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Integral of (x^2+y^2)^(-3/2)dy

  1. Sep 15, 2014 #1
    1. The problem statement, all variables and given/known data

    Hello, I came across an unusual integral problem in my Physics book that I could not solve. The book simply skips the work and says to check the integral tables, but I wasn't able to find one that satisfied this problem.

    2. Relevant equations

    This is the problem: integral of (x^2+y^2)^(-3/2)dy

    3. The attempt at a solution

    Tried u-sub, but the lack of y makes it impossible. Couldn't do trig sub because of the negative exponent.

    When I plugged it into WolframAlpha, it gave me a tidy solution, but the solutions are unavailable even in their app. It loads for a bit and shows "solution unavailable."

    The proper solution should be y/[(x^2)*√(x^2+y^2)]+constant.

    Here is the actual URL:
    http://www.wolframalpha.com/input/?i=integral+of+(x^2+y^2)^(-3/2)dy&incTime=true

    This is the second time coming across this integral in this chapter and I cannot solve it for the life of me. Any insights would be greatly appreciated. Thanks!
     
  2. jcsd
  3. Sep 15, 2014 #2
    The x can be regarded as a constant, right? Is it part of a double integral? Can you use IBP?
     
    Last edited: Sep 15, 2014
  4. Sep 15, 2014 #3
    Yes, x could be regarded as a constant, and no, not a part of a double integral. The physics parts of the problem were constants, so I just uploaded the core variables.

    IBP didn't work for me because when I used tabular integration because the derivative and integral of each part would continue indefinitely if that makes sense.

    Also if it helps, I worked through my problem using this guide I found:
    http://faculty.wwu.edu/vawter/PhysicsNet/Topics/ElectricForce/LineChargeDer.html [Broken]

    I only switched the x and y because my problem was a horizontal version of this example.
     
    Last edited by a moderator: May 6, 2017
  5. Sep 15, 2014 #4

    B3NR4Y

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    Gold Member

    I'm not sure what you mean by trig not working due to the negative exponent. I took a shot at it and got the same result as wolfram alpha. Try to take [itex] y = x tan \theta [/itex]. so the integral becomes:

    [itex] \int \frac{1}{(x^{2} + y^{2})^{\frac{3}{2}}} dx = \int \frac{sec^{2} \theta d\theta}{(x^{2} + x^{2}tan^{2} \theta)^{\frac{3}{2}}} [/itex]

    With trigonometric identities the solution should be easy.

    Edit: wooah that LaTeX screw up, going to fix it.
     
  6. Sep 15, 2014 #5
    Yes trigoniometrics is the way to go here.
     
  7. Sep 15, 2014 #6
    Wow, that was easy. I dunno how that didn't spring to mind; I really need to brush up on my math haha.

    Thank you!!
     
  8. Sep 15, 2014 #7

    Ray Vickson

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    Science Advisor
    Homework Helper

    What do you mean that in Wolfram it gives a tidy solution, but the solutions are unavailable? When I click on your link it gives a solution.
     
  9. Sep 15, 2014 #8
    Oops, I left out a word there. Meant to say "solution steps." Usually in their app it shows the procedure of obtaining the solution. By "tidy" I meant that the end solution looked elegant and nothing like what I had on my paper at the time...
     
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