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After constructing an even extension, and evaluating over $-\pi \le x \le \pi$ $\frac \pi2a_n=\int_0^\pi x^3\cos nx\ dx $

Now I've tried to solve this, I wish I had the skills and the patience to mathjax-ify my steps here but it'll be too much to do. I'll write my result after evaluating the integral at least:

$a_n = \frac{2(3(\pi^2n^2-2)\cos(\pi n) +6)}{\pi n^4}$

Now I tried to use the fact that $\cos n\pi$ follows a pattern, but I still couldn't get it to match the answer I'm expected to derive (which is again, beyond my mathjax skills). It's a cosine series where $ 0\le x \le \pi$. I'm way off, not even close.

So yeah, basically I'm asking for some tips on how to proceed from here. If possible, a full solution would be brilliant so that I can once and for all understand such problems.

EDIT: Here's the answer I'm supposed to get: enter image description here

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    I know it's not my place to judge, but honestly attempting to break your answer into some neat, clever form (if that form above could be consider better at all) will be a total waste of time on an exam. Get the right answer and move on. Spend your time on the content of the problems where the points come from and not on some superficial detail like this. If your teacher takes off for something like this, then they aren't a good teacher.2012-12-04

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The book just regrouped the terms:

$a_n=\frac{2(3(\pi^2n^2-2)\cos(\pi n) +6)}{\pi n^4}=\frac{6\pi^2n^2\cos(n\pi)}{\pi n^4}+\frac{12-12\cos(n\pi)}{\pi n^4}$

Since $\cos(n\pi)=(-1)^n$, the right term is $0$ for $n$ even and $24/\pi n^4$ when $n$ is odd.