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Integral using hyperbolic substitution

  1. Aug 14, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]\int\left(1+x^{2}\right)^{\frac{3}{2}}dx[/tex]

    2. Relevant equations
    The hyperbolic functions.

    3. The attempt at a solution
    We've been going over hyperbolic substitutions in class so I assume I'm meant to use one of those, but I'm just not sure how to choose which one. Any help greatly appreciated.
     
  2. jcsd
  3. Aug 14, 2011 #2

    SammyS

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    sinh2(u) + 1 = cosh2(u)

    So, I would try letting x = sinh(u) → dx = cosh(u) du
     
  4. Aug 15, 2011 #3
    [tex]\begin{eqnarray*}
    \int\left(1+x^{2}\right)^{\frac{3}{2}}dx & = & \int\left(1+\sinh^{2}\left(u\right)\right)^{\frac{3}{2}}\cosh\left(u\right)\, du\\
    & = & \int\sqrt{\left(\cosh^{2}\left(u\right)\right)^{3}}\cosh\left(u\right)\, du\\
    & = & \int\sqrt{\cosh^{6}\left(u\right)}\cosh\left(u\right)\, du\\
    & = & \int\cosh^{4}\left(u\right)\, du
    \end{eqnarray*}[\tex]
     
    Last edited: Aug 15, 2011
  5. Aug 15, 2011 #4
    Oops, that was meant to be nice neat LaTeX...

    Well, I tried doing that but ended up with int(cosh^4(u))du... not sure if there is a nice little trick for solving that or if I've ended up with a harder problem than I started with. Also, why didn't that tex code just work?
     
  6. Aug 15, 2011 #5
    Use the identity

    [tex]cosh^2 (t) = \frac{1 + cosh(2t)}{2}[/tex]


    Your tex didn't work because you used \ instead of /

    Omg i helped someone!!!
     
  7. Aug 15, 2011 #6

    SammyS

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    Changed '\' to '/' in the /tex tag so we can read your LaTeX.
     
  8. Aug 15, 2011 #7
    You could evaluate this integral without using the hyperbolic trig functions and just using the regular ones.
     
  9. Aug 15, 2011 #8

    SammyS

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    By The Way: [itex]u = \sinh^{-1}(x) = \ln \left( x+\sqrt{x^{2}+1} \right)[/itex]
     
  10. Aug 15, 2011 #9
    Aha! I was wondering how it got to that step.

    Thanks very much for your help!
     
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