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Integration By Substitution It's Been A While

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data

    It's been god knows how long since I've had to use integration by substitution. I've totally forgotten it. I am trying to integrate to solve for the value of an electric field at a given point. The integral I am trying to solve is:

    (2kz/4(pi)(epsilon_0)*1/(z^2+x^2)^(3/2) dx.

    I know the answer is (2kz/4(pi)(epsilon_0)*(x/[z^2(z^2+x^2)^(1/2)]

    2. Relevant equations

    I'm sure I have to set u=(z^2+x^2). This makes du = 2x.

    3. The attempt at a solution

    I'm confused as to what to do now. The equation I'm integrating doesn't have an x in it anywhere. I don't think I can say du/2x=dx because I will have x's and u's in the integral, which is no good. However, I can't just ignore it.

    Also, how did that z^2 get in there on the bottom? z is a constant in this integration and since u = z^2+x^2, the z-term drops right out. I feel terribly lost.
  2. jcsd
  3. Feb 1, 2009 #2


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    That's really confusing. Why don't you just post the actual problem you are working on and how you tried to solve it?
  4. Feb 1, 2009 #3
    Sure thing.

    I'm trying to find the electric field at an arbitrary distance z above a straight line segment, where the arbitrary distance z is measured above one of the endpoints of the line segment.

    Relevant Equations:

    We are given that the electric field of a line charge is [tex]\frac{1}{4 \pi \epsilon_0} \int_P \frac{\lambda (R)}{r^2}dl[/tex].

    Attempt At Solution.

    A little element of the electric field is going to be pointed in two directions. One will be in the z-direction. The other will be in the direction parallel to the line. Using the geometry of the problem, I found that

    [tex] dE=\frac{1}{4 \pi \epsilon_0}\frac{\lambda dx}{r^2}(cos(\theta) \textbf{z}+sin(\theta)\textbf{x})=\frac{\lambda}{4 \pi \epsilon_0}\frac{\lamda dx}{r^3}(z\textbf{z}+x\textbf{x})[/tex], where the bold indicates unit vectors.

    I split this up into two integrals. This is where I have to integrate by parts, and where I get stuck. I have for instance, one integral which is [tex]\frac{1}{4 \pi \epsilon_0} \int_0^L \frac{2 \lambda z}{(z^2+x^2)^{3/2}}}dx[/tex].

    I set my u = (z^2+x^2) and my du is then 2xdx. I am confused because there is no x in my numerator, and I can't just say du/2x=dx because then I am going to have both x's and u's in my equation when I integrate it.
  5. Feb 2, 2009 #4


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    Well, if [itex]u = (z^2+x^2)[/itex], then [itex]2x=2\sqrt{u-z^2}[/itex] right?....but I don't think that makes the integral any easier!

    Try the substitution [tex]u=\frac{x}{\sqrt{x^2+z^2}}[/tex] instead :wink:
  6. Feb 2, 2009 #5
    Ah, I got it, I think.

    If I use the u-substitution you suggest, I get [tex]du=\frac{1}{\sqrt{x^2+z^2}}-\frac{x^2}{(x^2+z^2)^{3/2}}dx[/tex], which, getting a common denominator yields:


    So [tex]\frac{du}{z^2}=\frac{dx}{(x^2+z^2}dx[/tex]

    My integral is then just [tex]\frac{\lambda z}{4 \pi \epsilon_0}\int_a^b \frac{du}{z^2}[/tex] which, after re-substituting for u back into x's, and plugging in the bounds 0 and L, I get

    [tex] \frac{1}{4 \pi \epsilon_0} \frac{\lambda z L}{z^2(z^2+x^2)^{3/2}}[/tex].

    Thanks, though I still wonder how I would have thought of that particular u-substitution on my own!
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