If $$I_{m}=\int^{2\pi}_{0}\cos x \cos (2x) \cos (3x)\cdots\cos(mx) \, dx.$$Then $m$ for which $I_m \neq 0$ is, where $m$ is any natural number
Options $(a)\; 5\;\; (b)\; 6\;\; (c)\; 7\;\; (d)\; 8$
Try: $$I_m = \frac 1 {2^m}\int^{2\pi}_0 \bigg[(e^{ix}+e^{-ix})(e^{2ix} + e^{-2ix}) \cdots (e^{imx}+e^{-mix})\bigg] \, dx$$
Could some help me to solve it , thanks in advance
Since $\int_{0}^{2\pi}e^{mi\theta}\,d\theta = 2\pi\delta(m)$ for any $m\in\mathbb{Z}$, $2^m I_m$ is given by $2\pi$ times the number of solutions of
$$ \pm 1 \pm 2 \pm 3\ldots \pm m = 0 $$
for a suitable choice of signs $\pm$. Both $I_5$ and $I_6$ are zero since the LHS is odd for any choice of $\pm$, but
$$ \color{green}{+1+2-3}\color{blue}{+4-5-6+7} = 0 $$
$$ \color{green}{+1-2-3+4}\color{blue}{+5-6-7+8}=0 $$
imply that both $I_7$ and $I_8$ are non-zero.
It is not difficult to prove by induction that $I_m\neq 0$ iff $m\pmod{4}\in\{0,3\}$.
You've made the right start. Using $\int_0^{2\pi}e^{ikx}dx=2\pi\delta_{k0}$ for any integer $k$, the challenge is finding $m$ so an $e^{0ix}$ term appears. Equivalently, there must be a way to write $1\pm 2\pm\cdots\pm m=0$, where each $\pm$ is independent. I'll leave it to you to verify by working modulo $2$ that this requires $m$ to be $0$ or $-1$ modulo $4$. I'll also leave it to you to show how to pull it off in each of those cases so (c) & (d) are correct.
