(i) To express \((\sqrt{3}) \cos x + \sin x\) in the form \(R \cos(x - \alpha)\), we use the identity:

\(R \cos(x - \alpha) = R \cos \alpha \cos x + R \sin \alpha \sin x\).

Comparing coefficients, we have:

\(R \cos \alpha = \sqrt{3}\) and \(R \sin \alpha = 1\).

Solving for \(R\), we use:

\(R^2 = (R \cos \alpha)^2 + (R \sin \alpha)^2 = 3 + 1 = 4\).

Thus, \(R = 2\).

To find \(\alpha\), we use:

\(\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{1}{\sqrt{3}}\).

Therefore, \(\alpha = \frac{1}{6}\pi\).

(ii) We need to evaluate:

\(\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi} \frac{1}{((\sqrt{3}) \cos x + \sin x)^2} \, dx\).

Substitute \((\sqrt{3}) \cos x + \sin x = 2 \cos(x - \frac{1}{6}\pi)\).

The integral becomes:

\(\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi} \frac{1}{4 \cos^2(x - \frac{1}{6}\pi)} \, dx\).

This is equivalent to:

\(\frac{1}{4} \int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi} \sec^2(x - \frac{1}{6}\pi) \, dx\).

The integral of \(\sec^2(x - \frac{1}{6}\pi)\) is \(\tan(x - \frac{1}{6}\pi)\).

Thus, the integral evaluates to:

\(\frac{1}{4} \left[ \tan(x - \frac{1}{6}\pi) \right]_{\frac{1}{6}\pi}^{\frac{1}{2}\pi}\).

Calculating the limits:

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

\(\tan(\frac{1}{6}\pi - \frac{1}{6}\pi) = \tan(0) = 0\).

Thus, the integral is:

\(\frac{1}{4} (\sqrt{3} - 0) = \frac{1}{4}\sqrt{3}\).

(i) To express \(4 \cos \theta + 3 \sin \theta\) in the form \(R \cos(\theta - \alpha)\), we use the identity:

\(R \cos(\theta - \alpha) = R \cos \alpha \cos \theta + R \sin \alpha \sin \theta\).

Matching coefficients, we have:

\(R \cos \alpha = 4\) and \(R \sin \alpha = 3\).

Solving for \(R\), we get:

\(R = \sqrt{4^2 + 3^2} = 5\).

Then, \(\cos \alpha = \frac{4}{5}\) and \(\sin \alpha = \frac{3}{5}\).

\(\alpha = \arctan\left(\frac{3}{4}\right) \approx 0.6435\).

(ii) (a) Solve \(4 \cos \theta + 3 \sin \theta = 2\):

Using \(R \cos(\theta - \alpha) = 2\), we have:

\(5 \cos(\theta - 0.6435) = 2\).

\(\cos(\theta - 0.6435) = \frac{2}{5}\).

\(\theta - 0.6435 = \arccos\left(\frac{2}{5}\right)\).

\(\theta = 0.6435 + \arccos\left(\frac{2}{5}\right) \approx 1.80\).

Also, \(\theta = 0.6435 + 2\pi - \arccos\left(\frac{2}{5}\right) \approx 5.77\).

(b) Find \(\int \frac{50}{(4 \cos \theta + 3 \sin \theta)^2} \, d\theta\):

Express the integrand as \(k \sec^2(\theta - 0.6435)\) where \(k = 2\).

\(\int 2 \sec^2(\theta - 0.6435) \, d\theta = 2 \tan(\theta - 0.6435) + C\).

(i) Let \(u = \\sin 2x\), then \(du = 2 \\cos 2x \, dx\) or \(dx = \frac{du}{2 \\cos 2x}\).

The integrand becomes \(\\sin^3 2x \\cos^3 2x = u^3 (1-u^2)^{3/2}\).

Substitute and simplify: \(\\int u^3 (1-u^2)^{3/2} \, du\).

Change limits: when \(x = 0, u = 0\) and when \(x = \frac{\\pi}{4}, u = 1\).

Integrate: \(\\int_0^1 \frac{1}{2} u^3 (1-u^2) \, du = \frac{1}{2} \\int_0^1 (u^3 - u^5) \, du\).

Calculate: \(\frac{1}{2} \left[ \frac{u^4}{4} - \frac{u^6}{6} \right]_0^1 = \frac{1}{2} \left( \frac{1}{4} - \frac{1}{6} \right) = \frac{1}{24}\).

Thus, \(A = \frac{1}{24}\).

