating the integral you wrote out above is certainly not a freshman exercise. Are you sure that you have copied out your integral correctly? Or is this something that arose in more advanced work?

 

[bcrowell]

Is this homework? If so, then see https://www.physicsforums.com/help/homeworkhelp/ .

One way to approach this would be to express the cosine using complex exponentials, then use the obvious substitution to get the integral into the form $\int du/P(u)$, where P is a polynomial. This might then be doable using partial fractions.

 

[Emmanuel_Euler]

Ok. Judging from Emmanuel_Euler's Mathematica solution, evaluating the integral you wrote out above is certainly not a freshman exercise. Are you sure that you have copied out your integral correctly? Or is this something that arose in more advanced work?

yes,i copied correctly. why??

 

[certainly]

yes,i copied correctly. why??

I believe the question was intended for the OP.

 

[Arash.]

Thanks you every body ,
But yet i couldn't find the right way , and the integral i wrote is correct.
wolframalpha also doesn't show step by step guide.

 

[certainly]

wolframalpha also doesn't show step by step guide.

It does usually, but with this one it seems to take forever to load.
Anyway, you saw the answer, are you comfortable with that sort of an answer ?
[EDIT:- Ah, I see what you mean, after loading it said the solution was not avaliable, this can happen sometimes with hard integrals like the one you provided.]