(i) Let \(x = 2 \sin \theta\), then \(dx = 2 \cos \theta \, d\theta\).

Substitute into the integral:

\(I = \int_0^1 \frac{(2 \sin \theta)^2}{\sqrt{4 - (2 \sin \theta)^2}} \, 2 \cos \theta \, d\theta\).

\(= \int_0^{\frac{\pi}{6}} \frac{4 \sin^2 \theta}{\sqrt{4 - 4 \sin^2 \theta}} \, 2 \cos \theta \, d\theta\).

\(= \int_0^{\frac{\pi}{6}} \frac{4 \sin^2 \theta}{2 \cos \theta} \, 2 \cos \theta \, d\theta\).

\(= \int_0^{\frac{\pi}{6}} 4 \sin^2 \theta \, d\theta\).

(ii) Use the identity \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\).

\(I = \int_0^{\frac{\pi}{6}} 4 \left( \frac{1 - \cos 2\theta}{2} \right) \, d\theta\).

\(= 2 \int_0^{\frac{\pi}{6}} (1 - \cos 2\theta) \, d\theta\).

\(= 2 \left[ \theta - \frac{1}{2} \sin 2\theta \right]_0^{\frac{\pi}{6}}\).

\(= 2 \left( \frac{\pi}{6} - \frac{1}{2} \sin \frac{\pi}{3} \right)\).

\(= 2 \left( \frac{\pi}{6} - \frac{1}{2} \times \frac{\sqrt{3}}{2} \right)\).

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

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