(ii) Given \(\\int_0^{k\\pi} |\\sin^3 2x \\cos^3 2x| \, dx = 40A\), substitute \(A = \frac{1}{24}\).

\(40 \times \frac{1}{24} = \frac{5}{3}\).

Since the integral over one period \([0, \frac{\\pi}{2}]\) is \(\frac{1}{24}\), and \(k\) periods are \(\frac{5}{3}\), solve for \(k\):

\(k \times \frac{1}{24} = \frac{5}{3} \Rightarrow k = 10\).

(i) Let \(y = \sec x = \frac{1}{\cos x}\). Differentiate using the quotient rule: \(\frac{dy}{dx} = \frac{0 \cdot \cos x - (-\sin x) \cdot 1}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x \tan x\).

(ii) Multiply numerator and denominator by \(\sec x + \tan x\): \(\frac{1}{\sec x - \tan x} \cdot \frac{\sec x + \tan x}{\sec x + \tan x} = \frac{\sec x + \tan x}{\sec^2 x - \tan^2 x}\). Using \(\sec^2 x - \tan^2 x = 1\), we have \(\sec x + \tan x\).

(iii) Substitute \(\sec x + \tan x\) into \(\frac{1}{(\sec x - \tan x)^2}\): \(\frac{1}{(\sec x - \tan x)^2} = (\sec x + \tan x)^2 = \sec^2 x + 2 \sec x \tan x + \tan^2 x\). Using \(\sec^2 x - \tan^2 x = 1\), we get \(2 \sec^2 x - 1 + 2 \sec x \tan x\).

(iv) Integrate \(\int_0^{\frac{1}{4}\pi} (2 \tan x - x + 2 \sec x) \, dx\). The integral of \(2 \tan x\) is \(-2 \ln |\cos x|\), the integral of \(x\) is \(\frac{x^2}{2}\), and the integral of \(2 \sec x\) is \(2 \ln |\sec x + \tan x|\). Evaluate from 0 to \(\frac{1}{4}\pi\) to obtain \(\frac{1}{4}(8\sqrt{2} - \pi)\).

(i) Let \(u = \tan x\), then \(\frac{du}{dx} = \sec^2 x\), so \(dx = \frac{du}{\sec^2 x}\). The integrand becomes \(u^{n+2} + u^n\). The limits of integration change from \(x = 0\) to \(x = \frac{\pi}{4}\), which correspond to \(u = 0\) to \(u = 1\). Thus,

\(\int_0^{\frac{\pi}{4}} (\tan^{n+2} x + \tan^n x) \, dx = \int_0^1 (u^{n+2} + u^n) \, du.\)

Integrating, we get:

\(\int_0^1 u^{n+2} \, du = \frac{u^{n+3}}{n+3} \bigg|_0^1 = \frac{1}{n+3},\)

\(\int_0^1 u^n \, du = \frac{u^{n+1}}{n+1} \bigg|_0^1 = \frac{1}{n+1}.\)

Thus,

\(\int_0^1 (u^{n+2} + u^n) \, du = \frac{1}{n+3} + \frac{1}{n+1} = \frac{1}{n+1}.\)

(ii) (a) Use \(\sec^2 x = 1 + \tan^2 x\) to rewrite the integrand:

\(\sec^4 x - \sec^2 x = (1 + \tan^2 x)^2 - (1 + \tan^2 x) = \tan^4 x + \tan^2 x.\)

Apply the result from part (i):

\(\int_0^{\frac{\pi}{4}} (\tan^4 x + \tan^2 x) \, dx = \frac{1}{3}.\)

(b) Rewrite the integrand:

\(\tan^9 x + 5 \tan^7 x + 5 \tan^5 x + \tan^3 x = t^9 + 5t^7 + 5t^5 + t^3.\)

Apply the result from part (i) to each term:

\(\int_0^{\frac{\pi}{4}} t^9 \, dt = \frac{1}{10},\)

\(5 \int_0^{\frac{\pi}{4}} t^7 \, dt = 5 \times \frac{1}{8} = \frac{5}{8},\)

\(5 \int_0^{\frac{\pi}{4}} t^5 \, dt = 5 \times \frac{1}{6} = \frac{5}{6},\)

\(\int_0^{\frac{\pi}{4}} t^3 \, dt = \frac{1}{4}.\)

Summing these results gives:

\(\frac{1}{10} + \frac{5}{8} + \frac{5}{6} + \frac{1}{4} = \frac{25}{24}.\)