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Integral of 1/(something)

  1. Feb 21, 2010 #1
    [tex]\int[/tex][tex]\stackrel{du}{\sqrt{u^{2}-4}}[/tex] = ln(u + sqrt.(u2-4))

    My answer key makes the jump directly from that integral to that natural log fxn, which would make sense to me if only the "u +" weren't inside the brackets of the ln fxn.
    Does anyone know where it comes from?
     
    Last edited: Feb 21, 2010
  2. jcsd
  3. Feb 21, 2010 #2
    I'm sorry about the latex. I edited it but the changes aren't showing up. Hope you can still decipher it.
    The latex program is ridiculously glitchy, but I'm sure I'm not the first person to say that.
     
    Last edited: Feb 21, 2010
  4. Feb 21, 2010 #3
    Um, that's the correct antiderivative? Have you differentiated it to check? Perhaps you forgot the chain rule.
     
  5. Feb 21, 2010 #4
    Try trig substitution
     
  6. Feb 21, 2010 #5
    If [itex]u = \cosh{t}[/itex], then [itex]u^2-1 = \sinh^2{t}[/itex], and [itex]du = \sinh{t} dt[/itex].
     
  7. Feb 22, 2010 #6

    HallsofIvy

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    Actually, that shouldn't make sense.
    [tex]\int \frac{1}{f(x)}dx[/tex]
    is NOT, in general, equal to ln(f(x))!

    hamster143 has suggested you use u= cosh(t) because [itex]cosh^2(t)- 1= sinh^2(t)[/itex] and you can get rid of the square root.

    It is also true that [itex]sec^2(t)- 1= csc^2(t)[/itex] so the substitution u= sec(t) would also work. It might be a little more complicated because then du= sec(t)tan(t)dt.

    (This board has a slight problem with its LaTex. When you edit LaTex, always click the "refresh" button on your screen (looks like two little 'circulating' green arrows). That should fix it.)
     
  8. Feb 23, 2010 #7
    Really? the equality I posted is straight from my prof's answer key. ....

    Also, I'm not sure I understand what you mean by cosh^2(t). Is that another way of expressing cos^2(ht)?

    In that case I thought cos^2(x) + sin^2(x) = 1,
    so how could cos^2(x) - 1 = sin^2(x) ? I was thinking it would equal -sin^2(x) , and so I was getting confused because it's under a sqrt. sign so it can't be -ve, so I resorted to the answer key only to find my prof's weird equality ... -_-

    (thanks for the LaTex tip)
     
  9. Feb 23, 2010 #8
    Cosh and sinh are hyperbolic sine and cosine, defined as

    [tex]\sinh{t} = (e^t-e^{-t})/2[/tex]
    [tex]\cosh{t} = (e^t+e^{-t})/2[/tex]
     
  10. Feb 24, 2010 #9
    Ya your answer isn't right. Remember that when integrating, derivatives are your best friend to check your work. If I were to take a derivative of that I would get a u + sqrt.(u2-4) on the bottom * whatever the derivative of that is on the top.

    You need to use a trig substitution here of some sort.
     
  11. Feb 24, 2010 #10
    woahhhh no youre answer is definetly right.

    [tex]\frac{d}{dx}(ln(x+\sqrt{x^{2}-4})=\frac{1}{x+\sqrt{x^{2}-4}}*(1+\frac{x}{\sqrt{x^{2}-4}})=\frac{1}{x+\sqrt{x^{2}-4}}*\frac{x+\sqrt{x^{2}-4}}{\sqrt{x^{2}-4}}=\frac{1}{\sqrt{x^{2}-4}}[/tex]
     
  12. Feb 24, 2010 #11
    use x=2*sec(u) and dx=2*sec(u)*tan(u)

    so the problem would become something like [tex]\int \frac{2*\sec u *\tan u}{2*\sqrt{(\sec x)^{2}-1}}du[/tex]

    you'll get [tex]\int \sec u[/tex]
     
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