Integral using hyperbolic substitution

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  • #1
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Homework Statement


[tex]\int\left(1+x^{2}\right)^{\frac{3}{2}}dx[/tex]

Homework Equations


The hyperbolic functions.

The Attempt at a Solution


We've been going over hyperbolic substitutions in class so I assume I'm meant to use one of those, but I'm just not sure how to choose which one. Any help greatly appreciated.
 

Answers and Replies

  • #2
SammyS
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sinh2(u) + 1 = cosh2(u)

So, I would try letting x = sinh(u) → dx = cosh(u) du
 
  • #3
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[tex]\begin{eqnarray*}
\int\left(1+x^{2}\right)^{\frac{3}{2}}dx & = & \int\left(1+\sinh^{2}\left(u\right)\right)^{\frac{3}{2}}\cosh\left(u\right)\, du\\
& = & \int\sqrt{\left(\cosh^{2}\left(u\right)\right)^{3}}\cosh\left(u\right)\, du\\
& = & \int\sqrt{\cosh^{6}\left(u\right)}\cosh\left(u\right)\, du\\
& = & \int\cosh^{4}\left(u\right)\, du
\end{eqnarray*}[\tex]
 
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  • #4
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[tex]\begin{eqnarray*}
\int\left(1+x^{2}\right)^{\frac{3}{2}}dx & = & \int\left(1+\sinh^{2}\left(u\right)\right)^{\frac{3}{2}}\cosh\left(u\right)\, du\\
& = & \int\sqrt{\left(\cosh^{2}\left(u\right)\right)^{3}}\cosh\left(u\right)\, du\\
& = & \int\sqrt{\cosh^{6}\left(u\right)}\cosh\left(u\right)\, du\\
& = & \int\cosh^{4}\left(u\right)\, du
\end{eqnarray*}[\tex]

Oops, that was meant to be nice neat LaTeX...

Well, I tried doing that but ended up with int(cosh^4(u))du... not sure if there is a nice little trick for solving that or if I've ended up with a harder problem than I started with. Also, why didn't that tex code just work?
 
  • #5
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Use the identity

[tex]cosh^2 (t) = \frac{1 + cosh(2t)}{2}[/tex]


Your tex didn't work because you used \ instead of /

Omg i helped someone!!!
 
  • #6
SammyS
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[tex]\int\left(1+x^{2}\right)^{\frac{3}{2}}dx = \int\left(1+\sinh^{2}\left(u\right)\right)^{\frac{3}{2}}\cosh\left(u\right)\, du[/tex]
[tex]=\int\sqrt{\left(\cosh^{2}\left(u\right)\right)^{3}}\cosh\left(u\right)\, du[/tex]
[tex]=\int\sqrt{\cosh^{6}\left(u\right)} \cosh\left(u\right)\, du[/tex]
[tex]=\int\cosh^{4}\left(u\right)\, du
[/tex]
Changed '\' to '/' in the /tex tag so we can read your LaTeX.
 
  • #7
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You could evaluate this integral without using the hyperbolic trig functions and just using the regular ones.
 
  • #8
SammyS
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sinh2(u) + 1 = cosh2(u)

So, I would try letting x = sinh(u) → dx = cosh(u) du
By The Way: [itex]u = \sinh^{-1}(x) = \ln \left( x+\sqrt{x^{2}+1} \right)[/itex]
 
  • #9
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By The Way: [itex]u = \sinh^{-1}(x) = \ln \left( x+\sqrt{x^{2}+1} \right)[/itex]

Aha! I was wondering how it got to that step.

Thanks very much for your help!
 

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