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Second degree function under root - integral

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Integral:

    [tex]\int\sqrt{1+\frac{1}{a^2+x^2}}\,\text dx[/tex]

    2. Relevant equations



    3. The attempt at a solution

    I don't know any method at the moment. Maybe Euler substitution? But this integral is [tex]\int\sqrt{\frac{1+a^2+x^2}{a^2+x^2}}[/tex] after making some calculation, but it is still not similar to any other I have seen in the past.
     
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  3. Feb 10, 2013 #2

    Dick

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    That doesn't look like an elementary integral to me. More like an elliptic integral. I don't think the usual techniques will get you very far.
     
  4. Feb 10, 2013 #3

    LCKurtz

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    Indeed. Maple gives a rather complicated expression in terms of ellipitic functions.
     
  5. Feb 10, 2013 #4
    The original problem was: calculate y coordinate of mass center of line: [tex]y=\text{arsinh}\,\frac xa[/tex]

    I need to compute:

    [tex]y_a=\frac{\int_Ly\,\text dl}{\int_L\text dl}[/tex]

    and I got this integral from first post. Any other method for mass center?

    Thanks for answers by the way :)
     
  6. Feb 10, 2013 #5

    SammyS

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    What is the wording of the original problem?
     
  7. Feb 10, 2013 #6
    It's other language... in translation, I need to compute y coordinate of mass center.
     
  8. Feb 10, 2013 #7

    SammyS

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    So, are you to compute the center of mass of a uniform wire lying along the curve, [itex]\displaystyle \ \ y=\text{arcsinh}\,\frac xa\ ?[/itex]

    What are the endpoints?
     
  9. Feb 10, 2013 #8
    Yeah. [tex]x\in[0,b][/tex]

    Thanks for help :)
     
  10. Feb 10, 2013 #9

    SammyS

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    If [itex]\displaystyle \ \ y=\text{arcsinh}\,\frac xa\,,\ [/itex] then [itex]\displaystyle \ \ x=a\, \sinh(y)\ .[/itex]

    You can compute the line integral in terms of either x or y -- or some other parameter for that matter.

    [itex]\displaystyle d\ell^2=dx^2+dy^2\ .[/itex]

    [itex]\displaystyle d\ell=\sqrt{1+\left(dy/dx\right)^2\ }\ dx=\sqrt{1+\left(dx/dy\right)^2\ }\ dy[/itex]
     
  11. Feb 11, 2013 #10
    Thanks for help:) but then I have [tex]\sqrt{1+\cosh^2x}[/tex] and nothing to do with it...
     
  12. Feb 11, 2013 #11

    LCKurtz

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    As has been pointed out before, the original integral gives a formula in terms of an elliptic function. A simple change of variable isn't going to change the nature of the problem.
     
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