Integral from zero to 2pi of sqrt(2+2cost)
Integral from zero to 2pi of sqrt(2-2cost)
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).
I kept hanging arround, just not as frequent as I once did (though that didn't last long either).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
I don't think the limits for u will stay the same...Originally Posted by galactus
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