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Dy/dx = 1/(x.sqrt(x^2-1))

  1. Apr 1, 2005 #1
    Hi,

    Can anyone help me here!

    dy/dx = 1/(x.sqrt(x^2-1)) , x>0 Note y = pi when x = 2

    Here is what I did:

    Let u = x^2 - 1
    x^2 = u + 1

    Substituting this in the above yields:

    dy/dx = 1/x.sqrt(u)

    Differentiating: u = x^2-1

    du = 2x.dx

    x = du/2.dx

    Then,

    dy/dx= 2/sqrt(u).du

    I don't know if this is rite.
    Plz help me!
     
  2. jcsd
  3. Apr 1, 2005 #2

    dextercioby

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    What are u trying to do?Find the antiderivative of [itex] \frac{1}{x\sqrt{x^{2}-1}} [/tex] ...?If so,then use a substitution involving [itex] \cosh u [/itex]...

    Daniel.
     
  4. Apr 1, 2005 #3
    Why is it that everyone has been taught about hyperbolic trig functions when we skipped it in high school and calc 1-3?
     
  5. Apr 1, 2005 #4
    I was trying to simplify the Right Hand side, before, I could integrate , by making a substitution.

    Anyway what would be the hyperbolic sub, for this involving coshu.

    I don't have my text book with me. I am at work, and I am trying to finish my homework.
     
  6. Apr 1, 2005 #5

    dextercioby

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    Need i say that your initial approach is incorrect...?

    It also works with [itex] \sec u [/itex]...

    Daniel.
     
  7. Apr 1, 2005 #6

    dextercioby

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    I wouldn't know.Self-taught hyperbolic trigonometry in HS (XII-th grade) when i had to evaluate antiderivatives using substitution...

    Daniel.
     
  8. Apr 1, 2005 #7
    Ok, then,

    Let x = secu

    dx/du = secu.tanu

    sec^2u-tan^2u = 1

    dy/dx = 1/secu.tanu


    Then what, to do!
     
  9. Apr 1, 2005 #8

    dextercioby

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    [tex] \int \frac{\sec u \ \tan u \ du}{\sec u \ \tan u} =\int du = u+C = \mbox{arcsec} \ x +C [/tex]

    Daniel.
     
  10. Apr 1, 2005 #9
    Ok, I applied the initial conditions:

    y = pi when x = 2

    and I got,

    y = sec-1(x) + pi - sec-1(2)

    can we write this as:

    y = sec-1 (x/2) + pi
     
  11. Apr 1, 2005 #10

    dextercioby

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    "sec" (as "cosine") is multivalued.But your could take

    [tex] \sec^{-1} 2=\frac{\pi}{3} [/tex]

    Daniel.
     
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