First, split the fraction:

\(\frac{x(x+1)}{x^2+4} = 1 + \frac{x-4}{x^2+4}\)

Integrate the first part:

\(\int 1 \, dx = x\)

For the second part, use partial fraction decomposition:

\(\int \frac{x-4}{x^2+4} \, dx = \frac{1}{2} \ln(x^2+4) - 2 \arctan\left(\frac{x}{2}\right)\)

Evaluate from 0 to 6:

\(\left[ x + \frac{1}{2} \ln(x^2+4) - 2 \arctan\left(\frac{x}{2}\right) \right]_{0}^{6}\)

Calculate at the bounds:

At \(x = 6\):

\(6 + \frac{1}{2} \ln(36+4) - 2 \arctan 3\)

At \(x = 0\):

\(0 + \frac{1}{2} \ln 4 - 2 \arctan 0\)

Subtract the results:

\(\left( 6 + \frac{1}{2} \ln 40 - 2 \arctan 3 \right) - \left( \frac{1}{2} \ln 4 \right)\)

Simplify:

\(6 + \frac{1}{2} \ln 10 - 2 \arctan 3\)

(i) The expansion of \(\cos(3x + x)\) is \(\cos 3x \cos x - \sin 3x \sin x\).

The expansion of \(\cos(3x - x)\) is \(\cos 3x \cos x + \sin 3x \sin x\).

Adding these, we get:

\(\cos(3x + x) + \cos(3x - x) = 2 \cos 3x \cos x\).

Thus, \(\frac{1}{2}(\cos 4x + \cos 2x) = \cos 3x \cos x\).

(ii) We have \(\int \cos 3x \cos x \, dx = \frac{1}{2} \int (\cos 4x + \cos 2x) \, dx\).

Integrating, we get:

\(\frac{1}{2} \left( \frac{1}{4} \sin 4x + \frac{1}{2} \sin 2x \right) = \frac{1}{8} \sin 4x + \frac{1}{4} \sin 2x\).

Substituting the limits \(-\frac{1}{6}\pi\) to \(\frac{1}{6}\pi\), we find:

\(\left[ \frac{1}{8} \sin 4x + \frac{1}{4} \sin 2x \right]_{-\frac{1}{6}\pi}^{\frac{1}{6}\pi} = \frac{3}{8}\sqrt{3}\).

(i) Start with the given equation: \(x = \ln(1-y) - \ln y\).

This can be rewritten using properties of logarithms: \(x = \ln \left( \frac{1-y}{y} \right)\).

Exponentiating both sides gives: \(e^x = \frac{1-y}{y}\).

Rearranging for \(y\), we get: \(y = \frac{1}{1 + e^x}\).

Substitute \(e^x = \frac{1}{e^{-x}}\) to obtain: \(y = \frac{e^{-x}}{1 + e^{-x}}\).

(ii) We need to evaluate \(\int_0^1 y \, dx\).

Using the expression for \(y\), we have \(y = \frac{e^{-x}}{1 + e^{-x}}\).

Let \(u = 1 + e^{-x}\), then \(du = -e^{-x} \, dx\), so \(dx = -\frac{du}{e^{-x}}\).

Substitute into the integral: \(\int y \, dx = \int \frac{e^{-x}}{u} \left(-\frac{du}{e^{-x}}\right) = -\int \frac{1}{u} \, du\).

This evaluates to \(-\ln |u| + C = -\ln(1 + e^{-x}) + C\).

Evaluate from 0 to 1:

At \(x = 1\), \(-\ln(1 + e^{-1})\).

At \(x = 0\), \(-\ln(1 + e^0) = -\ln(2)\).

Thus, \(\int_0^1 y \, dx = -\ln(1 + e^{-1}) + \ln(2)\).

Simplifying gives: \(\ln \left( \frac{2}{1 + e^{-1}} \right) = \ln \left( \frac{2e}{e+1} \right)\).

(i) Start with the left-hand side: \(\tan 2\theta - \tan \theta\).

Using the identity \(\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}\), substitute:

\(\frac{2\tan \theta}{1 - \tan^2 \theta} - \tan \theta\).

Combine into a single fraction:

\(\frac{2\tan \theta - \tan \theta (1 - \tan^2 \theta)}{1 - \tan^2 \theta} = \frac{2\tan \theta - \tan \theta + \tan^3 \theta}{1 - \tan^2 \theta}\).

