(i) To prove the identity \(\cos 3\theta \equiv 4 \cos^3 \theta - 3 \cos \theta\), we start with the triple angle formula:

\(\cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos \theta - \sin 2\theta \sin \theta\).

Using \(\cos 2\theta = 2\cos^2 \theta - 1\) and \(\sin 2\theta = 2\sin \theta \cos \theta\), we have:

\(\cos 3\theta = (2\cos^2 \theta - 1)\cos \theta - (2\sin \theta \cos \theta)\sin \theta\).

\(\cos 3\theta = 2\cos^3 \theta - \cos \theta - 2\sin^2 \theta \cos \theta\).

Using \(\sin^2 \theta = 1 - \cos^2 \theta\), we get:

\(\cos 3\theta = 2\cos^3 \theta - \cos \theta - 2(1 - \cos^2 \theta)\cos \theta\).

\(\cos 3\theta = 2\cos^3 \theta - \cos \theta - 2\cos \theta + 2\cos^3 \theta\).

\(\cos 3\theta = 4\cos^3 \theta - 3\cos \theta\).

(ii) Using the identity, we find the integral:

\(\int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \cos^3 \theta \, d\theta = \int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \frac{1}{4}(\cos 3\theta + 3\cos \theta) \, d\theta\).

\(= \frac{1}{4} \left[ \int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \cos 3\theta \, d\theta + 3 \int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \cos \theta \, d\theta \right]\).

The integral of \(\cos 3\theta\) is \(\frac{1}{3} \sin 3\theta\), and the integral of \(\cos \theta\) is \(\sin \theta\).

\(= \frac{1}{4} \left[ \frac{1}{3} (\sin(3 \times \frac{1}{2}\pi) - \sin(3 \times \frac{1}{3}\pi)) + 3 (\sin(\frac{1}{2}\pi) - \sin(\frac{1}{3}\pi)) \right]\).

\(= \frac{1}{4} \left[ \frac{1}{3} (0 - \frac{\sqrt{3}}{2}) + 3 (1 - \frac{1}{2}) \right]\).

\(= \frac{1}{4} \left[ -\frac{\sqrt{3}}{6} + \frac{3}{2} \right]\).

\(= \frac{1}{4} \left[ \frac{3}{2} - \frac{\sqrt{3}}{6} \right]\).

\(= \frac{1}{4} \left[ \frac{9}{6} - \frac{\sqrt{3}}{6} \right]\).

\(= \frac{1}{4} \times \frac{9 - \sqrt{3}}{6}\).

\(= \frac{9 - \sqrt{3}}{24}\).

\(= \frac{2}{3} - \frac{3}{8} \sqrt{3}\).