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Basic Integral help

  1. Feb 18, 2009 #1
    the problem is integrate 1/(9-x^2)dx.

    For some reason I'm just drawing a blank. I've tried to fit arccot, ln, and trig sub and get nowhere. Is it part frac decomp? I don't know i think I'm just making it harder than it is.
    any insight is appreciated.
     
  2. jcsd
  3. Feb 18, 2009 #2

    Tom Mattson

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    Let's see what you've tried for a trig sub (that's how I'd do it).
     
  4. Feb 18, 2009 #3

    Mark44

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    Partial fractions would also work.
     
  5. Feb 18, 2009 #4

    Tom Mattson

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    Yes it would, but I didn't want to torture him by demanding that he type out 2 attempts. So I picked my favorite method. :D
     
  6. Feb 18, 2009 #5
    I'm using u= asinθ so sinθ= x/3 & dx = 3cosθ dθ & 3cosθ = 9-x2

    so subing in I get the integral 3cosθ/3cosθ dθ which reduces to one and equates to x after integration.

    The thing is this is really an improper integral problem
    Integral from 0 to 3 of 1/(9-x2)
    so if I get x when I integrate I have

    lim [x]a0
    a-->3-

    so the lim is 3 but this doesn't seem right
     
  7. Feb 18, 2009 #6

    Tom Mattson

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    Sure, that's just what I would have done.

    No, that's not right. It should be the following:

    [tex]3\cos(\theta)=\sqrt{9-x^2}[/tex]

    This implies the following.

    [tex]9\cos^2(\theta)=9-x^2[/tex]

    Do you see why?
     
  8. Feb 18, 2009 #7
    oooooooohhhhh it doesn't fit the form the way i was trying to use it. i probably should have looked up the form before i got so frustrated. oh well ill know next time thanks for the help:)
     
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