Simplify the numerator:

\(\frac{\tan \theta (1 + \tan^2 \theta)}{1 - \tan^2 \theta}\).

Recognize that \(\sec 2\theta = \frac{1}{\cos 2\theta} = \frac{1}{1 - \tan^2 \theta}\).

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

Therefore, \(\tan 2\theta - \tan \theta = \tan \theta \sec 2\theta\).

(ii) Consider the integral \(\int_{0}^{\frac{1}{6}\pi} \tan \theta \sec 2\theta \, d\theta\).

Substitute \(\sec 2\theta = \frac{1}{\cos 2\theta}\), so the integral becomes:

\(\int_{0}^{\frac{1}{6}\pi} \frac{\tan \theta}{\cos 2\theta} \, d\theta\).

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

Integrate to obtain:

\(-\frac{1}{2} \ln(\cos 2\theta) + \ln(\cos \theta)\).

Evaluate from \(0\) to \(\frac{1}{6}\pi\):

\(-\frac{1}{2} \ln(\cos(\frac{1}{3}\pi)) + \ln(\cos(\frac{1}{6}\pi))\).

Substitute the values: \(\cos(\frac{1}{3}\pi) = \frac{1}{2}\) and \(\cos(\frac{1}{6}\pi) = \frac{\sqrt{3}}{2}\).

Resulting in:

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

(a) Use the identity \(\tan^2 2x = \sec^2 2x - 1\) to rewrite the integral as \(\int (4 + \sec^2 2x - 1) \, dx = \int (3 + \sec^2 2x) \, dx\).

The integral of \(3 \, dx\) is \(3x\), and the integral of \(\sec^2 2x \, dx\) is \(\frac{1}{2} \tan 2x\) (using substitution \(u = 2x\), \(du = 2 \, dx\)).

Thus, the integral is \(3x + \frac{1}{2} \tan 2x\).

(b) Use the identity \(\sin x \cos \frac{1}{2}\pi + \cos x \sin \frac{1}{6}\pi\) to simplify the integrand to \(\cos \frac{1}{6}\pi + \frac{\cos x \sin \frac{1}{6}\pi}{\sin x}\).

The integral becomes \(\int_{\frac{1}{4}\pi}^{\frac{1}{2}\pi} \left( \cos \frac{1}{6}\pi + \frac{\cos x \sin \frac{1}{6}\pi}{\sin x} \right) \, dx\).

Integrate to obtain \(\cos \frac{1}{6}\pi x + \sin \frac{1}{6}\pi \ln(\sin x)\).

Apply the limits and simplify to obtain \(\frac{1}{8} \pi \sqrt{3} - \frac{1}{2} \ln \left( \frac{1}{\sqrt{2}} \right)\).

(i) Start with the identity for \(\cot \theta\) and \(\tan \theta\):

\(\cot \theta = \frac{\cos \theta}{\sin \theta}\) and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).

Add them: \(\cot \theta + \tan \theta = \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}\).

Using the double angle identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), we have:

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

Thus, \(\cot \theta + \tan \theta \equiv 2 \csc 2\theta\).

(ii) We need to evaluate \(\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \csc 2\theta \, d\theta\).

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

\(\int \csc 2\theta \, d\theta = \int \frac{1}{\sin 2\theta} \, d\theta\).

Let \(u = 2\theta\), then \(du = 2 \, d\theta\) or \(d\theta = \frac{1}{2} \, du\).

The integral becomes \(\frac{1}{2} \int \csc u \, du\).

The integral of \(\csc u\) is \(\ln |\csc u - \cot u|\), so:

\(\frac{1}{2} \ln |\csc u - \cot u|\).

Substitute back \(u = 2\theta\):

\(\frac{1}{2} \ln |\csc 2\theta - \cot 2\theta|\).

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

\(\frac{1}{2} \left[ \ln |\csc \frac{2}{3}\pi - \cot \frac{2}{3}\pi| - \ln |\csc \frac{1}{3}\pi - \cot \frac{1}{3}\pi| \right]\).

Calculate the values:

\(\csc \frac{2}{3}\pi = \frac{1}{\sin \frac{2}{3}\pi} = \frac{2}{\sqrt{3}}\), \(\cot \frac{2}{3}\pi = \frac{1}{\tan \frac{2}{3}\pi} = -\frac{1}{\sqrt{3}}\).

