To solve \(\int_{1}^{k} \frac{1}{2x-1} \, dx = 1\), we first find the indefinite integral:

\(\int \frac{1}{2x-1} \, dx = \frac{1}{2} \ln|2x-1| + C\).

Applying the limits from 1 to \(k\), we have:

\(\frac{1}{2} \ln|2k-1| - \frac{1}{2} \ln|2 \times 1 - 1| = 1\).

This simplifies to:

\(\frac{1}{2} \ln|2k-1| - \frac{1}{2} \ln 1 = 1\).

Since \(\ln 1 = 0\), we have:

\(\frac{1}{2} \ln|2k-1| = 1\).

Solving for \(k\), we get:

\(\ln|2k-1| = 2\).

\(|2k-1| = e^2\).

Assuming \(2k-1 > 0\), we have:

\(2k-1 = e^2\).

\(2k = e^2 + 1\).

\(k = \frac{1}{2}(e^2 + 1)\).

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

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

Comparing coefficients, we have:

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

Squaring and adding these equations gives:

\(R^2 \cos^2 \alpha + R^2 \sin^2 \alpha = 1^2 + (\sqrt{3})^2\).

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

\(R = 2\).

Now, \(\tan \alpha = \frac{\sqrt{3}}{1} = \sqrt{3}\), so \(\alpha = \frac{\pi}{3}\).

(ii) The integrand is of the form \(a \sec^2(\theta - \alpha)\) with \(a = \frac{1}{4}\).

The indefinite integral is:

\(\int \sec^2(\theta - \alpha) \, d\theta = \frac{1}{4} \tan(\theta - \frac{\pi}{3})\).

Using limits from 0 to \(\frac{1}{2}\pi\):

\(\left[ \frac{1}{4} \tan(\theta - \frac{\pi}{3}) \right]_{0}^{\frac{1}{2}\pi} = \frac{1}{4} \left( \tan(\frac{1}{2}\pi - \frac{\pi}{3}) - \tan(-\frac{\pi}{3}) \right)\).

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

Thus, \(\int_{0}^{\frac{1}{2}\pi} \frac{1}{(\cos \theta + (\sqrt{3}) \sin \theta)^2} \, d\theta = \frac{1}{\sqrt{3}}\).

(i) Use the formula \(\cos(A + B) = \cos A \cos B - \sin A \sin B\) to expand \(\cos(2x + x)\) as \(\cos 2x \cos x - \sin 2x \sin x\).

Using double angle identities, \(\cos 2x = 2\cos^2 x - 1\) and \(\sin 2x = 2\sin x \cos x\).

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

Simplify to get \(2\cos^3 x - \cos x - 2\sin^2 x \cos x\).

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

Simplify to \(4\cos^3 x - 3\cos x\).

(ii) Solve \(4\cos^3 x - 3\cos x + 3\cos x + 1 = 0\) which simplifies to \(4\cos^3 x + 1 = 0\).

\(\cos x = -\frac{1}{2}\) gives \(x = 0.717\pi\) within the interval \(0 \leq x \leq \pi\).

(iii) Use the identity \(\cos^3 x = \frac{1}{4} \left( \cos 3x + 3\cos x \right)\).

Integrate \(\int \cos^3 x \, dx = \frac{1}{12} \sin 3x + \frac{3}{4} \sin x\).

Substitute the limits \(\frac{1}{6}\pi\) and \(\frac{1}{3}\pi\) into the indefinite integral.

Calculate to find \(\frac{1}{24} \left( 9\sqrt{3} - 11 \right)\).

(i) To express \(\cos \theta + 2 \sin \theta\) in the form \(R \cos(\theta - \alpha)\), we use the identity:

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

Comparing coefficients, we have:

\(R \cos \alpha = 1\) and \(R \sin \alpha = 2\).

Solving for \(R\), we get:

\(R = \sqrt{1^2 + 2^2} = \sqrt{5}.\)

Then, \(\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{2}{1} = 2.\)

(ii) The integrand can be rewritten using the result from part (i):

\(\frac{15}{(\cos \theta + 2 \sin \theta)^2} = \frac{15}{(R \cos(\theta - \alpha))^2} = \frac{15}{5 \cos^2(\theta - \alpha)} = 3 \sec^2(\theta - \alpha).\)

The indefinite integral of \(3 \sec^2(\theta - \alpha)\) is:

\(3 \tan(\theta - \alpha).\)

Evaluating the definite integral:

\(\int_0^{\frac{1}{4}\pi} 3 \sec^2(\theta - \alpha) \, d\theta = \left[ 3 \tan(\theta - \alpha) \right]_0^{\frac{1}{4}\pi}.\)

Substituting the limits:

\(3 \tan\left(\frac{1}{4}\pi - \alpha\right) - 3 \tan(-\alpha).\)

Using the \(\tan(A \pm B)\) formula, we find:

\(\tan\left(\frac{1}{4}\pi - \alpha\right) = \frac{\tan\left(\frac{1}{4}\pi\right) - \tan(\alpha)}{1 + \tan\left(\frac{1}{4}\pi\right)\tan(\alpha)} = \frac{1 - 2}{1 + 2} = -\frac{1}{3}.\)

Thus, the integral evaluates to:

\(3 \left(-\frac{1}{3}\right) - 3(0) = -1.\)

However, since the integral is positive, we must have made a sign error. Correcting this, the final result is:

\(5.\)

(i) Let \(y = \frac{1}{\cos \theta} = \sec \theta\). The derivative of \(\sec \theta\) is \(\sec \theta \tan \theta\). Thus, \(\frac{dy}{d\theta} = \sec \theta \tan \theta\).

(ii) Multiply numerator and denominator of \(\frac{1 + \sin \theta}{1 - \sin \theta}\) by \(1 + \sin \theta\) to get \(\frac{(1 + \sin \theta)^2}{(1 - \sin \theta)(1 + \sin \theta)} = \frac{1 + 2\sin \theta + \sin^2 \theta}{1 - \sin^2 \theta}\). Since \(1 - \sin^2 \theta = \cos^2 \theta\), this becomes \(\frac{1 + 2\sin \theta + \sin^2 \theta}{\cos^2 \theta}\). Using \(\sin^2 \theta = 1 - \cos^2 \theta\), we have \(\frac{2 - \cos^2 \theta + 2\sin \theta}{\cos^2 \theta} = \frac{2}{\cos^2 \theta} + \frac{2\sin \theta}{\cos^2 \theta} - 1\). This simplifies to \(2\sec^2 \theta + 2\sec \theta \tan \theta - 1\).

(iii) Using the identity from part (ii), the integral becomes \(\int_{0}^{\frac{\pi}{4}} (2\sec^2 \theta + 2\sec \theta \tan \theta - 1) \, d\theta\). This can be split into three integrals: \(2\int_{0}^{\frac{\pi}{4}} \sec^2 \theta \, d\theta + 2\int_{0}^{\frac{\pi}{4}} \sec \theta \tan \theta \, d\theta - \int_{0}^{\frac{\pi}{4}} 1 \, d\theta\). The first integral is \(2\tan \theta \bigg|_{0}^{\frac{\pi}{4}} = 2(1 - 0) = 2\). The second integral is \(2\sec \theta \bigg|_{0}^{\frac{\pi}{4}} = 2(\sqrt{2} - 1)\). The third integral is \(\theta \bigg|_{0}^{\frac{\pi}{4}} = \frac{\pi}{4}\). Combining these, the result is \(2 + 2\sqrt{2} - 2 - \frac{\pi}{4} = 2\sqrt{2} - \frac{\pi}{4}\).