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Need integral help, i think its trig sub., thanks!

  1. May 30, 2014 #1

    Hi all, I'm stuck on this one. Sure theres an easier way to do it. Don't have my calc book currently. Thanks!
  2. jcsd
  3. May 30, 2014 #2
    Put x2 = t.
  4. May 30, 2014 #3
    Got it with the trig sub [tex]x=z \tan{\theta}[/tex]
  5. May 30, 2014 #4
    No, that substitution isn't convenient,

    If you substitute x2 = t, you would get the numerator as dt (since dt= 2xdx)
  6. May 30, 2014 #5
    Thanks! That is slightly easier!
  7. May 30, 2014 #6


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    I would have used [itex]t= x^2+ z^2[/itex] so that [itex]dt= 2x dx[/itex], [itex](1/2)dt= xdx[/itex].

    The integral becomes
    [tex]\frac{1}{2}\int_{z^2}^{z^2+ a^2} t^{-3/2}dt[/tex]
  8. May 30, 2014 #7
    Thanks! Thats slightly easier still!
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