(a) Let \(x = 3 \tan \theta\). Then \(dx = 3 \sec^2 \theta \, d\theta\).

Substitute into the integral:

\(I = \int_0^3 \frac{27}{(9 + x^2)^2} \, dx = \int_0^{\frac{\pi}{4}} \frac{27}{(9 + 9 \tan^2 \theta)^2} \cdot 3 \sec^2 \theta \, d\theta\).

\(= \int_0^{\frac{\pi}{4}} \frac{81 \sec^2 \theta}{81 \sec^4 \theta} \, d\theta\).

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

(b) The integral \(\int \cos^2 \theta \, d\theta\) can be expressed as \(\int \frac{1 + \cos 2\theta}{2} \, d\theta\).

\(= \frac{1}{2} \int (1 + \cos 2\theta) \, d\theta\).

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

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

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

\(= \frac{1}{8}(\pi + 2)\).

Let \(u = 1 + 3 \tan x\). Then, \(\frac{du}{dx} = 3 \sec^2 x\), so \(du = 3 \sec^2 x \, dx\).

Rearranging gives \(dx = \frac{du}{3 \sec^2 x}\).

Substitute into the integral:

\(\int \frac{\sqrt{u}}{\cos^2 x} \cdot \frac{du}{3 \sec^2 x} = \int \frac{\sqrt{u}}{3} \, du.\)

This simplifies to:

\(\frac{1}{3} \int u^{\frac{1}{2}} \, du.\)

Integrate:

\(\frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}} = \frac{2}{9} u^{\frac{3}{2}}.\)

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

When \(x = 0\), \(u = 1 + 3 \tan 0 = 1\).

When \(x = \frac{\pi}{4}\), \(u = 1 + 3 \tan \frac{\pi}{4} = 4\).

Substitute these limits into the integrated function:

\(\frac{2}{9} \left[ 4^{\frac{3}{2}} - 1^{\frac{3}{2}} \right] = \frac{2}{9} (8 - 1) = \frac{2}{9} \times 7 = \frac{14}{9}.\)

Let \(u = 3x + 1\). Then, \(\frac{du}{dx} = 3\), so \(dx = \frac{1}{3} du\).

Substitute into the integral:

\(\int \frac{3x}{3x+1} \, dx = \int \left( \frac{1}{3} - \frac{1}{3u} \right) du\).

Integrate each term separately:

\(\int \frac{1}{3} \, du = \frac{1}{3} u\).

\(\int -\frac{1}{3u} \, du = -\frac{1}{3} \ln|u|\).

Combine the results:

\(\frac{1}{3} u - \frac{1}{3} \ln|u| + c\).

Substitute back \(u = 3x + 1\):

\(\frac{1}{3}(3x+1) - \frac{1}{3} \ln(3x+1) + c\).

Using the substitution \(x = \sqrt{3} \tan \theta\), we have \(dx = \sqrt{3} \sec^2 \theta \, d\theta\).

The integral becomes:

\(\int \frac{1}{\sqrt{3 + 3 \tan^2 \theta}} \cdot \sqrt{3} \sec^2 \theta \, d\theta = \int \sec \theta \, d\theta.\)

We need to change the limits of integration. When \(x = 1\), \(\sqrt{3} \tan \theta = 1\), so \(\tan \theta = \frac{1}{\sqrt{3}}\), giving \(\theta = \frac{\pi}{6}\).

When \(x = 3\), \(\sqrt{3} \tan \theta = 3\), so \(\tan \theta = \sqrt{3}\), giving \(\theta = \frac{\pi}{3}\).

Thus, the integral becomes:

\(\int_{\pi/6}^{\pi/3} \sec \theta \, d\theta.\)

The integral of \(\sec \theta\) is \(\ln |\sec \theta + \tan \theta|\).

Evaluating from \(\theta = \frac{\pi}{6}\) to \(\theta = \frac{\pi}{3}\):

\(\ln |\sec \frac{\pi}{3} + \tan \frac{\pi}{3}| - \ln |\sec \frac{\pi}{6} + \tan \frac{\pi}{6}|.\)

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

Thus, the expression becomes:

\(\ln (2 + \sqrt{3}) - \ln \left( \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} \right) = \ln (2 + \sqrt{3}) - \ln \left( \frac{3}{\sqrt{3}} \right).\)

\(= \ln (2 + \sqrt{3}) - \ln \sqrt{3} = \ln \left( \frac{2 + \sqrt{3}}{\sqrt{3}} \right).\)

Let \(u = \sin 4x\), then \(\frac{du}{dx} = 4\cos 4x\) or \(du = 4\cos 4x \, dx\).

