$\displaystyle \int\cos ^ {2004}x\cos 2004x\ dx $

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- Aug 7th 2012, 08:54 AMayushdadhwalintegration
$\displaystyle \int\cos ^ {2004}x\cos 2004x\ dx $

- Aug 7th 2012, 10:24 AMJohnDMalcolmRe: integration
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). - Aug 18th 2012, 11:12 AMayushdadhwalRe: integration
any other way

- Aug 29th 2012, 08:20 AMscubastevezRe: integration