(i) Using the expansions of \(\cos(3x + x)\) and \(\cos(3x - x)\), show that \(\frac{1}{2}(\cos 4x + \cos 2x) = \cos 3x \cos x\).

(ii) Hence show that \(\int_{-\frac{1}{6}\pi}^{\frac{1}{6}\pi} \cos 3x \cos x \, dx = \frac{3}{8}\sqrt{3}\).