\(\csc \frac{1}{3}\pi = \frac{1}{\sin \frac{1}{3}\pi} = 2\), \(\cot \frac{1}{3}\pi = \frac{1}{\tan \frac{1}{3}\pi} = \sqrt{3}\).

Substitute these back:

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

\(\frac{1}{2} \left[ \ln \left( \frac{3}{\sqrt{3}} \right) - \ln \left( 2 - \sqrt{3} \right) \right]\).

\(\frac{1}{2} \left[ \ln 3 - \ln 1 \right] = \frac{1}{2} \ln 3\).

(i) Express \(\cos 4\theta\) as \(2 \cos^2 2\theta - 1\) or \(\cos^2 2\theta - \sin^2 2\theta\) or \(1 - 2 \sin^2 2\theta\).

Express \(\cos 4\theta\) in terms of \(\cos \theta\).

Obtain \(8 \cos^4 \theta - 8 \cos^2 \theta + 1\).

Use \(\cos 2\theta = 2 \cos^2 \theta - 1\) to obtain given answer \(8 \cos^4 \theta - 3\).

(ii) (a) State or imply \(\cos^4 \theta = \frac{1}{2}\).

Obtain \(0.572\).

Obtain \(-0.572\).

(b) Integrate and obtain form \(k_1 \theta + k_2 \sin 4\theta + k_3 \sin 2\theta\).

Obtain \(\frac{3}{8} \theta + \frac{1}{32} \sin 4\theta + \frac{1}{4} \sin 2\theta\).

Obtain \(\frac{3}{32}\pi + \frac{1}{4}\) following completely correct work.

(i) To find \(f'(x)\), we first differentiate \(f(x) = 4 \cos^2 3x\) using the chain rule. Let \(u = \cos 3x\), then \(f(x) = 4u^2\). The derivative \(\frac{du}{dx} = -3 \sin 3x\).

Using the chain rule, \(\frac{df}{dx} = 8u \cdot \frac{du}{dx} = 8 \cos 3x (-3 \sin 3x) = -24 \cos 3x \sin 3x\).

Now, evaluate \(f'(\frac{1}{9}\pi)\):

\(\cos 3(\frac{1}{9}\pi) = \cos \frac{\pi}{3} = \frac{1}{2}\)

\(\sin 3(\frac{1}{9}\pi) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\)

Thus, \(f'(\frac{1}{9}\pi) = -24 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = -6\sqrt{3}\).

(ii) To integrate \(f(x) = 4 \cos^2 3x\), use the identity \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\).

\(f(x) = 4 \cdot \frac{1 + \cos 6x}{2} = 2 + 2 \cos 6x\).

Integrate term by term:

\(\int (2 + 2 \cos 6x) \, dx = \int 2 \, dx + \int 2 \cos 6x \, dx\)

\(= 2x + \frac{2}{6} \sin 6x + C = 2x + \frac{1}{3} \sin 6x + C\).

(i) To prove the identity \(\cos 3\theta \equiv 4 \cos^3 \theta - 3 \cos \theta\), we start with the triple angle formula:

\(\cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos \theta - \sin 2\theta \sin \theta\).

Using \(\cos 2\theta = 2\cos^2 \theta - 1\) and \(\sin 2\theta = 2\sin \theta \cos \theta\), we have:

\(\cos 3\theta = (2\cos^2 \theta - 1)\cos \theta - (2\sin \theta \cos \theta)\sin \theta\).

\(\cos 3\theta = 2\cos^3 \theta - \cos \theta - 2\sin^2 \theta \cos \theta\).

Using \(\sin^2 \theta = 1 - \cos^2 \theta\), we get:

\(\cos 3\theta = 2\cos^3 \theta - \cos \theta - 2(1 - \cos^2 \theta)\cos \theta\).

\(\cos 3\theta = 2\cos^3 \theta - \cos \theta - 2\cos \theta + 2\cos^3 \theta\).

\(\cos 3\theta = 4\cos^3 \theta - 3\cos \theta\).

(ii) Using the identity, we find the integral:

\(\int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \cos^3 \theta \, d\theta = \int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \frac{1}{4}(\cos 3\theta + 3\cos \theta) \, d\theta\).

\(= \frac{1}{4} \left[ \int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \cos 3\theta \, d\theta + 3 \int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \cos \theta \, d\theta \right]\).

The integral of \(\cos 3\theta\) is \(\frac{1}{3} \sin 3\theta\), and the integral of \(\cos \theta\) is \(\sin \theta\).

