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

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

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- Sep 4th 2006, 01:52 PMTexasGirlIntegral of sqrt(2+2cost)
Integral from zero to 2pi of sqrt(2+2cost)

Integral from zero to 2pi of sqrt(2-2cost) - Sep 4th 2006, 02:37 PMTD!
Switching to the the half angle using the right formula will make this a piece of cake:

$\displaystyle

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

$

The 2-2 will now disappear to get:

$\displaystyle

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

$

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

$\displaystyle

\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|

$

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

$\displaystyle

\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

$

$\displaystyle

\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

$

These too should be doable now, I'll let you finish. To check: you should find 8 (twice). - Sep 4th 2006, 02:40 PMThePerfectHacker
Welcome back TD!

- Sep 4th 2006, 02:43 PMTD!Quote:

Originally Posted by**ThePerfectHacker**

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 :) - Sep 4th 2006, 02:53 PMTD!Quote:

Originally Posted by**galactus**

*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 ;) - Sep 4th 2006, 03:12 PMgalactus
Yes, I know. An oversight on my part.