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Math Help - Integral of sqrt(2+2cost)

  1. #1
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    Integral of sqrt(2+2cost)

    Integral from zero to 2pi of sqrt(2+2cost)

    Integral from zero to 2pi of sqrt(2-2cost)
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  2. #2
    TD!
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    Switching to the the half angle using the right formula will make this a piece of cake:

    <br />
\sqrt {2 + 2\cos t}  = \sqrt {2 + 2\left( {2\cos ^2 \left( {\frac{t}{2}} \right) - 1} \right)} <br />

    The 2-2 will now disappear to get:

    <br />
\sqrt {4\cos ^2 \left( {\frac{t}{2}} \right)}  = 2\left| {\cos \left( {\frac{t}{2}} \right)} \right|<br />

    In the same way, but using the other formula, you'll find:

    <br />
\sqrt {2 - 2\cos t}  = \sqrt {2 - 2\left( {1 - 2\sin ^2 \left( {\frac{t}{2}} \right)} \right)}  = 2\left| {\sin \left( {\frac{t}{2}} \right)} \right|<br />

    Be careful with the absolute values though, when using this for definite integration:

    <br />
\int\limits_0^{2\pi } {\sqrt {2 + 2\cos t} } dt = 2\int\limits_0^{2\pi } {\left| {\cos \left( {\frac{t}{2}} \right)} \right|} dt<br />

    <br />
\int\limits_0^{2\pi } {\sqrt {2 - 2\cos t} } dt = 2\int\limits_0^{2\pi } {\left| {\sin \left( {\frac{t}{2}} \right)} \right|} dt<br />

    These too should be doable now, I'll let you finish. To check: you should find 8 (twice).
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  3. #3
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    Welcome back TD!
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  4. #4
    TD!
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    Quote Originally Posted by ThePerfectHacker
    Welcome back TD!
    I kept hanging arround, just not as frequent as I once did (though that didn't last long either).

    When I come arround, the threads I'd usually reply to are already answered (I only check a few forums briefly actually) - this one wasn't
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  5. #5
    TD!
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    Quote Originally Posted by galactus
    This gives:

    \int_{0}^{2{\pi}}\sqrt{2+2(\frac{1-u^{2}}{1+u^{2}})}\cdot\frac{2}{1+u^{2}}du

    Which simplifies to:

    4\int_{0}^{2{\pi}}\frac{1}{(u^{2}+1)^{\frac{3}{2}}  }du
    I don't think the limits for u will stay the same...

    Also: this method is fine for indefinite integration, but you have to be carefull when using it for definite integrals. This substitution is only valid on (-pi,pi) and the interval of integration is [0,2pi]. This can be resolved by noticing the symmetry in the function, but it makes it a bit more complicated of course
    Last edited by TD!; September 4th 2006 at 03:04 PM.
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  6. #6
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    Yes, I know. An oversight on my part.
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