\(= \frac{1}{4} \left[ \frac{1}{3} (\sin(3 \times \frac{1}{2}\pi) - \sin(3 \times \frac{1}{3}\pi)) + 3 (\sin(\frac{1}{2}\pi) - \sin(\frac{1}{3}\pi)) \right]\).

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

\(= \frac{1}{4} \left[ -\frac{\sqrt{3}}{6} + \frac{3}{2} \right]\).

\(= \frac{1}{4} \left[ \frac{3}{2} - \frac{\sqrt{3}}{6} \right]\).

\(= \frac{1}{4} \left[ \frac{9}{6} - \frac{\sqrt{3}}{6} \right]\).

\(= \frac{1}{4} \times \frac{9 - \sqrt{3}}{6}\).

\(= \frac{9 - \sqrt{3}}{24}\).

\(= \frac{2}{3} - \frac{3}{8} \sqrt{3}\).

(i) Use the angle subtraction and addition formulas:

\(\cos(3x-x) = \cos 2x = \cos 3x \cos x + \sin 3x \sin x\)

\(\cos(3x+x) = \cos 4x = \cos 3x \cos x - \sin 3x \sin x\)

Subtract these equations:

\(\cos 2x - \cos 4x = 2 \sin 3x \sin x\)

Thus, \(\frac{1}{2}(\cos 2x - \cos 4x) = \sin 3x \sin x\).

(ii) The integral becomes:

\(\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \sin 3x \sin x \, dx = \int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \left( \frac{1}{4} \sin 2x - \frac{1}{8} \sin 4x \right) \, dx\)

Integrate each term:

\(\int \sin 2x \, dx = -\frac{1}{2} \cos 2x\)

\(\int \sin 4x \, dx = -\frac{1}{4} \cos 4x\)

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

\(\left[ -\frac{1}{8} \cos 2x + \frac{1}{32} \cos 4x \right]_{\frac{1}{6}\pi}^{\frac{1}{3}\pi}\)

Calculate the definite integral:

\(= \left( -\frac{1}{8} \cos \frac{2}{3}\pi + \frac{1}{32} \cos \frac{4}{3}\pi \right) - \left( -\frac{1}{8} \cos \frac{1}{3}\pi + \frac{1}{32} \cos \frac{2}{3}\pi \right)\)

\(= \frac{1}{8}\sqrt{3}\)

(i) To prove the identity \(\cos 4\theta - 4 \cos 2\theta + 3 \equiv 8 \sin^4 \theta\), we start by using the double angle formulas:

\(\cos 2\theta = 1 - 2\sin^2 \theta\)

\(\cos 4\theta = 2\cos^2 2\theta - 1 = 2(1 - 2\sin^2 \theta)^2 - 1\)

Expanding \(\cos 4\theta\):

\(\cos 4\theta = 2(1 - 4\sin^2 \theta + 4\sin^4 \theta) - 1\)

\(= 2 - 8\sin^2 \theta + 8\sin^4 \theta - 1\)

\(= 1 - 8\sin^2 \theta + 8\sin^4 \theta\)

Now substitute into the original expression:

\(\cos 4\theta - 4\cos 2\theta + 3 = (1 - 8\sin^2 \theta + 8\sin^4 \theta) - 4(1 - 2\sin^2 \theta) + 3\)

\(= 1 - 8\sin^2 \theta + 8\sin^4 \theta - 4 + 8\sin^2 \theta + 3\)

\(= 8\sin^4 \theta\)

Thus, the identity is proven.

(ii) Using the result from part (i), we find the integral:

\(\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \sin^4 \theta \, d\theta\)

Using the identity \(\sin^4 \theta = \frac{1}{8}(\cos 4\theta - 4\cos 2\theta + 3)\), the integral becomes:

\(\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \frac{1}{8}(\cos 4\theta - 4\cos 2\theta + 3) \, d\theta\)

\(= \frac{1}{8} \left[ \int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \cos 4\theta \, d\theta - 4 \int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \cos 2\theta \, d\theta + 3 \int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} 1 \, d\theta \right]\)

Calculate each integral:

\(\int \cos 4\theta \, d\theta = \frac{1}{4} \sin 4\theta\)

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

\(\int 1 \, d\theta = \theta\)

