Integrals containing (x^2+a^2-2xa cos(theta))^(-1/2)

  • Thread starter ShayanJ
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  • #1
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Integrals containing [itex] \frac{1}{\sqrt{x^2+a^2-2xa \cos{\theta}}} [/itex] occure frequently in physics but I still have problem solving them. Is there a general method for dealing with them?(Either w.r.t. x or [itex] \theta [/itex])
Thanks
 

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  • #2
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These two integrals are commonly expressed thanks to logarithm and elliptic functions :
 

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  • #3
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Integrals containing [itex] \frac{1}{\sqrt{x^2+a^2-2xa \cos{\theta}}} [/itex] occure frequently in physics but I still have problem solving them. Is there a general method for dealing with them?(Either w.r.t. x or [itex] \theta [/itex])
Thanks
For respect to x:

Perform a trig substitution (not theta, a different variable) by first completing the square under the square root. A nice simplification will occur. Then proceed as you normally would after a trig substitution to get the first answer provided in the image provided in the post above mine.
 
  • #4
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This is how I did it :
[itex]
\int \frac{dx}{\sqrt{x^2+a^2-2xa\cos\theta}}=\int \frac{dx}{\sqrt{(x-a\cos\theta)^2+a^2\sin^2\theta}}=\frac{1}{a\sin\theta}\int \frac{dx}{\sqrt{1+(\frac{x-a\cos\theta}{a\sin\theta})}}=\sinh^{-1} \frac{x-a\cos\theta}{a\sin\theta}
[/itex]
Anyway...thanks both!
 
  • #5
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I'm positive it is a typo, but there is a missing exponent in the second last step. I haven't worked with hyperbolic functions since college so I missed that neat shortcut. Nice work.
 

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