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I have to integrate int_0^1 sqrt(t^2-1)dt

  1. Oct 21, 2005 #1
    I was wondering how to solve this integral:

    [tex]\int_{0}^{1}\sqrt{t^2-1}\,dt[/tex]

    When I punch it into mathematica, it gives:

    [tex] 1/2 t\sqrt{-1+t^2}-1/2\log{(t+\sqrt{-1+t^2})} [/tex]

    I was wondering what steps are done to get this result

    I suppose I forgot to enter it in as a definate integral, but still...
     
    Last edited: Oct 21, 2005
  2. jcsd
  3. Oct 21, 2005 #2

    benorin

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    Note that I used maple to find this substitution.

    Start with [tex]\int\sqrt{t^2-1}\,dt[/tex] and apply the substitution [tex]t=\sec(\theta)\Rightarrow dt=\tan(\theta)\sec(\theta)d\theta[/tex] which gives [tex]\sqrt{t^2-1}=\tan(\theta)[/tex] so that the integral becomes [tex]\int\tan^{2}(\theta)\sec(\theta)d\theta[/tex], you should be able to work it from there.
     
  4. Oct 22, 2005 #3

    VietDao29

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    Or you can try another way:
    Let:
    [tex]\sqrt{t ^ 2 - 1} = x - t[/tex]. Differentiate both sides gives:
    [tex]\frac{t dt}{\sqrt{t ^ 2 - 1}} = dx - dt[/tex]
    [tex]\Leftrightarrow \left( \frac{t}{\sqrt{t ^ 2 - 1}} + 1 \right) dt = dx[/tex]
    [tex]\Leftrightarrow \left( \frac{t + \sqrt{t ^ 2 - 1}}{\sqrt{t ^ 2 - 1}} \right) dt = dx[/tex]
    But you should note that:
    [tex]\sqrt{t ^ 2 - 1} + t = x[/tex]
    That gives:
    [tex]\Leftrightarrow \left( \frac{x}{\sqrt{t ^ 2 - 1}} \right) dt = dx[/tex]
    [tex]\Leftrightarrow \frac{dt}{\sqrt{t ^ 2 - 1}} = \frac{dx}{x}[/tex]
    Integrate both sides gives:
    [tex]\int \frac{dt}{\sqrt{t ^ 2 - 1}} = \int \frac{dx}{x} = \ln x = \ln (\sqrt{t ^ 2 - 1} + t)[/tex].
    -----------------
    In general, you can show that:
    [tex]\int \frac{dt}{\sqrt{t ^ 2 + \alpha}} = \ln (\sqrt{t ^ 2 + \alpha} + t)[/tex].
    -----------------
    You can try to integrate [tex]\int \sqrt{t ^ 2 - 1} dt[/tex] by parts, then you may need to use [tex]\int \frac{dt}{\sqrt{t ^ 2 + \alpha}} = \ln (\sqrt{t ^ 2 + \alpha} + t)[/tex].
    Viet Dao,
     
    Last edited: Oct 22, 2005
  5. Oct 22, 2005 #4

    arildno

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    The simplest is possibly to use the substitution [itex]t=Cosh(u)[/itex]
    However, your limits can't be right, unless you're dealing with a complex integral, rather than a real one.
     
  6. Oct 22, 2005 #5
    hmm... all these help a lot, thanks people!
     
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