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Help needed to solve an integral

  1. Apr 21, 2010 #1
    hi, i was readin through a text book on electrodynamics and this integral has stumped me - can someone please tell me how to solve it-thanks!


    where z is an independent variable, which can be treated as a constant
  2. jcsd
  3. Apr 21, 2010 #2
    thats meant to be the integral of 1 over (z^2+x^2)^(3/2) wrt x. dont no how to mak it bigger!!!
  4. Apr 21, 2010 #3


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    Use the fact that [itex]1+ tan^2(\theta)= sec^2(\theta)[/itex] so that [itex]\sqrt{1+ tan^2(\theta)}= sec(\theta)[/itex] to get rid of the square root.

    Specifically, let [itex]x= z tan(\theta)[/itex] so that [itex]z^2+ x^2= z^2+ z^2tan^2(\theta)= z^2(1+ tan^2(\theta))[/itex] and then [itex]\sqrt{z^2+ x^2}= z sec(\theta)[/itex].

    Of course, you will also need to set [itex]dx= z sec^2(\theta)d\theta[/itex] so the integral becomes
    [tex]\int \frac{1}{z^3 sec^3theta)} (z sec^2(\theta)d\theta= \frac{1}{z^2}\int \frac{1}{sec(\theta)}d\theta[/tex]

    Now, since [itex]sec(\theta)= 1/cos(\theta)[/itex], that integral becomes
    [tex]\frac{1}{z^2}\int cos(\theta)d\theta=\frac{1}{z^2} sin(\theta)+ C[/tex].

    Since [itex]tan(\theta)= x/z[/itex], you can imagine a right triangle with "opposite side" x and "near side" z. Then the hypotenuse has length [itex]\sqrt{x^2+ z^2}[/itex], [itex]sin(\theta)[/itex]= "opposite side over hypotenuse" or [itex]x/\sqrt{x^2+ z^2}[/itex] so the integral is
    [tex]\frac{xz^2}{x^2+ z^2}+ C[/tex]
  5. Apr 21, 2010 #4
    \int \bigl(z^{2} + x^{2}\bigr)^{-3/2} d x = \frac{x}{\sqrt{z^{2} + x^{2}} z^{2}}+C
  6. Apr 21, 2010 #5
    The above answer (from HallsofIvy) is not correct which can be verified by taking the derivative and checking.

    It's not clear if you want the answer or the method. The answer is

    [tex]{{x}\over{z^2 \sqrt{z^2+x^2}}} + C[/tex]

    Knowing the answer, you can prove it by taking the derivative and verifying. Reversing the steps gives you a method to solve, although seeing that method beforehand would not be easy and there are more elegant methods I'm sure. Perhaps there is a simple mistake in the above method and it can be salvaged.
  7. Apr 21, 2010 #6
    Just to add a comment here. I did check using the substitution x=ztan(u) as suggested by HallsofIvy. This works and leads to the correct answer. Although I didn't check the above derivation, I have to assume that there is a simple mistake. The basic suggested approach appears to be a good one.
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