Thus, \(dx = \frac{du}{4\cos 4x}\).

Substitute into the integral:

\(\int \cos^3 4x \, dx = \int \cos^3 4x \cdot \frac{du}{4\cos 4x} = \frac{1}{4} \int \cos^2 4x \, du\).

Since \(\cos^2 4x = 1 - \sin^2 4x = 1 - u^2\), the integral becomes:

\(\frac{1}{4} \int (1 - u^2) \, du\).

Integrate to obtain:

\(\frac{1}{4} \left( u - \frac{u^3}{3} \right) + C\).

Evaluate from \(u = \sin 4(0) = 0\) to \(u = \sin 4\left(\frac{1}{24}\pi\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\).

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

\(= \frac{1}{4} \left( \frac{1}{2} - \frac{1}{24} \right)\).

\(= \frac{1}{4} \cdot \frac{11}{24}\).

\(= \frac{11}{96}\).

(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 \(\theta = \frac{\pi}{4}\):

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

(a) To find the stationary point, we need to find \(\frac{dy}{dx}\) and set it to zero. Using the quotient rule, \(\frac{dy}{dx} = \frac{(9-x)(-\frac{1}{2}x^{-3/2}) - \sqrt{x}}{(9-x)^2 x}\).

Simplifying, \(\frac{dy}{dx} = \frac{-\frac{1}{2}(9-x)x^{-3/2} - x^{-1/2}}{(9-x)^2}\).

Setting \(\frac{dy}{dx} = 0\), we solve for \(x\) to get \(x = 3\).

(b) Using the substitution \(u = \sqrt{x}\), then \(x = u^2\) and \(dx = 2u \, du\).

The integral becomes \(\int_0^2 \frac{1}{(9-u^2)u} \, 2u \, du = \int_0^2 \frac{2}{9-u^2} \, du\).

This can be integrated using partial fractions or a standard integral formula to obtain \(\frac{1}{3} \ln \left( \frac{3+u}{3-u} \right)\).

Evaluating from \(u = 0\) to \(u = 2\), we get \(\frac{1}{3} \ln 5\).

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

The limits change from \(x = 3\) to \(u = \sqrt{3}\) and \(x = \infty\) to \(u = \infty\).

Substitute into the integral:

\(\int_{\sqrt{3}}^{\infty} \frac{1}{(u^2 + 1)u} \, 2u \, du = \int_{\sqrt{3}}^{\infty} \frac{2}{u^2 + 1} \, du\).

The integral \(\int \frac{2}{u^2 + 1} \, du\) is \(2 \arctan(u)\).

Evaluate from \(\sqrt{3}\) to \(\infty\):

\(2 \left[ \arctan(\infty) - \arctan(\sqrt{3}) \right] = 2 \left[ \frac{\pi}{2} - \frac{\pi}{3} \right]\).

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

(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 \(\frac{1}{3}\pi\):

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

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

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

Differentiate to find \(dx\):

\(\frac{du}{dx} = \frac{1}{2\sqrt{x}} = \frac{1}{2u}\)

Thus, \(du = \frac{1}{2u} \, dx\) or \(dx = 2u \, du\).

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

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

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

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

Change the limits of integration:

When \(x = 1\), \(u = \sqrt{1} = 1\).

When \(x = 4\), \(u = \sqrt{4} = 2\).

Thus, \(I = \int_{1}^{2} \frac{u - 1}{u + 1} \, du\).

(i) Let \(u = 1 + x^2\). Then \(du = 2x \, dx\), so \(dx = \frac{du}{2x}\).

Substitute \(x^2 = u - 1\) into the integral:

\(I = \int_0^1 \frac{x^5}{(1+x^2)^3} \, dx = \int_1^2 \frac{(u-1)^{5/2}}{u^3} \cdot \frac{du}{2\sqrt{u-1}}\).

Simplifying, we get:

\(I = \int_1^2 \frac{(u-1)^2}{2u^3} \, du\).

(ii) To find the exact value of \(I\), integrate:

\(\int \frac{(u-1)^2}{2u^3} \, du = \int \left( \frac{1}{2u} - \frac{1}{u^2} + \frac{1}{4u^2} \right) \, du\).

This gives:

\(\frac{1}{2} \ln u + \frac{1}{u} - \frac{1}{4u^2}\).

Evaluate from 1 to 2:

\(\left[ \frac{1}{2} \ln u + \frac{1}{u} - \frac{1}{4u^2} \right]_1^2 = \left( \frac{1}{2} \ln 2 + \frac{1}{2} - \frac{1}{16} \right) - \left( 0 + 1 - \frac{1}{4} \right)\).

