Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Evaluating indefinite integral - toughie!

  1. Oct 27, 2005 #1
    Evaluating indefinite integral -- toughie!

    I have the velocity function v(x) = [(k*x^2)/(2*m)] + v0
    I need to integrate this to get position as a function of time.
    So v = dx/dt.
    Separating variables, I get t = Integral [2m/(2mv0 + kx^2)]
    Here's where I'm stuck...If i pull out the 2m, then I get something that resembles the integral which equals arc tangent.
    Am I on the right track?
    Can someone check my though process?
  2. jcsd
  3. Oct 27, 2005 #2
    I assume m and k are constants. If they are, this integral seems easy.

    [tex]x= \frac{k}{2m}\int x^2 +v_{0}dx[/tex]

    EDIT: Mistake. Sorry. Look at the below post.
    Last edited: Oct 27, 2005
  4. Oct 27, 2005 #3


    User Avatar
    Science Advisor

    Jameson: You understand, don't you, that your response makes no sense? You have x on one side and an integral with respect to x on the other! What happened to t?

    We are given
    [tex]v(x)= \frac{dx}{dt}= \frac{kx^2}{2m}+ v_0[/tex]
    That gives, as don_anon25 says
    [tex] \int\frac{dx}{\frac{kx^2}{2m}+ v_0}= \int dt[/tex]
    Yes, that does look like an arctangent. What's wrong with that?
    If we factor out v0 on the left, we get
    [tex]\frac{1}{v_0}\int\frac{dx}{\frac{k}{2mv_0}x^2+ 1}= t+ C[/tex]
    Now make the substitution
    [tex] u= \sqrt{\frac{k}{2mv_0}}x[/tex]
    and the integral on the left becomes
    [tex]\sqrt{\frac{k}{2mv_0^3}}\int{\frac{du}{u^2+ 1}[/tex]
    which is, indeed, an arctangent.
    Last edited by a moderator: Oct 27, 2005
  5. Oct 27, 2005 #4
    Sorry bout that. Wasn't thinking obviously. Thanks for the correction :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook