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

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

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

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

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

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

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

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

(i) Let \(x = 2 \tan \theta\). Then \(dx = 2 \sec^2 \theta \, d\theta\).

Substitute into the integral:

\(\int_0^2 \frac{8}{(4+x^2)^2} \, dx = \int_0^{\frac{\pi}{4}} \frac{8}{(4 + (2 \tan \theta)^2)^2} \cdot 2 \sec^2 \theta \, d\theta.\)

Simplify the expression:

\(4 + (2 \tan \theta)^2 = 4 + 4 \tan^2 \theta = 4 \sec^2 \theta.\)

Thus, the integral becomes:

\(\int_0^{\frac{\pi}{4}} \frac{8 \cdot 2 \sec^2 \theta}{(4 \sec^2 \theta)^2} \, d\theta = \int_0^{\frac{\pi}{4}} \cos^2 \theta \, d\theta.\)

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

Integrate:

\(\int_0^{\frac{\pi}{4}} \cos^2 \theta \, d\theta = \int_0^{\frac{\pi}{4}} \frac{1}{2}(1 + \cos 2\theta) \, d\theta = \frac{1}{2} \left[ \theta + \frac{1}{2} \sin 2\theta \right]_0^{\frac{\pi}{4}}.\)

Evaluate the limits:

\(\frac{1}{2} \left( \frac{\pi}{4} + \frac{1}{2} \cdot 1 - (0 + 0) \right) = \frac{1}{8}(\pi + 2).\)

Let \(u = \sqrt{x}\), then \(x = u^2\).

Differentiating both sides, \(\frac{du}{dx} = \frac{1}{2\sqrt{x}} = \frac{1}{2u}\), so \(dx = 2u \, du\).

Substitute \(x = u^2\) and \(dx = 2u \, du\) into the integral:

\(I = \int_1^4 \frac{1}{u^2(4-u)} \, (2u \, du)\).

This simplifies to:

\(I = \int_1^2 \frac{2}{u(4-u)} \, du\).

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

Substitute into the integral:

\(\int \sqrt{\left( \frac{x}{1-x} \right)} \, dx = \int \sqrt{\left( \frac{\sin^2 \theta}{1-\sin^2 \theta} \right)} \, 2 \sin \theta \cos \theta \, d\theta\).

Since \(1 - \sin^2 \theta = \cos^2 \theta\), the expression simplifies to:

\(\int \sqrt{\left( \frac{\sin^2 \theta}{\cos^2 \theta} \right)} \, 2 \sin \theta \cos \theta \, d\theta = \int 2 \sin^2 \theta \, d\theta\).

(ii) Use the result from part (i):

\(\int_0^{\frac{1}{4}} \sqrt{\left( \frac{x}{1-x} \right)} \, dx = \int_0^{\frac{\pi}{6}} 2 \sin^2 \theta \, d\theta\).

Using the identity \(\sin^2 \theta = \frac{1}{2}(1 - \cos 2\theta)\), the integral becomes:

\(\int_0^{\frac{\pi}{6}} (1 - \cos 2\theta) \, d\theta = \left[ \theta - \frac{1}{2} \sin 2\theta \right]_0^{\frac{\pi}{6}}\).

Evaluating the limits:

\(\left( \frac{\pi}{6} - \frac{1}{2} \sin \frac{\pi}{3} \right) - \left( 0 - 0 \right) = \frac{\pi}{6} - \frac{1}{4} \sqrt{3}\).

(i) Let \(x = \tan \theta\). Then \(dx = \sec^2 \theta \, d\theta\).

Substitute into the integral:

\(\int \frac{1-x^2}{(1+x^2)^2} \, dx = \int \frac{1-\tan^2 \theta}{(1+\tan^2 \theta)^2} \sec^2 \theta \, d\theta.\)

Using the identity \(1 + \tan^2 \theta = \sec^2 \theta\), the integral becomes:

\(\int \frac{1-\tan^2 \theta}{\sec^4 \theta} \sec^2 \theta \, d\theta = \int \frac{\sec^2 \theta - \tan^2 \theta}{\sec^4 \theta} \sec^2 \theta \, d\theta.\)

Simplify to:

\(\int \frac{\sec^2 \theta - \tan^2 \theta}{\sec^2 \theta} \, d\theta = \int (1 - \sin^2 \theta) \, d\theta = \int \cos 2\theta \, d\theta.\)

(ii) From part (i), we have:

\(\int_0^1 \frac{1-x^2}{(1+x^2)^2} \, dx = \int_0^{\frac{\pi}{4}} \cos 2\theta \, d\theta.\)

The integral of \(\cos 2\theta\) is \(\frac{1}{2} \sin 2\theta\).

Evaluate from \(\theta = 0\) to \(\theta = \frac{\pi}{4}\):

\(\frac{1}{2} \sin 2\theta \bigg|_0^{\frac{\pi}{4}} = \frac{1}{2} (\sin \frac{\pi}{2} - \sin 0) = \frac{1}{2} (1 - 0) = \frac{1}{2}.\)