# Integral of sqrt(2+2cost)

• Sep 4th 2006, 02:52 PM
TexasGirl
Integral 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, 03:37 PM
TD!
Switching to the the half angle using the right formula will make this a piece of cake:

$
\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:

$
\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:

$
\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:

$
\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
$

$
\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, 03:40 PM
ThePerfectHacker
Welcome back TD!
• Sep 4th 2006, 03:43 PM
TD!
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 :)
• Sep 4th 2006, 03:53 PM
TD!
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 ;)
• Sep 4th 2006, 04:12 PM
galactus
Yes, I know. An oversight on my part.