$\displaystyle \int\cos ^ {2004}x\cos 2004x\ dx $
This can be solved using a bunch of integration by parts with a little help by a useful trig identity.
$\displaystyle \int_{a}^{b} u \frac{dv}{dx} dx = \left. uv \right|_{x=a}^{b} - \int_{a}^{b} \frac{du}{dx} v dx $
Let $\displaystyle u = \cos^{2004}{x} \ , \ \frac{dv}{dx} = \cos{2004x} \Rightarrow v = \frac{1}{2004} \sin{2004x} $
$\displaystyle \int \cos^{2004}{x} \cos{2004x} dx = C + \frac{2004}{2004} \int \sin^{2003}{x} \sin{2004x} dx $
Continue integrating by parts until you notice a pattern. This pattern should take you all the way down to an integrand that you can split up using a handy trig identity. (One that takes the product of two trig functions and gives the sum of two other trig functions).