(i) To express \((\sqrt{3}) \cos x + \sin x\) in the form \(R \cos(x - \alpha)\), we use the identity:

\(R \cos(x - \alpha) = R \cos \alpha \cos x + R \sin \alpha \sin x\).

Comparing coefficients, we have:

\(R \cos \alpha = \sqrt{3}\) and \(R \sin \alpha = 1\).

Solving for \(R\), we use:

\(R^2 = (R \cos \alpha)^2 + (R \sin \alpha)^2 = 3 + 1 = 4\).

Thus, \(R = 2\).

To find \(\alpha\), we use:

\(\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{1}{\sqrt{3}}\).

Therefore, \(\alpha = \frac{1}{6}\pi\).

(ii) We need to evaluate:

\(\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi} \frac{1}{((\sqrt{3}) \cos x + \sin x)^2} \, dx\).

Substitute \((\sqrt{3}) \cos x + \sin x = 2 \cos(x - \frac{1}{6}\pi)\).

The integral becomes:

\(\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi} \frac{1}{4 \cos^2(x - \frac{1}{6}\pi)} \, dx\).

This is equivalent to:

\(\frac{1}{4} \int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi} \sec^2(x - \frac{1}{6}\pi) \, dx\).

The integral of \(\sec^2(x - \frac{1}{6}\pi)\) is \(\tan(x - \frac{1}{6}\pi)\).

Thus, the integral evaluates to:

\(\frac{1}{4} \left[ \tan(x - \frac{1}{6}\pi) \right]_{\frac{1}{6}\pi}^{\frac{1}{2}\pi}\).

Calculating the limits:

\(\tan(\frac{1}{2}\pi - \frac{1}{6}\pi) = \tan(\frac{1}{3}\pi) = \sqrt{3}\).

\(\tan(\frac{1}{6}\pi - \frac{1}{6}\pi) = \tan(0) = 0\).

Thus, the integral is:

\(\frac{1}{4} (\sqrt{3} - 0) = \frac{1}{4}\sqrt{3}\).