1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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 integrator, 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?
     
  2. jcsd
  3. Apr 1, 2006 #2

    TD

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

  6. Apr 1, 2006 #5
    Got it done. Thanks!
     
  7. Apr 2, 2006 #6

    TD

    User Avatar
    Homework Helper

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

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Yet another motion integral
Loading...