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Evaluation trig sub

  1. Nov 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Having trouble evaluating:
    [tex]\int_{\infty}^{-\infty} \frac{dz}{(z^2+x^2)^(3/2)}[/tex]

    2. Relevant equations
    Trig sub
    [tex] z=xtan(\theta)[/tex]

    3. The attempt at a solution
    comes down to:
    [tex]\int_{x\frac{\pi}{2}}^{-x\frac{\pi}{2}} cos(\theta) d\theta [/tex]
    goes to
    [tex] sin(\theta)[/tex] from [tex]x\frac{\pi}{2} \longrightarrow -x\frac{\pi}{2}[/tex]
    Last edited: Nov 4, 2007
  2. jcsd
  3. Nov 4, 2007 #2

    Gib Z

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    Homework Helper

    You can change the bounds differently. Think about it, as long x is some constant, and z approaching negative infinity and positive infinity, the tan theta has to be...
  4. Nov 4, 2007 #3
    mm...just [tex]\frac{\pi}{2}[/tex]
  5. Nov 4, 2007 #4
    is that supposed to be to the 3/2 power or is that a constant multiplying the denominator?
  6. Nov 4, 2007 #5
    Sorry it is that quantity raised to the (3/2)
  7. Nov 4, 2007 #6
    Basically I am trying to figure out how this book got
    [tex] \frac{z}{x^2\sqrt(z^2+x^2)}[/tex] with bounds from infinity to -infinity.
    They then go on to get [1--1]=2.

    I have to do this for a similar problem except my bounds will be L to -L.
    the answer is: [tex] \frac{L}{x^2 \sqrt(L^2+x^2)}[/tex] I do not know how.
    Trying to remember how to do bounds on trig subs.
  8. Nov 4, 2007 #7
  9. Nov 4, 2007 #8
    maybe divide top and bottom by z. then plug in the limits from infinity to -infinity

    I get zero??? I get (1/x^2)-(1/X^2)
  10. Nov 5, 2007 #9

    Gib Z

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    Homework Helper

    Integrating with respect to z or x? State the exact question given to you please.
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