# Definite integral over a period

• Sep 2nd 2010, 10:57 PM
Mollier
Definite integral over a period
Hi,

how do I integrate,

$\displaystyle \int^{\pi}_0 (8-8cos(2t))^{1/2}dt$?

I've tried by defining

$\displaystyle u=8-8cos(2t)$ and then

$\displaystyle \frac{du}{dt}=16sin(2t) \Rightarrow dt=\frac{1}{16sin(2t)}du,$

but that doesn't get me very far.

I think I remember that there are some trick when dealing with definite integrals over periods and when the function is trigonometric. I don't remember how it works though..
• Sep 2nd 2010, 11:06 PM
TheCoffeeMachine
$\displaystyle 8-8\cos{2t} = 8(1-\cos{2t}) = 16\sin^2{t}.$
• Sep 2nd 2010, 11:07 PM
Prove It
Remember that $\displaystyle \cos{2t} = 1 - 2\sin^2{t}$, so

$\displaystyle \int{\sqrt{8 - 8\cos{2t}}\,dt} = \int{\sqrt{8 - 8(1 - 2\sin^2{t})}\,dt}$

$\displaystyle = \int{\sqrt{8 - 8 + 16\sin^2{t}}\,dt}$

$\displaystyle = \int{\sqrt{16\sin^2{t}}\,dt}$

$\displaystyle = \int{4\sin{t}\,dt}$

$\displaystyle = -4\cos{t} + C$.

Therefore

$\displaystyle \int_0^\pi{\sqrt{8 - 8\cos{2t}}\,dt} = \left[-4\cos{t}\right]_0^\pi$

$\displaystyle = \left[-4\cos{\pi}\right] - \left[-4\cos{0}\right]$

$\displaystyle = -4(-1) + 4(1)$

$\displaystyle = 4 + 4$

$\displaystyle = 8$.
• Sep 2nd 2010, 11:48 PM
Mollier
Ah, all those trig identities!!

Thank you both very much.