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

\(\frac{1}{8} \left[ \frac{1}{4} (\sin \frac{4}{3}\pi - \sin \frac{2}{3}\pi) - 4 \times \frac{1}{2} (\sin \frac{2}{3}\pi - \sin \frac{1}{3}\pi) + 3 \left( \frac{1}{3}\pi - \frac{1}{6}\pi \right) \right]\)

\(= \frac{1}{8} \left[ 0 - 2(\sqrt{3}/2 - 1/2) + \frac{\pi}{2} \right]\)

\(= \frac{1}{8} \left[ -\sqrt{3} + \frac{\pi}{2} \right]\)

\(= \frac{1}{32} (2\pi - \sqrt{3})\)

(a) The expansion of \(\sin(3x + 2x)\) is \(\sin 5x = \sin 3x \cos 2x + \cos 3x \sin 2x\).

The expansion of \(\sin(3x - 2x)\) is \(\sin x = \sin 3x \cos 2x - \cos 3x \sin 2x\).

Adding these, we get \(\sin 5x + \sin x = 2 \sin 3x \cos 2x\).

Thus, \(\frac{1}{2}(\sin 5x + \sin x) = \sin 3x \cos 2x\).

(b) We have \(\int \sin 3x \cos 2x \, dx = \frac{1}{2} \int (\sin 5x + \sin x) \, dx\).

This integrates to \(\frac{1}{2} \left( -\frac{1}{5} \cos 5x - \cos x \right)\).

Evaluating from 0 to \(\frac{1}{4}\pi\), we get:

\(\left[ -\frac{1}{10} \cos 5x - \frac{1}{2} \cos x \right]_0^{\frac{1}{4}\pi}\).

Substituting the limits, we find:

\(-\frac{1}{10} \cos \left(\frac{5}{4}\pi\right) - \frac{1}{2} \cos \left(\frac{1}{4}\pi\right) + \frac{1}{10} \cos 0 + \frac{1}{2} \cos 0\).

This simplifies to \(\frac{1}{5}(3 - \sqrt{2})\).

(i) To prove the identity \(\sin^2 \theta \cos^2 \theta \equiv \frac{1}{8}(1 - \cos 4\theta)\), we start by using the double angle identities:

\(\sin 2\theta = 2 \sin \theta \cos \theta\) and \(\cos 2\theta = 2 \cos^2 \theta - 1\).

Thus, \(\sin^2 \theta \cos^2 \theta = \left( \frac{1}{2} \sin 2\theta \right)^2 = \frac{1}{4} \sin^2 2\theta\).

Using \(\sin^2 x = \frac{1}{2}(1 - \cos 2x)\), we have:

\(\sin^2 2\theta = \frac{1}{2}(1 - \cos 4\theta)\).

Therefore, \(\sin^2 \theta \cos^2 \theta = \frac{1}{4} \cdot \frac{1}{2}(1 - \cos 4\theta) = \frac{1}{8}(1 - \cos 4\theta)\).

(ii) To find \(\int_{0}^{\frac{1}{3}\pi} \sin^2 \theta \cos^2 \theta \, d\theta\), substitute the identity:

\(\int_{0}^{\frac{1}{3}\pi} \frac{1}{8}(1 - \cos 4\theta) \, d\theta = \frac{1}{8} \left[ \int_{0}^{\frac{1}{3}\pi} 1 \, d\theta - \int_{0}^{\frac{1}{3}\pi} \cos 4\theta \, d\theta \right]\).

The first integral is:

\(\int_{0}^{\frac{1}{3}\pi} 1 \, d\theta = \left[ \theta \right]_{0}^{\frac{1}{3}\pi} = \frac{1}{3}\pi\).

The second integral is:

\(\int_{0}^{\frac{1}{3}\pi} \cos 4\theta \, d\theta = \left[ \frac{1}{4} \sin 4\theta \right]_{0}^{\frac{1}{3}\pi} = \frac{1}{4} \left( \sin \left( \frac{4}{3}\pi \right) - \sin 0 \right)\).

Since \(\sin \left( \frac{4}{3}\pi \right) = -\frac{\sqrt{3}}{2}\), we have:

\(\frac{1}{4} \left( -\frac{\sqrt{3}}{2} \right) = -\frac{\sqrt{3}}{8}\).

Thus, the integral evaluates to:

\(\frac{1}{8} \left( \frac{1}{3}\pi + \frac{\sqrt{3}}{8} \right)\).

