# Stuck on definite integral of root

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• Feb 18th 2009, 11:26 PM
Canadian0469
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...
• Feb 19th 2009, 02:24 AM
tom@ballooncalculus
$\sqrt{2 + t^2}$

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

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

The first fraction is the derivative of inverse sinh and the second is good for parts - will post a pic...
• Feb 19th 2009, 02:31 AM
DeMath
Quote:

Originally Posted by Canadian0469
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

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

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

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

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

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

$= \frac{{\pi \sqrt {{\pi ^2} + 2} }}{2} + \ln \left( {\pi + \sqrt {{\pi ^2} + 2} } \right) - \frac{{\ln 2}}{2} \approx 6.95025$
• Feb 19th 2009, 03:28 AM
tom@ballooncalculus
Just in case a picture helps...

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

http://www.ballooncalculus.org/asy/p...tsRootQuad.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
• Feb 19th 2009, 05:55 AM
Krizalid
tom, please reduce the horizontal size of your pic, it distorts the size of the forum.