Ta có:
$\int\limits_0^1\cos x\cos2x\cos3xdx$
$=\int\limits_0^1\dfrac{\cos2x(\cos4x+\cos2x)}{2}dx$
$=\int\limits_0^1\dfrac{\cos2x\cos4x+\cos^22x)}{2}dx$
$=\int\limits_0^1\dfrac{\cos2x+\cos6x+1+\cos4x)}{4}dx$
$=\left(\dfrac{\sin2x}{8}+\dfrac{\sin6x}{24}+\dfrac{x}{4}+\dfrac{\sin4x}{16}\right)\left|\begin{array}{l}1\\0\end{array}\right.$
$=\dfrac{\sin2}{8}+\dfrac{\sin6}{24}+\dfrac{1}{4}+\dfrac{\sin4}{16}$