(i) To prove \(\cot x - \cot 2x \equiv \csc 2x\), use the identities:

\(\cot 2x = \frac{\cos 2x}{\sin 2x}\) and \(\csc 2x = \frac{1}{\sin 2x}\).

Express \(\cot x\) as \(\frac{\cos x}{\sin x}\).

Then, \(\cot x - \cot 2x = \frac{\cos x}{\sin x} - \frac{\cos 2x}{\sin 2x}\).

Using the double angle identity \(\cos 2x = 2\cos^2 x - 1\) and \(\sin 2x = 2\sin x \cos x\), simplify:

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

Combine the fractions:

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

Thus, \(\cot x - \cot 2x = \csc 2x\).

(ii) The indefinite integral of \(\cot x\) is \(\ln |\sin x|\).

Evaluate \(\int_{\frac{1}{6}\pi}^{\frac{1}{4}\pi} \cot x \, dx = \left[ \ln |\sin x| \right]_{\frac{1}{6}\pi}^{\frac{1}{4}\pi}\).

Substitute the limits:

\(\ln |\sin \frac{1}{4}\pi| - \ln |\sin \frac{1}{6}\pi| = \ln \frac{\sin \frac{1}{4}\pi}{\sin \frac{1}{6}\pi}\).

\(\sin \frac{1}{4}\pi = \frac{\sqrt{2}}{2}\) and \(\sin \frac{1}{6}\pi = \frac{1}{2}\).

\(\ln \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = \ln \sqrt{2} = \frac{1}{2} \ln 2\).

(iii) The indefinite integral of \(\csc 2x\) is \(\frac{1}{2} \ln |\tan x|\).

Evaluate \(\int_{\frac{1}{6}\pi}^{\frac{1}{4}\pi} \csc 2x \, dx = \left[ \frac{1}{2} \ln |\tan x| \right]_{\frac{1}{6}\pi}^{\frac{1}{4}\pi}\).

Substitute the limits:

\(\frac{1}{2} \ln |\tan \frac{1}{4}\pi| - \frac{1}{2} \ln |\tan \frac{1}{6}\pi| = \frac{1}{2} \ln \frac{\tan \frac{1}{4}\pi}{\tan \frac{1}{6}\pi}\).

\(\tan \frac{1}{4}\pi = 1\) and \(\tan \frac{1}{6}\pi = \frac{1}{\sqrt{3}}\).

\(\frac{1}{2} \ln \frac{1}{\frac{1}{\sqrt{3}}} = \frac{1}{2} \ln \sqrt{3} = \frac{1}{4} \ln 3\).

(a) Express \(\csc 2\theta\) and \(\cot 2\theta\) in terms of \(\sin 2\theta\) and \(\cos 2\theta\):

\(\csc 2\theta = \frac{1}{\sin 2\theta}\)

\(\cot 2\theta = \frac{\cos 2\theta}{\sin 2\theta}\)

Thus, \(\csc 2\theta - \cot 2\theta = \frac{1 - \cos 2\theta}{\sin 2\theta}\).

Using the double angle identity \(\cos 2\theta = 1 - 2\sin^2 \theta\), we have:

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

(b) The integral becomes:

\(\int_{\frac{1}{4}\pi}^{\frac{3}{4}\pi} (\csc 2\theta - \cot 2\theta) \, d\theta = \int_{\frac{1}{4}\pi}^{\frac{3}{4}\pi} \tan \theta \, d\theta\).

The integral of \(\tan \theta\) is \(-\ln |\cos \theta|\).

Evaluate from \(\frac{1}{4}\pi\) to \(\frac{3}{4}\pi\):

\(-\ln |\cos \frac{3}{4}\pi| + \ln |\cos \frac{1}{4}\pi|\).

\(\cos \frac{3}{4}\pi = -\frac{1}{\sqrt{2}}\) and \(\cos \frac{1}{4}\pi = \frac{1}{\sqrt{2}}\).

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

Therefore, the result is \(\frac{1}{2} \ln 2\).

(a) Use the double angle formula for cosine: \(\cos 2\theta = 1 - 2\sin^2 \theta\).

Substitute into the expression: \(\frac{1 - (1 - 2\sin^2 \theta)}{1 + (1 - 2\sin^2 \theta)} = \frac{2\sin^2 \theta}{2\cos^2 \theta} = \tan^2 \theta\).

(b) Express \(\tan^2 \theta\) in terms of \(\sec^2 \theta\): \(\tan^2 \theta = \sec^2 \theta - 1\).

