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Indefinite integral problem

  1. Dec 25, 2007 #1
    how to evaluate the indefinite integral

    ∫(x^2-1)^(-1/2) dx

    and

    ∫x^(-1)*(1-x^2)^(-1/2) dx
     
  2. jcsd
  3. Dec 25, 2007 #2

    arildno

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    Try to think like this:
    Why are these integrals hard?

    First hurdle:
    There are ugly square roots here!

    So, the next question is naturally:
    What is the simplest way to "eliminate" a square root sign?
    Answer:
    Ensuring that the radicand is a perfect square!
    Then, the root and the square would eliminate each other.

    Agreed?

    Thus, tackle the first one:
    If [tex]x^{2}-1=y^{2}[/tex]
    then [tex]x^{2}-y^{2}=1[/tex]

    Do you know of any functions that behaves like x and y here?

    Possibly, you are familiar with the hyperbolic functions, Cosh(u) and Sinh(u) which satisfies for all u the identity: [tex]Cosh^{2}u-Sinh^{2}(u)=1[/tex]

    In terms of exponentials, we have the relations:
    [tex]Cosh(u)=\frac{e^{u}+e^{-u}}{2},Sinh(u)=\frac{e^{u}-e^{-u}}{2}[/tex]

    Thus, we set [tex]x=Cosh(u)[/tex], whereby we transform the integral as follows:
    [tex]\int\frac{dx}{\sqrt{x^{2}-1}}=\int\frac{1}{|Sinh(u)|}\frac{dx}{du}du[/tex]

    Now, we have that [tex]\frac{d}{du}Cosh(u)=Sinh(u)[/tex], whereby the integral reduces to:
    [tex]sgn(Sinh(u))\int{du}=sgn(Sinh(u))u=|u|[/tex]
    since Sinh(u) and u has matching sign domains.

    Okay so far?
     
  4. Dec 25, 2007 #3

    Defennder

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    Actually, arildno I don't think we need to introduce hyperbolic functions. They might work, but you're thinking too much. I could solve both of them by using normal (non-hyperbolic) trigo substitutions.

    @OP:

    You might find this trigo identity useful:

    (tan θ)^2 = (sec θ)^2 - 1

    Once you figured out the proper trigo subst for the first one from the clue above, you're done.

    For the 2nd one, you need a different trigo substitution. The following identity is useful:

    (sin θ)^2 = 1 - (cos θ)^2
     
  5. Dec 25, 2007 #4

    arildno

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    I LIKE hyperbolic functions, that's why I use them. :smile:

    I am perfectly well aware of the sec substitution, but the result is more elegantly written in terms of hyperbolics/their inverses.
    That's why I like them..
     
  6. Dec 25, 2007 #5

    HallsofIvy

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    Defennnder, I also would tend to use trig functions, perhaps because that is what I first learned, but I noticed that arildno tends to use hyperbolic functions. It's certainly NOT a matter of "thinking too much"- one is as good as the other. One advantage I will conced to hyperbolic functions- you can use sinh and cosh rather than sec and tan which have more complicated integration and differentiation rules than sinh and cosh.
     
  7. Dec 25, 2007 #6
    Thanks arildno, i haven't thought out the hyperbolic solution to the first problem.

    yet to solve the 2nd problem,
    is there any other way rather than using substitution x=sin u ?
    I've thought out another possible substitution u=1/x, and then we could use the result of the firtst problem.
    am i right?
     
  8. Dec 25, 2007 #7

    arildno

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    That should work nicely, yes.
     
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