How to prove that $\int_{-\pi}^{+\pi}\cos{(2x)}\cos{(3x)}\cos{(4x)}\cdots\cos{(2005x)}dx$ is positive 
show that
  $$I=\int_{-\pi}^{+\pi}\cos{(2x)}\cos{(3x)}\cos{(4x)}\cdots\cos{(2005x)}dx>0$$

This problem is my frend ask me,
My try:

$$I=2\int_{0}^{\pi}\cos{(2x)}\cos{(3x)}\cos{(4x)}\cdots\cos{(2005x)}dx$$
  and I think maybe use 
  $$2\cos{x}=e^{ix}+e^{-ix}$$
  $$\Longleftrightarrow \dfrac{1}{2^{2003}}\int_{0}^{\pi}(e^{2ix}+e^{-2ix})\cdots(e^{-2005ix}+e^{2005ix})dx>0$$
  and then I can't works,and I think this is nice inequality,Thank you 

By the way: I have ask my frend where is from this problem,He tell me is from mathlinks
:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=549449&p=3252846#p3252846
 A: Keep the integral on $(-\pi,\pi)$ and use your idea of replacing each $\cos(kx)$ by $\frac12(\mathrm e^{\mathrm ikx}+\mathrm e^{-\mathrm ikx})$. This yields
$$
2^{2004}I=\int_{-\pi}^\pi\prod_{k=2}^{2005}(\mathrm e^{\mathrm ikx}+\mathrm e^{-\mathrm ikx})\,\mathrm dx.
$$
Expanding the product in the integral yields a linear combination of $\mathrm e^{\mathrm inx}$ for $n$ integer, and this linear combination has nonnegative integer coefficients. Now, for every integer $n$,
$$
\int_{-\pi}^\pi\mathrm e^{\mathrm inx}\,\mathrm dx=2\pi\,\mathbf 1_{n=0},
$$
hence $2^{2004}I$ is $2\pi$ times the coefficient of $\mathrm e^{\mathrm i0x}$ in the expansion of the product. This coefficient is the size of the set $S$ of $(x_n)$ in $\{-1,+1\}^{2004}$ such that $2x_2+3x_3+\cdots+2005x_{2005}=0$.
Finally, either $S=\varnothing$, then $I=0$, or $S\ne\varnothing$, then $I=2\pi\cdot\#S/2^{2004}\geqslant2\pi/2^{2004}\gt0$. Since $2005-1=4\cdot501$ is a multiple of $4$ and $n-(n+1)-(n+2)+(n+3)=0$ for every $n$, the sequence $1(-1)(-1)1$ repeated $501$ times is in $S$, hence $I\gt0$.
The argument works replacing $(2,3,4,\ldots,2005)$ by any $(n+1,n+2,\ldots,n+4k)$.
Edit: The change of variable $x\to x+\pi$ and the remark that $\cos(n(x+\pi))=(-1)^n\cos(nx)$ for every $n$, shows that the analogue of $I$ based on $(n+1,n+2,\ldots,n+i)$ is $0$ when $ni+\frac12i(i+1)$ is odd. In the question $n=1$ hence $I=0$ when $\frac12i(i+3)$ is odd, that is, for every $i$ such that $i=2\pmod{4}$ or $i=3\pmod{4}$ (this was observed by @Jean-Sébastien in a comment). 
When $i=0\pmod{4}$, $I\gt0$ by the above. 
When $i=1\pmod{4}$, note that the five first cosines can be grouped along the signs $+2+3-4+5-6=0$, and each group of four after them along the usual signs $+--+$. This yields $I\gt0$ for $n=1$ and every $i=1\pmod{4}$, $i\geqslant5$ (also observed by @Jean-Sébastien), except that $I=0$ when $i=1$.
A: Idea: Note that when you have a polynomials $p(x)$, we have $$\int_{-\pi}^\pi p(e^{it}) \cdot e^{-n\cdot it} dt = \int_0^{2\pi} p(e^{it}) \cdot e^{-n\cdot it}dt = 2\pi \times [x^n]\{p(x)\} $$
where $[x^n]\{p(x)\}$ is the coefficient of $x^n$ in $p(x)$.