Simplifying, we find:

\(\frac{1}{2} \ln 2 - \frac{5}{16}\).

(i) Let \(x = \sqrt{3} \tan \theta\). Then \(dx = \sqrt{3} \sec^2 \theta \, d\theta\).

Substitute into the integral:

\(I = \int_0^1 \frac{9}{(3 + x^2)^2} \, dx = \int_0^{\frac{\pi}{6}} \frac{9}{(3 + 3 \tan^2 \theta)^2} \cdot \sqrt{3} \sec^2 \theta \, d\theta\).

\(= \sqrt{3} \int_0^{\frac{\pi}{6}} \frac{9 \sec^2 \theta}{(3 \sec^2 \theta)^2} \, d\theta\).

\(= \sqrt{3} \int_0^{\frac{\pi}{6}} \cos^2 \theta \, d\theta\).

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

\(I = \sqrt{3} \int_0^{\frac{\pi}{6}} \left( \frac{1}{2} \cos 2\theta + \frac{1}{2} \right) \, d\theta\).

\(= \sqrt{3} \left[ \frac{1}{4} \sin 2\theta + \frac{1}{2} \theta \right]_0^{\frac{\pi}{6}}\).

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

\(= \sqrt{3} \left( \frac{1}{4} \cdot \frac{\sqrt{3}}{2} + \frac{\pi}{12} \right)\).

\(= \frac{1}{12} \sqrt{3} \pi + \frac{3}{8}\).

Let \(u = 4 - 3 \cos x\). Then, \(du = 3 \sin x \, dx\).

Using the identity \(\sin 2x = 2 \sin x \cos x\), we have:

\(\int \frac{9 \sin 2x}{\sqrt{4 - 3 \cos x}} \, dx = \int \frac{18 \sin x \cos x}{\sqrt{u}} \, dx\).

Substitute \(dx = \frac{du}{3 \sin x}\), so:

\(\int \frac{18 \sin x \cos x}{\sqrt{u}} \cdot \frac{du}{3 \sin x} = \int \frac{6 \cos x}{\sqrt{u}} \, du\).

Since \(\cos x = \frac{4 - u}{3}\), the integral becomes:

\(\int \frac{6 \left(\frac{4 - u}{3}\right)}{\sqrt{u}} \, du = \int \frac{8 - 2u}{\sqrt{u}} \, du\).

Integrate to obtain:

\(\int \frac{8}{\sqrt{u}} \, du - \int \frac{2u}{\sqrt{u}} \, du = 16u^{\frac{1}{2}} - \frac{4}{3}u^{\frac{3}{2}}\).

Apply the limits from \(x = 0\) to \(x = \frac{1}{2}\pi\):

When \(x = 0, u = 4 - 3 \cdot 1 = 1\).

When \(x = \frac{1}{2}\pi, u = 4 - 3 \cdot 0 = 4\).

Evaluate:

\(\left[ 16u^{\frac{1}{2}} - \frac{4}{3}u^{\frac{3}{2}} \right]_1^4 = \left( 16 \cdot 2 - \frac{4}{3} \cdot 8 \right) - \left( 16 \cdot 1 - \frac{4}{3} \cdot 1 \right)\).

\(= (32 - \frac{32}{3}) - (16 - \frac{4}{3})\).

\(= \frac{64}{3} - \frac{44}{3} = \frac{20}{3}\).

(i) Let \(u = 2 - \sqrt{x}\). Then \(\sqrt{x} = 2 - u\) and \(x = (2-u)^2\).

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

Thus, \(dx = -2\sqrt{x} \, du = -2(2-u) \, du\).

Substitute into the integral:

\(I = \int_0^1 \frac{\sqrt{x}}{2 - \sqrt{x}} \, dx = \int_1^2 \frac{2(2-u)^2}{u} \, du\).

(ii) Integrate \(\int_1^2 \frac{2(2-u)^2}{u} \, du\):

\(\int \frac{2(2-u)^2}{u} \, du = \int \left( \frac{8}{u} - 8 + u \right) \, du\).

\(= 8 \ln u - 8u + \frac{u^2}{2} + C\).

Evaluate from 1 to 2:

\(\left[ 8 \ln u - 8u + \frac{u^2}{2} \right]_1^2 = \left( 8 \ln 2 - 16 + 2 \right) - \left( 0 - 8 + \frac{1}{2} \right)\).

\(= 8 \ln 2 - 14 + 8 - \frac{1}{2}\).

\(= 8 \ln 2 - 5\).