Originally Posted by
Prove It Actually, I don't think it's that simple...
Remember that $\displaystyle x = r\cos{\theta}$.
Would that not therefore mean that
$\displaystyle \cos{(m\theta)}$ and $\displaystyle \cos{(n\theta)}$ are functions of $\displaystyle x$, not constants?
Only if the OP had said so and he didn't. The use of theta doesn't automatically mean we're using polar coordinates and, deducing from the question, there's no a priori relation between theta and x. In fact, it'd be rather odd to have a problem with x and theta: if we already changed coordinates, what's x doing there? Why wasn't it changed into r cos(theta)? And why does the OP explicitly asks for the integral wrt x if he already changed to polar coordinates?
Anyway, that's something for the OP to clear out and not for us, imo. I assumed that theta and x are independient and under this understanding my answer is, I think, correct. If the OP meant something else he should write it down clearly.
Tonio
In fact:
$\displaystyle x = r\cos{\theta} \implies \cos{\theta} = \frac{x}{r}$
and
$\displaystyle y = r\sin{\theta} \implies \sin{\theta} = \frac{y}{r}$.
$\displaystyle \cos{(n\theta)} = \sum_{k = 0}^n{{{n}\choose{k}}\cos^k{\theta}\sin^{n - k}{\theta}\sin{\left[\frac{1}{2}(n - k)\pi\right]}}$
$\displaystyle = \sum_{k = 0}^n{{{n}\choose{k}}\left[\frac{x}{r}\right]^k\left[\frac{y}{r}\right]^{n - k}\sin{\left[\frac{1}{2}(n - k)\pi\right]}}$.
$\displaystyle \cos{(m\theta)} = \sum_{k = 0}^m{{{m}\choose{k}}\cos^k{\theta}\sin^{m - k}{\theta}\sin{\left[\frac{1}{2}(m - k)\pi\right]}}$
$\displaystyle = \sum_{k = 0}^m{{{m}\choose{k}}\left[\frac{x}{r}\right]^k\left[\frac{y}{r}\right]^{m - k}\sin{\left[\frac{1}{2}(m - k)\pi\right]}}$.
These are definitely not constants...