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Math Help - Stuck on definite integral of root

  1. #1
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    Stuck on definite integral of root

    Hi,

    I'm taking an engineering course and in an example problem my teaching assistant solved on the board he solved an integral and I can't remember how it was done. The integral is \int^\pi_0 \sqrt{2+t^2}\,dt

    Can someone please help walk me through how to solve this integral, my cal's a little rusty...
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  2. #2
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    \sqrt{2 + t^2}

    = \frac{2 + t^2}{\sqrt{2 + t^2}}

    =\frac{2}{\sqrt{2 + t^2}}\ +\ \frac{t^2}{\sqrt{2 + t^2}}

    The first fraction is the derivative of inverse sinh and the second is good for parts - will post a pic...
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Canadian0469 View Post
    Hi,

    I'm taking an engineering course and in an example problem my teaching assistant solved on the board he solved an integral and I can't remember how it was done. The integral is \int^\pi_0 \sqrt{2+t^2}\,dt

    Can someone please help walk me through how to solve this integral, my cal's a little rusty...
    Use hyperbolic substitution

    \int\limits_0^\pi  {\sqrt {2 + {t^2}} dt}  = \left\{ \begin{gathered}t = \sqrt 2 \sinh \left( x \right), \hfill \\dt = \sqrt 2 \cosh \left( x \right)dx \hfill \\ \end{gathered}  \right\} = 2\int\limits_0^{{\text{arcsinh}}\left( {\frac{{\pi \sqrt 2 }}{2}} \right)} {{{\cosh }^2}\left( x \right)dx}  =

    = 2\int\limits_0^{{\text{arcsinh}}\left( {\frac{{\pi \sqrt 2 }}{2}} \right)} {{{\cosh }^2}\left( x \right)dx}  = \int\limits_0^{{\text{arcsinh}}\left( {\frac{{\pi \sqrt 2 }}{2}} \right)} {\left( {\cosh \left( {2x} \right) + 1} \right)dx}  =

    = \left. {\left( {\frac{1}{2}\sinh \left( {2x} \right) + x} \right)} \right|_0^{{\text{arcsinh}}\left( {\frac{{\pi \sqrt 2 }}{2}} \right)} = \frac{1}{2}\sinh \left( {2{\text{arcsinh}}\left( {\frac{{\pi \sqrt 2 }}{2}} \right)} \right) + {\text{arcsinh}}\left( {\frac{{\pi \sqrt 2 }}{2}} \right) =

    = \sinh \left( {{\text{arcsinh}}\left( {\frac{{\pi \sqrt 2 }}{2}} \right)} \right)\cosh \left( {{\text{arcsinh}}\left( {\frac{{\pi \sqrt 2 }}{2}} \right)} \right) + {\text{arcsinh}}\left( {\frac{{\pi \sqrt 2 }}{2}} \right) =

    = \frac{{\pi \sqrt 2 }}{2}\sqrt {{{\left( {\frac{{\pi \sqrt 2 }}{2}} \right)}^2} + 1}  + \ln \left( {\frac{{\pi \sqrt 2 }}{2} + \sqrt {{{\left( {\frac{{\pi \sqrt 2 }}{2}} \right)}^2} + 1} } \right) =

    = \frac{{\pi \sqrt {{\pi ^2} + 2} }}{2} + \ln \left( {\pi  + \sqrt {{\pi ^2} + 2} } \right) - \frac{{\ln 2}}{2} \approx 6.95025
    Last edited by DeMath; February 19th 2009 at 02:43 AM.
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  4. #4
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    Just in case a picture helps...

    Click here for the horizontal... http://www.ballooncalculus.org/asy/p...sRootQuad1.png





    Straight lines differentiate down / integrate up with respect to t, so that the triangular network satisfies the chain rule for differentiation.

    Now just solve the top row for \int \sqrt{2+t^2}\ dt

    Don't integrate - balloontegrate!

    http://www.ballooncalculus.org/examples
    Last edited by tom@ballooncalculus; February 19th 2009 at 09:49 AM. Reason: typo
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  5. #5
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    tom, please reduce the horizontal size of your pic, it distorts the size of the forum.
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