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..