find :
Either using $\displaystyle \int_{a}^{b}f(x)\;{dx} = \int_{a}^{b}f(a+b-x)\;{dx}$ or putting $\displaystyle x \mapsto x-\pi$ gives:
$\displaystyle \displaystyle \begin{aligned} I & = \int_{0}^{\pi} \cos{x}\cos{3x}\cos{5x} \\& = \int_{0}^{\pi} \cos(\pi-x)\cos(3\pi-3x)\cos(5\pi-5x) \\& = -\int_{0}^{\pi} \cos{x}\cos{3x}\cos{5x}. \end{aligned}$
Hence $\displaystyle 2I = 0$, thus $\displaystyle I = 0$.
More generally, if $\displaystyle n$ is even, then:
$\displaystyle \displaystyle \begin{aligned} & \int_{0}^{\pi}\prod_{0\le k \le n}\cos\left[(2k+1)x\right]\;{dx} \\& = \int_{0}^{\pi}\prod_{0\le k \le n}\cos\left[(2k+1)\pi-(2k+1)x\right]\;{dx} \\& = -\int_{0}^{\pi}\prod_{0\le k \le n}\cos\left[(2k+1)x\right]\;{dx}. \end{aligned} $
Thus $\displaystyle \displaystyle \int_{0}^{\pi}\prod_{0\le k \le n}\cos[(2k+1)x]\;{dx} = 0.$
this was a question of http://www.mathhelpforum.com/math-he...st-169383.html