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

[NoobixCube]

Hi all,
I have checked many integral tables and done a lot of working, but I can't seem to derive a solution for:

<br /> \int \frac{1}{(1+a cos(\theta - \phi))^2} d\theta<br />

where a is constant.
Any help would be most appreciated

 

[n0_3sc]

It would be good for us too see an attempt at the problem...there aren't many people that are kind enough to give you the answer straight up.

 

[n0_3sc]

Possibly Helpful Hint:
1) expand the denominator
2) look up the derivative of tan^{-1}{(\theta + \phi)}

See if you can link the two together...

 

[NoobixCube]

Bet you cant

 

[n0_3sc]

hahahaha .
nice.

 

[grmnsplx]

yeah I've come across this before in an "applied mathematics" course.
the integral is going to be really big and ugly and involve a lot of hyperbolic arctangents

the real crappy part is that while you're writing out reams of expressions you'll be thinking the whole time that there must be some cute simplification for all this. I mean it really looks like there should be, but there just isn't.

use some software to solve this one. that's why we invented it.

 

[grmnsplx]

oh, by the way the derivative of tan^{-1}{(\theta + \phi)} = 1/(1 +(\theta - \phi)^2)

p.s. sorry, my latex skills are weak

 

[mrandersdk]

tried it in Mathematica it is a nasty solution if it is correct

-\frac{2 ArcTanh[\frac{(a-1)Tan(\frac{\Theta-\phi}{2})}{\sqrt{a^2-1}}]}{(a^2-1)^{3/2}} + \frac{a * Sin(\Theta-\phi)}{(a^2-1)(1+a*Cos(\Theta-\phi))}

 

[grmnsplx]

wow! that's pretty if you ask me

 

[NoobixCube]

Thanks for all the input guys, I used MATLAB and Mathematica, both give different solutions...
I have heard rumors that Mathematica can sometimes give wrong answers, can anyone support this claim?

 

[NoobixCube]

Oh, by the by, maybe I should have said that a<1 ... this may affect the first term of the solution from Mathematica

 

[n0_3sc]

at least a tan^{-1}{(...)} appears in the answer...too bad I can't figure out how or why.

 

[John Creighto]

at least a tan^{-1}{(...)} appears in the answer...too bad I can't figure out how or why.

Can you do a partial fraction expansion?

 

[ice109]

Thanks for all the input guys, I used MATLAB and Mathematica, both give different solutions...
I have heard rumors that Mathematica can sometimes give wrong answers, can anyone support this claim?

just differentiate each answer and see which one is the right one

 

[n0_3sc]

Can you do a partial fraction expansion?

I thought about that too. I think I did attempt it and got stuck somewhere...I can't remember anymore.

 

[NoobixCube]

I will opt for differentiating both
thanks

 

[NoobixCube]

Would anyone know an easy method to invert this function


<br /> t(\theta)=-\frac{2 ArcTanh[\frac{(a-1)Tan(\frac{\Theta-\phi}{2})}{\sqrt{a^2-1}}]}{(a^2-1)^{3/2}} + \frac{a * Sin(\Theta-\phi)}{(a^2-1)(1+a*Cos(\Theta-\phi))}<br />

to
<br /> \theta(t)<br /> <br />
?

 

[n0_3sc]

Construct a Matrix of t(\theta), do the inverse then the determinant. - i think that's how its done...its been too long.

 

[NoobixCube]

what would be the elements of the matrix t(\theta)?
I have never heard of this method before...

 

[ice109]

Construct a Matrix of t(\theta), do the inverse then the determinant. - i think that's how its done...its been too long.

? that's for a matrix not for a function

what would be the elements of the matrix t(\theta)?

why do you need to invert that answer? your integrand in the original post is in terms of theta?

 

[n0_3sc]

oh wait a sec...
I don't think that method works here because the theta's are in the sin/cos and tan terms...Maybe i'll just be quite and wait for someone smarter to help you.

 

[n0_3sc]

? that's for a matrix not for a function

yeah I remember doing it for multiple equations with multiple variables...

 

[ice109]

yeah I remember doing it for multiple equations with multiple variables...

except there's only variable here...

here is my mathematica notebook

 

[NoobixCube]

? that's for a matrix not for a function


why do you need to invert that answer? your integrand in the original post is in terms of theta?

Well the original question was a solution to an ODE, but I need the inverted form, so the solution can be of use to me

 

[ice109]

meh use a numerical nonlinear solver.

 

[mrandersdk]

how precise do you need it? If you only need an approximation, plot the answer in a (t(theta),theta) coordinatesystem and make a fit.

I doubt you will find an analytic inverse.

 

[ice109]

give me values for a and phi and i'll make you a plot

 

[NoobixCube]

a=0.927
and
\phi = 278 degrees

 

[NoobixCube]

how precise do you need it? If you only need an approximation, plot the answer in a (t(theta),theta) coordinatesystem and make a fit.

I doubt you will find an analytic inverse.

What is the usual course of action if there isn't an analytical inverse?

 

[n0_3sc]

What is the usual course of action if there isn't an analytical inverse?

Ignore the problem and move on.
Its not like this is part of a Masters Thesis Research...

 

[ice109]

What is the usual course of action if there isn't an analytical inverse?

depends on what you want to use the inverse for?

Ignore the problem and move on.
Its not like this is part of a Masters Thesis Research...

what range for theta?

 

[NoobixCube]

say from 0 4pi , or any range that might show the functions periodicy

 

[n0_3sc]

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.

 

[ice109]

take this as a hint; get mathematica

 

[n0_3sc]

What is the usual course of action if there isn't an analytical inverse?

Apparently its "Get Mathematica".

 

[Gib Z]

If you mean elementary anti derivative, then a) If you have limits of integration, numerical methods, or b) Define it as a new function ! =]

 

[NoobixCube]

Thanks for taking the time ice109

 

[coomast]

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.