Integrate: \(\int (\sec^2 \theta - 1) \, d\theta = \tan \theta - \theta\).

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

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

Calculate: \(\tan \frac{1}{3}\pi = \sqrt{3}\), \(\tan \frac{1}{6}\pi = \frac{1}{\sqrt{3}}\).

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

(a) To find \(f'(x)\), use the quotient rule: \(\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\).

Let \(u = \cos x\) and \(v = 1 + \sin x\). Then \(u' = -\sin x\) and \(v' = \cos x\).

So, \(f'(x) = \frac{-\sin x (1 + \sin x) - \cos x (\cos x)}{(1 + \sin x)^2}\).

Simplify: \(f'(x) = \frac{-\sin x - \sin^2 x - \cos^2 x}{(1 + \sin x)^2}\).

Using \(\sin^2 x + \cos^2 x = 1\), we have \(f'(x) = \frac{-\sin x - 1}{(1 + \sin x)^2}\).

Since \(-\sin x - 1 < 0\) and \((1 + \sin x)^2 > 0\) for \(-\frac{1}{2}\pi < x < \frac{3}{2}\pi\), \(f'(x) < 0\).

(b) To find \(\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi} f(x) \, dx\), note that \(f(x) = \frac{\cos x}{1 + \sin x}\).

Use substitution: let \(u = 1 + \sin x\), then \(du = \cos x \, dx\).

The integral becomes \(\int \frac{1}{u} \, du = \ln |u| + C\).

Evaluate from \(x = \frac{1}{6}\pi\) to \(x = \frac{1}{2}\pi\):

\(u(\frac{1}{6}\pi) = 1 + \frac{1}{2} = \frac{3}{2}\) and \(u(\frac{1}{2}\pi) = 1 + 1 = 2\).

Thus, \(\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi} f(x) \, dx = \ln 2 - \ln \frac{3}{2} = \ln \frac{4}{3}\).

(i) Use the double angle formulas: \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\) and \(\sin 2\theta = 2\sin \theta \cos \theta\).

Substitute into \(f(\theta)\):

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

Simplify the expression:

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

\(f(\theta) = \frac{(\sin \theta + \cos \theta)^2}{(\cos \theta + \sin \theta)^2} = \tan \theta\).

(ii) The integral becomes:

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

Let \(u = \cos \theta\), then \(du = -\sin \theta \, d\theta\).

The integral becomes:

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

Evaluate from \(\frac{\pi}{6}\) to \(\frac{\pi}{4}\):

\(-\ln \left| \cos \frac{\pi}{4} \right| + \ln \left| \cos \frac{\pi}{6} \right|\).

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

\(= \ln \frac{\sqrt{3}}{2} + \ln \sqrt{2} = \ln \frac{\sqrt{3} \cdot \sqrt{2}}{2} = \ln \frac{\sqrt{6}}{2}\).

\(= \frac{1}{2} \ln \frac{3}{2}\).

(i) Start by expanding \(\sin(2x + x)\) using the angle addition formula: \(\sin(2x + x) = \sin 2x \cos x + \cos 2x \sin x\).

We know \(\sin 2x = 2 \sin x \cos x\) and \(\cos 2x = 1 - 2 \sin^2 x\).

Substitute these into the expansion: \(\sin 3x = (2 \sin x \cos x) \cos x + (1 - 2 \sin^2 x) \sin x\).

Simplify: \(\sin 3x = 2 \sin x \cos^2 x + \sin x - 2 \sin^3 x\).

Since \(\cos^2 x = 1 - \sin^2 x\), substitute: \(\sin 3x = 2 \sin x (1 - \sin^2 x) + \sin x - 2 \sin^3 x\).

Simplify further: \(\sin 3x = 2 \sin x - 2 \sin^3 x + \sin x - 2 \sin^3 x\).

Combine terms: \(\sin 3x = 3 \sin x - 4 \sin^3 x\).

(ii) Use the identity \(\sin^3 x = \frac{3}{4} \sin x - \frac{1}{12} \sin 3x\) to integrate.

Integrate: \(\int \sin^3 x \, dx = \int \left( \frac{3}{4} \sin x - \frac{1}{12} \sin 3x \right) \, dx\).

This becomes: \(\frac{3}{4} \int \sin x \, dx - \frac{1}{12} \int \sin 3x \, dx\).

