<s>[INST] Showing $\int_{0}^{2\pi}\cos(x)\cos(2x)\cos(3x)\,dx = \frac\pi2$ An integral from MIT Integration Bee:

Show that
$$I = \int_{0}^{2\pi}\cos(x)\cos(2x)
\cos(3x)\,dx = \frac\pi2$$

This integral appeared in the 2019 paper. Below is my own solution:
$$\begin{align}
I &= \int_{0}^{2\pi}\cos(x)\cos(2x)(\cos x\cos2x-\sin x\sin2x)\,dx \\[6pt]
&= \int_{0}^{2\pi}\cos^2(x)\cos^2(2x) \,dx -\int_{0}^{2\pi}\cos(x)\cos(2x)\sin(x)\sin(2x)\, dx
\end{align}$$
Replacing $\cos^2(x)= \frac{1+\cos(2x)}{2}$ for the first integral and  $\sin(x)\cos(x)= \sin(2x)/2 $ for the second, we get
$$\begin{align}
&\int_{0}^{2\pi}\frac{1+\cos(2x)}{2}\cos^2(2x) dx-\frac{1}{2}\int_{0}^{2\pi}\cos(2x)\sin^2(2x) dx \\[6pt]
=\; &\frac{1}{2}\left(\int_{0}^{2\pi}\cos^2(2x)dx \, + \int_{0}^{2\pi}\cos(2x)\cos(4x)dx  \right) \\[6pt]
=\; &\frac{1}{2} \left( \pi + 0\right) \qquad \text{$\because$ the orthogonality of $\cos(mx)$} \\[6pt]
=\; &\frac\pi2
\end{align}$$
This solution is rather awkward, and I'm sure there's a better and faster approach to this integral. Could anyone provide a more elegant solution(or a sketch of it)? Thanks.
 [/INST] Answer inspired by Aaron's answer.
Required formulas from handbook:
$\cos(2x) = 2\cos^2(x) - 1.$ 
$\cos(3x) = 4\cos^3(x) - 3\cos(x).$
$\cos^2(x) = \frac{1}{2}[\cos(2x) + 1].$ 
$\cos^4(x) = \frac{1}{8}[\cos(4x) + 4\cos(2x) + 3).$ 
$\cos^6(x) = \frac{1}{32}[\cos(6x) + 6\cos(4x) + 15\cos(2x) + 10].$
The first thing to notice is that when integrating $\cos^2(x), \cos^4(x),$ or $\cos^6(x)$, the indefinite integral will look like $A\sin(6x) + B\sin(4x) + C\sin(2x) + D.$
Since the integral is being taken from $(0)$ to $(2\pi)$, all but the constant term can be ignored.  That is, for example, at $(x = 2\pi), \sin(6x) = 0.$

Therefore
$\cos(x) \cos(2x) \cos(3x) = 
[2\cos^3(x) - \cos(x)] [4\cos^3(x) - 3\cos(x)]$
$= 8\cos^6(x) - 10\cos^4(x) + 3\cos^2(x).$
Therefore 
$\displaystyle I = \int_0^{2\pi} \left[\frac{8}{32}(10)
- \frac{10}{8}(3)
+ \frac{3}{2}(1)\right] dx.
$
This equals 
$\displaystyle (2\pi) \times \left[\frac{10}{4} - \frac{15}{4} + \frac{6}{4}\right] = (2\pi) \times \frac{1}{4} = \frac{\pi}{2}.$
</s>