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Homework Help: Integration help

  1. Mar 7, 2010 #1
    Hi,

    I would like some help with the following integral

    [tex]\int t^3\cdot \sqrt{1+t^2}~dt[/tex]

    I don't know what to do with the [tex]t^3[/tex] term.
    If it wasn't there it would be easy.

    Help appreciated.
     
  2. jcsd
  3. Mar 7, 2010 #2
    how about letting [tex]u=1+t^{2}[/tex]?
     
  4. Mar 7, 2010 #3
    Thats what i thought about too, at first.

    But it doesn't get me anywhere

    [tex]t^3\cdot\sqrt{u}[/tex]
     
  5. Mar 7, 2010 #4
    Well, before taking the integral, you have to have only one variable. What you need to do is to represent dt in terms of du, and t^3 in terms of u.
     
  6. Mar 7, 2010 #5
    Aha!
    I think i understand.

    So i need [tex]\int(u-1)\sqrt{u}du[/tex]

    How do i proceed now?
     
  7. Mar 7, 2010 #6
    Well, I think you are done. Don't you see it yet? Just expand and integrate.
     
  8. Mar 7, 2010 #7
    You made a mistake. since
    [tex]u=1+t^{2}[/tex]
    [tex]du=2tdt[/tex] or
    [tex]dt=du/(2t)[/tex]

    You should have 1/2 somewhere.
     
    Last edited: Mar 7, 2010
  9. Mar 7, 2010 #8
    Oh, dang i forgot about that. Thanks!

    [tex]\frac{1}{5}u^{5/2} -u^{3/2}[/tex]
     
  10. Mar 7, 2010 #9
    Don't forget 1/3
     
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