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Re: Integral involving square root of e^x

  1. Mar 30, 2008 #1
    [SOLVED] Re: Integral involving square root of e^x

    1. The problem statement, all variables and given/known data

    [tex]\int \sqrt{1-e^{-x}}[/tex]

    2. Relevant equations

    Sub rule.

    3. The attempt at a solution

    I realised that it's fairly obvious I can use u=e^-x/2 to give [itex]\sqrt {1-u^2}[/itex]

    but I'm kind of looking at the answers and I'm not seeing how I can get to them from here?

    The answer I have from my maths program looks a bit awkward? Not sure how they got here?

    [tex]2(\sqrt{1-e^x})-2\tanh^{-1}(\sqrt{1-e^x})+c[/tex]

    Are there any other useful subs anyone can think of, or tips for this one.

    Not a homework question as such but I thought this was the best place for it.
     
  2. jcsd
  3. Mar 30, 2008 #2

    cristo

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    Well, remember that the actual interview you are trying to determine is
    [tex]\int \sqrt{1-e^{-x}}dx[/tex]. Performing the substitution [itex]u=e^{-x/2}[/itex] gives [itex]du=-\frac{1}{2}e^{-x/2}\cdot dx[/itex], yielding[tex]-2\int\frac{\sqrt{1-u^2}}{u}\cdot du[/tex]. You can integrate this by another subsitution, u=cosy, say.
     
    Last edited: Mar 30, 2008
  4. Mar 30, 2008 #3
    Thanks Christo I'm still not sure here. I'm sure I'm just being thick but could you spell it out, I'm not seeing how I'm going to end up with something even remotely like that answer, even with your sub? It's probably the hyperbolic function, I've got to admit they weren't heavily dealt with in the course I did. I can see where the answer turns into what it is with the sub cos y but the arctanh, where does that come from exactly?
     
    Last edited: Mar 30, 2008
  5. Mar 30, 2008 #4
    Thanks Christo I'm still not sure here. I'm sure I'm just being thick but could you spell it out, I'm not seeing how I'm going to end up like that answer, even with your sub? It's probably the hyperbolic function, I've got to admit they weren't heavily dealt with in the course I did. I can see where the answer turns into what it is with the sub cos y but the arctanh, where does that come from exactly? Is it some clever trig identity I'm missing

    I can see where you get

    [tex]2(1-e^x)-2tanh^{-1}(1-e^x)+c[/tex]

    from

    [tex]-2\int\frac{\sqrt{1-u^2}}{u}\cdot du[/tex] but why the arctanh?
     
    Last edited: Mar 30, 2008
  6. Mar 30, 2008 #5

    cristo

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    Hmm, I don't really know to be honest. My substitution doesn't really work, since the fraction has a "u" in the denominator and not a "u^2," which is what I must have assumed.

    I guess we'll have to wait for someone else to come along, as I can't see how to integrate that by hand, at the moment.
     
  7. Mar 30, 2008 #6

    tiny-tim

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    Hi Schrodinger's Dog! :smile:

    Put u = sechv, du = -sechvtanhvdv.

    Then ∫(√(1 - u^2))/udu = -∫(tanh^2(v)sechv/sechv)dy

    = ∫(sech^2(v) - 1)dv

    = tanhv - v

    = √(1 - u^2)) - tanh^-1√(1 - u^2)

    = √(1 - e^-x) - tanh^-1√(1 - e^-x). :smile:

    (hmm … might have been quicker to start by putting e^(-x/2) = sechv :rolleyes:)
     
  8. Mar 30, 2008 #7
    Thanks tiny tim, when I get a spare moment I'll go through that thoroughly. :smile:
     
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