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Is there a u-substitution that works?

  1. Mar 8, 2008 #1
    1. The problem statement, all variables and given/known data
    What technique should I use to evaluate:

    [tex] \int_0^L \frac{1}{(x^2+z^2)^{3/2}} dx [/tex]

    Is there a u-substitution that works?

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Mar 8, 2008
  2. jcsd
  3. Mar 8, 2008 #2
    Trigonometric substitution

    [tex]dx=z\sec^{2}\theta d\theta[/tex]

    Assuming you're integrating with respects to x.
  4. Mar 8, 2008 #3
    With respect to what variable are you integrating? x or z?

    Try pulling out the variable you're integrating w.r.t. and you should be able to do a u-substitution then.
  5. Mar 8, 2008 #4
    Sorry. I fixed it.

    What do you mean by that?
  6. Mar 8, 2008 #5
    Pull out x from the ()^3/2 term, and do a you can do a u-substitution from there.
  7. Mar 8, 2008 #6
    I don't see how you can "pull out" the x from ()^(3/2) term. Anyway, I did the trig substitution that rocophysics recommended and that worked nicely.
  8. Mar 8, 2008 #7

    u-sub the 1+z^2.... and you'll get a 1/u^3/2 integral

    You get the same answer as with a trig substitution, but you will have to multiply by x/x in order to match the answer from trig substitution.
  9. Mar 8, 2008 #8
    Let [itex]u = 1+z^2 x^{-2}[/itex]. Then [itex]du = -2 z^2 x^{-3}[/itex]

    Then say the integral is from -L to L. What is wrong with this:

    [tex]\int_{-L}^L \frac{1}{(x^2+z^2)^{3/2}} dx = \int_{-L}^L \frac{x^{-3}}{(1+z^2x^{-2})^{3/2}} dx = \int_{1+z^2 L^{-2}}^{1+z^2 L^{-2}}\frac{u^{-3/2}}{-2 z^2} du [/tex]

    That integral has to be 0 because the limits are the same. But that makes absolutely no sense because the LHS integral is positive everywhere on the interval that you are integrating over! What in the world is going on here!!!!
    Last edited: Mar 8, 2008
  10. Mar 8, 2008 #9
    you changed the limits, your original integral is from 0 to L, so the new limits should be from 0 to 1+(z^2)/L^2
  11. Mar 8, 2008 #10

    Gib Z

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    [itex]x= z \sinh u[/itex] works out fastest here, I can do it in my head with that one.
  12. Mar 8, 2008 #11
    I know these are not the original limits. My question is now about the substitution rule. Please tell me why the equation I wrote in my last post is bogus.

    Do you really mean hyperbolic sine?
    Last edited: Mar 9, 2008
  13. Mar 9, 2008 #12

    Gib Z

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    When we use the use the substitution rule is the normal direction, x= f(u), it doesn't matter, but when doing it in reverse, u= g(x), we have to change the bounds into values of u from x, so we must ensure g(x) is one-to-one in the interval of integration, otherwise we run into exactly the same problem here. To do this, we note the integral is equal to two times the integral from 0 to L, and then we may continue since in this interval, our substitution is 1 to 1.

    And yes, i really meant Hyperbolic sine.
  14. Mar 9, 2008 #13
    How do you handle the fact that u blows up at x = 0? I think that u substitution fails because u has to be differentiable on the interval you are differentiating over and your u is not even defined on that interval.
    Last edited: Mar 9, 2008
  15. Mar 9, 2008 #14
    if you integrate that using that u-substitution you get:

    [tex]\int \frac{x^{-3} dx}{(1+z^2x^{-2})^{3/2}}=\frac{-1}{2z^2}\int\frac{du}{u^{3/2}}=\frac{1}{z^2}\frac{1}{\sqrt{1+z^2x^{-2}}} +C[/tex]

    multiply by 1, x/x and you get

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

    which is what you would get if you do the integral any other way.

    If you then use that final equation with the limits you will get the same answer doing it any other way.

    But you're right if you do a definite integral with 0 as one of the limits it will come out incorrectly because you need to multiply by x/x and with 0 you'd have to multiply by 0/1 so as not to multiply by 0/0...
  16. Mar 9, 2008 #15
    I cannot even do that on paper. It works out nicely until you half to back-substitute the limits of integration and calculate [itex]\tanh(u(L))-\tanh(u(0))[/tex].
  17. Mar 10, 2008 #16

    Gib Z

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    I usually find it easier to express the integral is terms of the original variable again, rather than change the limits of integration. Solving the substitution for u is quite straightforward:

    [tex]u= \asinh \left( \frac{x}{z} \right)[/tex].

    Putting the into the anti derivative we already had: [tex] \frac{1}{z^2} \tanh \left( \asinh \left( \frac{x}{z} \right) \right) + C[/tex]

    and now evaluating and subtracting the limits, the integral is equal to [tex]\frac{1}{z^2} \tanh \left( \asinh \left( \frac{L}{z} \right) \right)[/tex].

    This expression can be simplified further in a similar manner to how we would simplify the circular trig analogue.

    EDIT: I don't know why the arsinhs arent coming up in the latex, press the images to see what i meant.
    Last edited: Mar 10, 2008
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