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Integral involving hyperbolic functions

  1. Apr 15, 2012 #1

    sharks

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    The problem statement, all variables and given/known data
    Find [tex]\int \frac{x}{\sqrt{2x^2-2x+1}}\,dx[/tex]

    The attempt at a solution

    First, i complete the square for the quadratic expression:
    [tex]2x^2-2x+1=2((x-\frac{1}{2})^2+\frac{1}{4})[/tex]
    [tex]\int \frac{x}{\sqrt{2x^2-2x+1}}\,dx=\int \frac{x}{\sqrt 2 \sqrt{(x-\frac{1}{2})^2+\frac{1}{4}}}\,dx[/tex]

    Using substitution: Let [itex](x-\frac{1}{2})=\frac{1}{2}\sinh u[/itex]

    I eventually get this expression, after integrating:
    [tex]\frac{1}{2\sqrt 2}(\cosh (\sinh^{-1}(2x-1)+\sinh^{-1}(2x-1))[/tex]

    But i don't know how to simplify this any further, without getting a complicated answer. The final answer is: [itex]\frac{\sinh^{-1}1}{\sqrt 2}[/itex]
     
    Last edited: Apr 15, 2012
  2. jcsd
  3. Apr 15, 2012 #2

    tiny-tim

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    hi sharks! :smile:

    (you haven't said what the limits are)
    cosh(sinh-1x) = √(x2 + 1) :wink:

    (btw, you could have simplified the original integrand by rewriting the numerator as x - 1/2 + 1/2 :wink:)
     
  4. Apr 15, 2012 #3

    sharks

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    Hi tiny-tim! :smile:

    Congratulations on your PF award!

    Doh! I forgot to include the limits.

    [tex]\int^1_0 \frac{x}{\sqrt{2x^2-2x+1}}\,dx[/tex]

    If i evaluate over the limits:
    [tex]\frac{1}{2\sqrt 2}(\cosh (\sinh^{-1}(2x-1)+\sinh^{-1}(2x-1))[/tex]
    Indeed, i get the answer!

    I am trying to understand your suggestion:
    [tex]\cosh (\sinh^{-1}x) = \sqrt{(x^2 + 1)}[/tex]
    It's too complicated to prove? I can't get the R.H.S. from the L.H.S.

    Here is where i've reached...
    [tex]\frac{x^2+x\sqrt{x^2+1}+1}{x+\sqrt{x^2+1}}[/tex]
     
    Last edited: Apr 15, 2012
  5. Apr 15, 2012 #4

    tiny-tim

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    if y = sinh-1x

    then x = sinhy,

    so √(x2 + 1) = √(sinh2y + 1) = coshy = cosh(sinh-1x) :smile:

    (if you think about it, this is obvious! :wink:)
     
  6. Apr 15, 2012 #5

    sharks

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    I have also tried to solve this problem using your other suggestion - rewriting the numerator as x - 1/2 + 1/2, and i was able to get the answer.

    Thank you for your insightful suggestions, tiny-tim. :smile:
     
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