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How do I Integrate This?

  1. May 2, 2005 #1
    (x) sq. root of ((x^2) - 1)?
     
  2. jcsd
  3. May 2, 2005 #2
    notice that the derivative of the inside is the same power as the one outside...

    hmm, i'm thinking substitution... lol

    let u = x^2 - 1
    du = 2x dx
    dx = du/2x

    therefore you will then get (in terms of u)

    1/2* (integral of sqrt(u))

    [tex]1/2\int\sqrt{u}du[/tex]


    or somethign like that..

    hope you can do the rest

    PS, is this an ap calc question?
     
    Last edited: May 2, 2005
  4. May 2, 2005 #3

    Zurtex

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    I would use the wonderful substiution:

    [tex]x = \sec (u)[/tex]
     
  5. May 2, 2005 #4
    Or you could notate like this...

    [tex]u = x^2-1[/tex]

    [tex]du = 2x dx[/tex]

    [tex]\frac{du}{2} = x dx[/tex]
     
    Last edited: May 3, 2005
  6. May 2, 2005 #5
    can you explain why it is wrong?

    i only know through calc bc, got the ap tommorow... if that explains it or not...

    it would make sense to let dx = du/2x, so that i could just substitute it in

    can't i treat dx and x as seperate variables?
     
  7. May 2, 2005 #6
    I believe Jameson is just worried about your notation. du/(2x)=dx would be okay.

    As well, Zurtex' method will work, but it's not necessary as the other suggestion in this thread is easier.
     
    Last edited: May 2, 2005
  8. May 2, 2005 #7

    HallsofIvy

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    It's not wrong, either way works:

    xdx= du/2 so that [tex]x\sqrt{x^2-1)[/tex] becomes u1/2du/2 or

    dx= du/2x so that [tex]x\sqrt{x^2-1}[/tex] becomes xu1/2du/2x and the x's cancel.
     
  9. May 3, 2005 #8

    dextercioby

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    I would suggest the delicious substitution

    [tex]x=\cosh t [/tex]

    Daniel.
     
  10. May 3, 2005 #9
    Ah, sorry, your notation does work. It just takes one extra step of cancelling the x's. I wasn't paying close enough attention. I fixed my post. Sorry for the messup.

    Jameson
     
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