Integral of hyperbolic function

  • Thread starter DryRun
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  • #1
DryRun
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Homework Statement
[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]
 

Answers and Replies

  • #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:
 
  • #3
DryRun
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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:
  • #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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