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Integrating a complicated polynomial

  1. Aug 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi
    An integration question:
    t= 1/(1+r2)

    Can you please show me how to integrate with respect to r. Thank you.



    3. The attempt at a solution
    ∴t = (1+r2)-1
    I then tried substituting u = 1+r^2 but that didn't work!
    Is there a trick with this?
     
  2. jcsd
  3. Aug 11, 2012 #2
    Well I'll hint and say it has to do with an inverse trig function.
     
  4. Aug 11, 2012 #3

    eumyang

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    Just so I understand, you're trying to find the integral of this?
    [tex]\int \frac{1}{1+r^2}\,dr[/tex]
    BTW, what you have after the t is not a polynomial, but a rational expression.
    Yes, it's called trigonometric substitution. Try r = tan u.
     
  5. Aug 12, 2012 #4
    eumyang...thank you...yes thats what i meant.
     
  6. Aug 12, 2012 #5
    NO idea! I tried r= tan u
    1/(1+(tan u)^2)...
    Too me that just seems even more worse!

    Then (LET: < be integral sign/ integrand)
    [itex]
    = <1/(1+u) dr
    = <(1/(1+u)) (secu)^2 du
    = <((secu)^2/(1+u)) du [/itex]
    Stuck...

    and my latex is not working as well? Have I typed the wrong code?
     
    Last edited: Aug 12, 2012
  7. Aug 12, 2012 #6

    eumyang

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    No no, put the two attempts together:
    [tex]\int \frac{1}{1+r^2}\,dr = \int \frac{1}{1+\tan^2 u}\,\sec^2 u \, du[/tex]
    You know your trig identities, right?

    EDIT: Regarding LaTeX, that's because you're using the wrong tags. It should be tex or itex in the brackets, not latex.
     
  8. Aug 12, 2012 #7
    Delete this. See below
     
    Last edited: Aug 12, 2012
  9. Aug 12, 2012 #8
    Oh wait is the answer:
    u because 1+tan^2 u = sec^2 u
    then
    [tex] = ∫1 du
    =u + C [/tex]
    =tan(r)^-1 + C

    Thank you for your help.
     
    Last edited: Aug 12, 2012
  10. Aug 12, 2012 #9

    eumyang

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    If you mean
    [tex]\int 1[/tex]
    ...then you're missing something.
    Do you mean
    [itex]\tan r^{-1} + C[/itex]
    or
    [itex]\tan^{-1} r + C[/itex]?
     
  11. Aug 12, 2012 #10
    Opps... I meant [tex] tan-1 r+C [/tex]

    Thank you so much for your help. Sorry for being lazy.
     
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