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Homework Help: Solving an integral using a*sinh substitution

  1. Jun 29, 2010 #1
    1. The problem statement, all variables and given/known data
    The statement says: Calculate the next integrals using the adequate trigonometric substitution:

    [tex]\displaystyle\int_{}^{}x^2\sqrt[ ]{x^2+3}dx[/tex]


    2. Relevant equations
    [tex]ch^2(t)-sh^2(t)=1\Rightarrow{ch(t)=\sqrt[ ]{1+sh^2(t)}}[/tex]


    3. The attempt at a solution

    [tex]x=\sqrt[ ]{3}sh(t)[/tex]
    [tex]dx=\sqrt[ ]{3}ch(t)dt[/tex]

    [tex]u=ch(t)[/tex]
    [tex]du=sh(t)dt[/tex]

    [tex]dv=ch(t)dt[/tex]
    [tex]v=sh(t)[/tex]

    [tex]\displaystyle\int_{}^{}ch^2(t)dt=ch(t)sh(t)-\displaystyle\int_{}^{}sh(t)sh(t)dt[/tex]

    [tex]\displaystyle\int_{}^{}ch^2(t)dt=ch(t)sh(t)-\displaystyle\int_{}^{}sh^2(t)dt=ch(t)sh(t)-\displaystyle\int_{}^{}(ch^2(t)-1)dt[/tex]

    [tex]*[/tex] [tex]\displaystyle\int_{}^{}ch^2(t)dt=\displaystyle\frac{1}{2}(ch(t)sh(t)+t)[/tex]

    [tex]9\displaystyle\int_{}^{}sh^2(t)ch^2(t)dt=9(\displaystyle\frac{sh^2(t)}{2}ch(t)sh(t)+t)-\displaystyle\int_{}^{}(ch(t)sh(t)+t)sh(t)ch(t)dt=9(\displaystyle\frac{sh^2(t)}{2}ch(t)sh(t)+t)-\displaystyle\int_{}^{}ch(t)^2sh(t)^2dt-\displaystyle\int_{}^{}ch(t)sh(t)tdt[/tex]

    Well, if you see an easier way of solving this let me know :P

    Bye there!
     
    Last edited: Jun 29, 2010
  2. jcsd
  3. Jun 29, 2010 #2

    vela

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    You might want to try the substitution [itex]x=\sqrt{3}\tan \theta[/itex] instead.
     
  4. Jun 29, 2010 #3

    vela

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    Or you could use your substitution and use the identities

    [tex]\sinh 2x = 2\sinh x\cosh x[/tex]

    [tex]\sinh^2 x = \frac{\cosh 2x-1}{2}[/tex]

    instead of integration by parts.
     
  5. Jun 29, 2010 #4
    Thanks vela. I'll try both ways.
     
    Last edited: Jun 29, 2010
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