Definite integral over a period

**Mollier** · September 2nd 2010, 10:57 PM

Hi,

how do I integrate,

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

I've tried by defining

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

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

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..

**TheCoffeeMachine** · September 2nd 2010, 11:06 PM

$8-8\cos{2t} = 8(1-\cos{2t}) = 16\sin^2{t}.$

**Prove It** · September 2nd 2010, 11:07 PM

Remember that $\cos{2t} = 1 - 2\sin^2{t}$ , so

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

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

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

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

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

Therefore

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

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

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

$= 4 + 4$

$= 8$ .

**Mollier** · September 2nd 2010, 11:48 PM

Ah, all those trig identities!!

Thank you both very much.

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