(i) Let \(x = \cos^2 \theta\). Then \(dx = -2 \cos \theta \sin \theta \, d\theta = -\sin 2\theta \, d\theta\).

Substitute \(x = \cos^2 \theta\) into the integral:

\(\sqrt{\frac{x}{1-x}} = \sqrt{\frac{\cos^2 \theta}{1-\cos^2 \theta}} = \sqrt{\frac{\cos^2 \theta}{\sin^2 \theta}} = \frac{\cos \theta}{\sin \theta} = \cot \theta\).

Thus, the integral becomes:

\(I = \int \cot \theta (-\sin 2\theta \, d\theta) = \int -\cos 2\theta \, d\theta\).

Change the limits: when \(x = \frac{1}{4}\), \(\cos^2 \theta = \frac{1}{4}\), so \(\theta = \frac{1}{3}\pi\). When \(x = \frac{3}{4}\), \(\cos^2 \theta = \frac{3}{4}\), so \(\theta = \frac{1}{6}\pi\).

Therefore, \(I = \int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} 2 \cos^2 \theta \, d\theta\).

(ii) The integral \(\int 2 \cos^2 \theta \, d\theta = \int (1 + \cos 2\theta) \, d\theta = \theta + \frac{1}{2} \sin 2\theta\).

Evaluate from \(\frac{1}{6}\pi\) to \(\frac{1}{3}\pi\):

\(\left[ \theta + \frac{1}{2} \sin 2\theta \right]_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} = \left( \frac{1}{3}\pi + \frac{1}{2} \sin \frac{2}{3}\pi \right) - \left( \frac{1}{6}\pi + \frac{1}{2} \sin \frac{1}{3}\pi \right)\).

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