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Integrate 1/(x^2+y^2)^(3/2)

  1. Sep 4, 2008 #1
    Well, so I really want to integrate what's shown in the title:
    [tex]\int \frac{dx}{(x^2+y^2)^\frac{3}{2}}[/tex]

    Now, I know there are quite a few straightforward answers to this. But what I really want is how people who do math got this formula in the first place. I don't just want a formula that seems to have come from a serendipitous accident or something. Please tell me how to derive the answer.

    (You might have guessed this has something to do with electric fields)

    Thank you for helping.
    Last edited: Sep 4, 2008
  2. jcsd
  3. Sep 4, 2008 #2
    Re: 1/(x^2+y^2)^(3/2)

    Hmm, are we keeping y constant here? If so then a trig substitution should might be a good first step.
  4. Sep 4, 2008 #3

    Yes, y is a constant.

    Trig substitution? The final answer doesn't have any trig functions.
  5. Sep 4, 2008 #4

    That's because it probably involves an inverse trig function nested inside a trig function which of course would undo the trig function. Anyways you'll see what I mean if you try it.

    Anyways I would try the substitution x = y*tan(t) (Have you learned u-sub?).
    Last edited: Sep 4, 2008
  6. Sep 5, 2008 #5


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    [itex]sin^2(\theta)+ cos^2(\theta)= 1[/itex] so, dividing on both sides by [itex]cos(\theta)[/itex], [itex]tan^2(\theta)+ 1= sec^2(\theta)[/itex]. Thats why snipez90 suggested "tan(t)". Let [itex]x= ytan(\theta)[/itex]. In that case [itex]x^2+ y^2= y^2tan^2(\theta)+ y^2= y^2(tan^2(\theta)+ 1)= y^2sec^2(\theta)[/itex] so that [itex](x^2+ y^2)^{3/2}= (y^2sec^2(\theta))^{3/2}= y^3 sec^3(\theta)[/itex].

    Also, it [itex]x= y tan(\theta)[/itex], then [itex]dx= ysec^2(\theta)d\theta[/itex].

    So the integral, [itex]\int dx/(x^2+ y^2)^{3/2}[/itex] becomes
    [tex]\int \frac{y sec^2(\theta)d\theta}{y^3 sec^3(\theta)}= \int\frac{d\theta}{y^2 sec(\theta)}[/tex]
    [tex]= \frac{1}{y^2}\int cos(\theta)d\theta= \frac{1}{y^2}sin(\theta)+ C[/itex]

    Since [itex]tan(\theta)= x/y[/itex], imagine a right triangle having angle [itex]\theta[/itex], opposite side of length x, and near side of length y. Then the hypotenuse of that triangle has length [itex]\sqrt{x^2+ y^2}[/itex] and so [itex]cos(\theta)= y/\sqrt{x^2+ y^2}[/itex].

    That means that [itex](1/y^2)cos(\theta)+ C= (1/y^2)(y/\sqrt{x^2+ y^2})+ C= 1/(y\sqrt{x^2+ y^2})+ C[/itex]
  7. Sep 5, 2008 #6

    Yes I've learned substitution. If it's solvable by substitution, then I just have the problem of finding what to substitute.

    Ooh, that is one clever substitution. I never thought of that. What I don't get is why you are trying to find:

    when there's a sin(x) in the final equation. In that case, the answer would be:
    [tex]{1 \over y^2} sin(\theta) + C = {1 \over y^2} {x \over \sqrt{x^2+y^2}} + C[/tex]

    Thanks so much!

  8. Sep 5, 2008 #7

    To put it to you bluntly, this is in fact where most integration formulae come from. Things like the above proof usually only come after the discovery of the initial function.
  9. Sep 5, 2008 #8

    That's a depressing thought. :frown:

    (I digress - Is there a possibility that a Turing machine exists that calculates integrals?)
  10. Sep 6, 2008 #9

    Ben Niehoff

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    No, there is no algorithm for finding general integrals (contrast to derivatives, for which there are very clear-cut algorithms). Many integral formulas were found, as mentioned, by lucky guesses (although many of those are still arrivable deterministically, if one is clever enough).
  11. Sep 6, 2008 #10


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    In fact, this is generally true of "inverse" problems.

    If I were to define a function, say, f(x)= x5- 3x4+ 3x3- 7 x2- 4x + 5, and ask "What is f(7)?", you would only need to evaluate it-do the arithmetic- because you are already given the formula. But if I were ask "For what x is f(x)= 9" that would be a very difficult problem. It involves solving the equation and we know that, even for something as simple as a polynomial equation there may be several different solutions or no solution at all and, indeed, if the polynomial has degree 5 or more, there may be solutions that cannot be written in terms of algebraic operations.
  12. Sep 6, 2008 #11

    Oh OK. Thanks for the explanation.

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