Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integration Tricks

  1. Aug 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi all, as part of an electrodynamics-problem, I encountered:

    [itex]\int {\frac{x}{{{{({x^2} + 1)}^3}}} \cdot dx} [/itex]

    I know how to get the answer with my computer, but I wanted to know how to do it by hand. It’s getting to the point here in physics-grad-school that I need to know how to do this stuff “off the cuff”, so to speak.

    2. Relevant equations

    3. The attempt at a solution

    Well....this is part of a solution of a bigger problem. I know there's some sort of trig substitution or trig identity. It happens that x = r/d, where "r" is an in-plane polar distance (x^2 + y^2), and "d" is distance above the plane, and I rearranged the integral I got into getting the dimensionless length scale x = r/d ... and by the above geometry, if you consider theta to be the angle between "r" and "d", then r/d = tan(theta). But:

    dx = d(r/d) = d(tan(theta)) = (1 + tan(theta)^2)*d(theta)

    That's just a mess. :-p
  2. jcsd
  3. Aug 26, 2010 #2
    Am I missing something? Or does it really just work with u=x^2+1?
  4. Aug 26, 2010 #3
    The above coment is correct. This is a simple substitution of [tex]u=x^{2}+1[/tex]. Then the next step is a simple interation of a polynomial of degree -3.
  5. Aug 26, 2010 #4
    I assume the dot there is an accident then, it doesn't mean anything.
  6. Aug 26, 2010 #5
    Well it is obvious that the function does not define a vector field. I believe it means nothing.
  7. Aug 26, 2010 #6
    Wow....u = x^2 + 1 makes it easy-peezee, lemon squeezy. ::blush:: guess I missed it. I learned how to integrate stuff 7 years ago, and I've made the mistake of letting a computer do it for me most of the time. Bah.
  8. Aug 26, 2010 #7
    u = 1+x^2
    du = 2x dx

    -> integral = 0.5*(1/u^3)du
  9. Aug 26, 2010 #8


    Staff: Mentor

    It does mean something - it's the period at the end of the sentence.
  10. Aug 26, 2010 #9
    44 I think he was referring to the dot in OP's post, the one between his function and dx :-).
  11. Aug 26, 2010 #10
    I use dots to make scalar-multiplication clear. I recognize it as "it's ok to assume this kind of multiplication commutes", rather than the multiplication of things like operators and cross products.
  12. Aug 26, 2010 #11
    Why,by convention dot is used for other things other than regular scalar multiplication. It might prove to be confusing for some people. The truth is that I originally didn't notice it.
  13. Aug 26, 2010 #12


    Staff: Mentor

    I don't think so. kbaumen quoted your post (the first one above), not the OP with the integral in it.
  14. Aug 26, 2010 #13


    Staff: Mentor

    In this case, I believe kbaumen was referring to the period at the end of the sentence. Dots are often used for multiplication, especially with constants, as in
    [tex]3 \cdot 5 = 15[/tex].

    The first dot indicates multiplcation; the last one is the period at the end of the sentence.
  15. Aug 29, 2010 #14
    Sorry for not making it clear, but I was indeed referring to the dot between the function and dx.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook