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Integration of sqrt(x^2 - 9) ?

  1. Jan 28, 2005 #1
    This seems like it should be easy but I can't seem to wrap my brain around it right now. I'm integrating to find the area of a hyperbola cut off by the line x=4 (I assume just the nose of the hyperbola cut off by the line on the positive side)

    [tex]\frac{x^2}{9} - \frac{y^2}{4} = 1[/tex]

    Since it's cut off by the line x=4 I rearrange the equation in terms of x.
    [tex]y = \frac{2}{3}\sqrt{x^2-9}[/tex]

    I should just need [tex]\int{\sqrt{x^2-9}} dx[/tex] (mulltiplied by 2/3 and then solve for the area, of course) then, yes? The only way I've tried is substituting in x = 3sec(u) in which leaves you with (including the 2/3 in this):

    [tex]6\int\tan^2{u} sec{u} du[/tex]

    Which doesn't seem any easier. Any hints? This is not homework ask I'm self-teaching in my spare time. Hints rule but I'm not turning down a full answer either as this has been bugging me for a couple of days now. :mad:
  2. jcsd
  3. Jan 28, 2005 #2
    You can use the "integration by parts" formula.
  4. Jan 28, 2005 #3


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    How about writing it
    [tex] 3\int \sqrt{(\frac{x}{3})^{2}-1} \ dx [/tex]

    and then make the obvious substitution
    [tex] \frac{x}{3}=\cosh u [/tex]

  5. Jan 31, 2005 #4
    Hmm I've tried integration by parts a few different ways like tan^2 * sec (which ends up being too hard or circular and cancelling out), and also rearranging it in different ways first.

    And as for the cosh substitution. It requires taking the integral of sinh^2 though. I haven't been taught any way of doing that other than taking the integral of it in terms of e. When you take the integral and go to resubstitute x's back in for the u's, since u=cosh^(-1)[x/3), it seems overly complicated to do.

    Ah I don't know. I'd think my book wouldn't give me integrals that can't be solved at least somewhat normally (with what I've done so far I mean). I've searched through all my notes and I haven't had to do one like this before. The closest is substituting tan's or sin's in for sqrt(a^2-x^2) type functions or partial fractions and things.

    If anyone can lend a hand... :frown:
  6. Jan 31, 2005 #5


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    Use the substitution i prescribed and the (hyperbolic) trigonometric identity:

    [tex] \sinh^{2}x=\frac{1-\cosh 2x}{2} [/tex]

  7. Jan 31, 2005 #6
    For this integral...
    [tex]\frac{2}{3} * \int{\sqrt{x^2-9}} dx[/tex]

    Here is this answer that my TI-89 and Wolfram's Integrator give:

    [tex] \frac{2}{3} * (\frac{1}{2}x\sqrt{x^2-9} - \frac{9}{2}ln[\sqrt{x^2-9}+x])[/tex]
  8. Jan 31, 2005 #7
    Ah I will try that out. Offhand, do you know if it yields the answer in my previous post?
  9. Jan 31, 2005 #8


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    Of course it does...BTW to show that it ain't really pretty...You'll end up with the [itex] arg\cosh x [/itex] which you may want (it's not compulsory) to express it in terms of the natural logarithm...

  10. Jan 31, 2005 #9
    Thanks for the help. You are the man. BTW, What did you mean by

    do you mean for taking the integral of
    [tex]\frac{1-\cosh 2x}{2} [/tex]
    and expressing it as
    [tex] (\frac{1}{2}x\sqrt{x^2-9} - \frac{9}{2}ln[\sqrt{x^2-9}+x])[/tex]

    Excuse my newbness.
  11. Jan 31, 2005 #10


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    No,your integral,initially had the variable "x".After the substitution (cosh u),the new integration variable will be "u"...Evaluate that integral,write the antiderivative and then take the transformed (by substitution) limits,in one word (actually more :tongue2: ) apply the Fundamental Theorem of Calculus of Leibniz & Newton...

  12. Feb 1, 2005 #11
    Ahh I got it now. I guess I wasn't thinking straight when you were talking about expressing cosh in terms of it's natural log equivalent. I forgot about that identity.

    Your help is much appreciated
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