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Homework Help: Yet another motion integral

  1. Apr 1, 2006 #1
    [tex]\int \frac{dx} { \sqrt{ \frac{1}{x} - \frac{1}{b}} }[/tex]
    where b is a positive constant.
    Using http://integrals.wolfram.com" [Broken], I know the result:
    [tex]-\sqrt{bx} \sqrt{b-x} + b\sqrt{b} \tan^{-1}( \frac{\sqrt{x}} {\sqrt{b-x}})[/tex]
    but how do I calculate it from scratch?
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Apr 1, 2006 #2

    TD

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    I'd substitute

    [tex]
    \frac{1}{x} - \frac{1}{b} = y^2 \Leftrightarrow \frac{1}{x} = y^2 + \frac{1}{b} \Leftrightarrow x = \frac{b}{{by^2 + 1}} \Leftrightarrow dx = \frac{{ - b^2 y}}{{\left( {by^2 + 1} \right)^2 }}
    [/tex]

    It gets rid of the square root and returns a rational function in y.
     
  4. Apr 1, 2006 #3
    After a few conversions, I ended up with this
    [tex]C \int \frac{du}{(u^2 + 1)^2}[/tex]
    Which I couldn't solve. I tried switching
    [tex]u = \sinh(t), du = \cosh(t)dt[/tex]
    and
    [tex]\int \frac{dt}{\cosh^3(t)}[/tex]
    but got stuck again.
     
  5. Apr 1, 2006 #4

    TD

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    Last edited by a moderator: Apr 22, 2017
  6. Apr 1, 2006 #5
    Got it done. Thanks!
     
  7. Apr 2, 2006 #6

    TD

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    You're welcome!
     
  8. Apr 3, 2006 #7

    dextercioby

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    The way i see it, there's no need for hyperbolic trigonometry.

    [tex] \int \frac{\sqrt{bx}}{\sqrt{b-x}} \ dx [/tex]

    becomes under the substitution [tex] \sqrt{x} =\sqrt{b} \sin t [/tex]

    an integral ~ to [itex] \int \sin^{2} t \ dt [/itex].

    Daniel.
     
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