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Indefinite Integration

  1. Apr 6, 2005 #1
    [tex]\!\(∫x^2/Sqrt[1 - x^2] \[DifferentialD]x\)[/tex]

    I need to find the integral of
    (x^2)/ Sqrt(1-(x^2))
    if the above doesnt work properly
    integration by parts results in 0=0 how do i do this?
     
  2. jcsd
  3. Apr 6, 2005 #2

    dextercioby

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    You mean

    [tex] \int \frac{x^{2}}{\sqrt{1-x^{2}}} \ dx [/tex]

    How about the substitution [itex] x=\sin u [/tex] and then a nice trigonometrical identity involving a double angle...?

    Daniel.
     
  4. Apr 6, 2005 #3

    dextercioby

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    It can be done by parts,too.

    Daniel.
     
  5. Apr 6, 2005 #4
    ok, i got it using x= sinU
    can you give me a hint how to do it using integration by parts?
     
  6. Apr 6, 2005 #5
    Integration by parts uses

    [tex]\int u \ dv = uv - \int v \ du[/tex]

    choose [itex] u = x[/itex] and

    [tex] v = \frac{x}{\sqrt{1-x^2}}[/tex]

    Edit: That should be [itex]dv = (x / \sqrt{1-x^2}) \ dx[/itex]!
     
    Last edited: Apr 6, 2005
  7. Apr 6, 2005 #6

    dextercioby

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    Not really,Data.U needn't specify "u" & "v",but the factors in the LHS,"u" & "dv"...

    So

    [tex] u=x \ \mbox{and} \ dv=\frac{x}{\sqrt{1-x^{2}}} \ dx [/tex]

    Daniel.

    P.S.Data,u see the difference,right...?:uhh:
     
  8. Apr 6, 2005 #7
    indeed, silly me :tongue2:~
     
    Last edited: Apr 6, 2005
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