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Integral of hyperbolic function

  1. Apr 16, 2012 #1

    sharks

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    The problem statement, all variables and given/known data
    [tex]\int^4_3 \frac{1}{\sqrt{3x^2-6x+1}}\,.dx[/tex]

    The attempt at a solution
    I complete the square for the quadratic:
    [tex]\sqrt{3x^2-6x+1}
    \\=\sqrt{3(x^2-2x+\frac{1}{3})}
    \\=\sqrt 3 \times \sqrt{(x-1)^2-\frac{2}{3}}[/tex]
    [tex]\int^4_3 \frac{1}{\sqrt{3x^2-6x+1}}\,.dx
    \\=\frac{1}{\sqrt 3}\int^4_3 \frac{1}{\sqrt{(x-1)^2-\frac{2}{3}}}\,.dx
    \\=\frac{1}{\sqrt 3} \left[\cosh^{-1}\frac{\sqrt 3(x-1)}{\sqrt 2}\right]^{x=4}_{x=3}
    \\=\frac{1}{\sqrt 3} \left[\cosh^{-1}\frac{3\sqrt 3}{\sqrt 2}-\cosh^{-1}\sqrt 6\right][/tex]
    I already simplified it but it doesn't agree with the final answer:
    [tex]\frac{1}{\sqrt 3}\ln \left(\frac{15+\sqrt {219}}{12+\sqrt {138}}\right)[/tex]
     
  2. jcsd
  3. Apr 16, 2012 #2

    tiny-tim

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    hi sharks! :smile:

    cosh-1x = ln(ecosh-1x)

    = ln(cosh(cosh-1x) + sinh(cosh-1x))

    = ln (x + √(x2 - 1)) :wink:
     
  4. Apr 16, 2012 #3

    sharks

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    Hi tiny-tim :smile:

    The above expression is what i used to expand:
    [tex]\frac{1}{\sqrt 3} \left [\cosh^{-1}\frac{3\sqrt 3}{\sqrt 2}-\cosh^{-1}\sqrt 6\right]
    \\=\frac{1}{\sqrt 3} \ln \left[\frac {3\sqrt 3+5}{\sqrt{12}+\sqrt{10}}\right]=0.4625025064[/tex]
    But if i evaluate the answer from my notes, i get:
    [tex]\frac{1}{\sqrt 3}\ln \left (\frac{15+\sqrt {219}}{12+\sqrt {138}}\right)=0.1310541888[/tex]
    Since the answers are not the same, i'm thinking that maybe the answer in my notes is wrong?
     
    Last edited: Apr 16, 2012
  5. Apr 16, 2012 #4

    tiny-tim

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    well that's obviously …
    [tex]\frac{1}{\sqrt 3} \left[\cosh^{-1}\frac{\sqrt 3(x+1)}{\sqrt 2}\right]^{x=4}_{x=3}[/tex]

    … either the question or the answer is wrong
     
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