Integrate each term: \(-\frac{3}{4} \cos x - \frac{1}{36} \cos 3x\).

Evaluate from \(0\) to \(\frac{1}{3}\pi\):

\(\left[ -\frac{3}{4} \cos x - \frac{1}{36} \cos 3x \right]_0^{\frac{1}{3}\pi}\).

Calculate: \(\left( -\frac{3}{4} \cos \frac{1}{3}\pi - \frac{1}{36} \cos \pi \right) - \left( -\frac{3}{4} \cos 0 - \frac{1}{36} \cos 0 \right)\).

\(= \left( -\frac{3}{4} \times \frac{1}{2} + \frac{1}{36} \right) - \left( -\frac{3}{4} - \frac{1}{36} \right)\).

\(= \left( -\frac{3}{8} + \frac{1}{36} \right) - \left( -\frac{3}{4} - \frac{1}{36} \right)\).

\(= \left( -\frac{3}{8} + \frac{1}{36} + \frac{3}{4} + \frac{1}{36} \right)\).

\(= \frac{5}{24}\).

(i) To find the stationary point, we need to find \(\frac{dy}{dx}\) and set it to zero. Using the quotient rule:

\(\frac{dy}{dx} = \frac{-3 \sin x (2 + \sin x) - 3 \cos x \cos x}{(2 + \sin x)^2}\)

Equate the numerator to zero:

\(-3 \sin x (2 + \sin x) - 3 \cos^2 x = 0\)

\(-6 \sin x - 3 \sin^2 x - 3 (1 - \sin^2 x) = 0\)

\(-6 \sin x - 3 + 3 \sin^2 x = 0\)

\(3 \sin^2 x + 6 \sin x + 3 = 0\)

\(\sin x = -\frac{1}{2}\)

\(x = -\frac{\pi}{6}\)

Substitute back to find \(y\):

\(y = \frac{3 \cos(-\frac{\pi}{6})}{2 + \sin(-\frac{\pi}{6})} = \sqrt{3}\)

(ii) To find \(a\), we evaluate the integral:

\(\int \frac{3 \cos x}{2 + \sin x} \, dx = 3 \ln |2 + \sin x| + C\)

Substitute limits and equate to 1:

\(3 \ln(2 + \sin a) - 3 \ln 2 = 1\)

\(3 \ln \left( \frac{2 + \sin a}{2} \right) = 1\)

\(\ln \left( \frac{2 + \sin a}{2} \right) = \frac{1}{3}\)

\(\frac{2 + \sin a}{2} = e^{\frac{1}{3}}\)

\(2 + \sin a = 2e^{\frac{1}{3}}\)

\(\sin a = 2e^{\frac{1}{3}} - 2\)

\(a = \sin^{-1}(2e^{\frac{1}{3}} - 2) \approx 0.913\)

(i) Start with the left-hand side: \(\frac{2 \sin x - \sin 2x}{1 - \cos 2x}\).

Using the double angle formula, \(\sin 2x = 2 \sin x \cos x\) and \(\cos 2x = 2 \cos^2 x - 1\), substitute these into the expression:

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

Simplify the denominator: \(2 - 2 \cos^2 x = 2(1 - \cos^2 x) = 2 \sin^2 x\).

Thus, the expression becomes \(\frac{2 \sin x (1 - \cos x)}{2 \sin^2 x} = \frac{\sin x (1 - \cos x)}{\sin^2 x} = \frac{\sin x}{1 + \cos x}\), which matches the right-hand side.

(ii) The integral \(\int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \frac{2 \sin x - \sin 2x}{1 - \cos 2x} \, dx\) can be rewritten using the result from part (i) as \(\int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \frac{\sin x}{1 + \cos x} \, dx\).

Let \(u = 1 + \cos x\), then \(du = -\sin x \, dx\), so \(dx = -\frac{du}{\sin x}\).

The integral becomes \(-\int \frac{1}{u} \, du = -\ln |u| + C\).

Substitute back \(u = 1 + \cos x\): \(-\ln |1 + \cos x|\).

Evaluate from \(x = \frac{1}{3}\pi\) to \(x = \frac{1}{2}\pi\):

\(-\ln |1 + \cos \frac{1}{2}\pi| + \ln |1 + \cos \frac{1}{3}\pi|\).

\(\cos \frac{1}{2}\pi = 0\) and \(\cos \frac{1}{3}\pi = \frac{1}{2}\), so the expression becomes:

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