Frustrating Integral

1. Mar 10, 2008

NoobixCube

Hi all,
I have checked many integral tables and done alot of working, but I cant seem to derive a solution for:

$$\int \frac{1}{(1+a cos(\theta - \phi))^2} d\theta$$

where a is constant.
Any help would be most appreciated :yuck:

2. Mar 10, 2008

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.

3. Mar 10, 2008

n0_3sc

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

See if you can link the two together...

4. Mar 10, 2008

NoobixCube

Bet you cant

5. Mar 10, 2008

n0_3sc

hahahaha :rofl: .
nice.

6. Mar 10, 2008

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.

7. Mar 10, 2008

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

Last edited: Mar 10, 2008
8. Mar 11, 2008

mrandersdk

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

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

9. Mar 11, 2008

grmnsplx

wow! that's pretty if you ask me

10. Mar 11, 2008

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?

11. Mar 11, 2008

NoobixCube

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

12. Mar 11, 2008

n0_3sc

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

13. Mar 12, 2008

John Creighto

Can you do a partial fraction expansion?

14. Mar 12, 2008

ice109

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

15. Mar 12, 2008

n0_3sc

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

16. Mar 12, 2008

NoobixCube

I will opt for differentiating both
thanks

17. Mar 12, 2008

NoobixCube

Would anyone know an easy method to invert this function

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

to
$$\theta(t)$$
?

18. Mar 12, 2008

n0_3sc

Construct a Matrix of $$t(\theta)$$, do the inverse then the determinant. - i think thats how its done...its been too long.

19. Mar 12, 2008

NoobixCube

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

Last edited: Mar 12, 2008
20. Mar 12, 2008

ice109

???? thats 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?

21. Mar 12, 2008

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.

22. Mar 12, 2008

n0_3sc

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

23. Mar 12, 2008

ice109

except there's only variable here...

here is my mathematica notebook

Attached Files:

• Untitled-1.nb.pdf
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24. Mar 12, 2008

NoobixCube

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
:tongue:

25. Mar 12, 2008

ice109

meh use a numerical nonlinear solver.