Ta có:
$\int\limits_0^{\frac{\pi}{2}}\cos x\cos2x\cos3xdx$
$=\dfrac{1}{2}\int\limits_0^{\frac{\pi}{2}}\cos2x(\cos4x+\cos2x)dx$
$=\dfrac{1}{4}\int\limits_0^{\frac{\pi}{2}}(\cos6x+\cos4x+\cos2x+1)dx$
$=\left(\dfrac{\sin6x}{24}+\dfrac{\sin4x}{16}+\dfrac{\sin2x}{8}+\dfrac{x}{4}\right)\left|\begin{array}{l}\dfrac{\pi}{2}\\0\end{array}\right.=\dfrac{\pi}{8}$