(i) Let \(y = \frac{\cos x}{\sin x} = \cot x\). Differentiate using the quotient rule:

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

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

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

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

(ii) Integrate by parts: Let \(u = x\) and \(dv = \csc^2 x \, dx\).

Then \(du = dx\) and \(v = -\cot x\).

\(\int x \csc^2 x \, dx = -x \cot x + \int \cot x \, dx\)

\(= -x \cot x + \ln |\sin x|\)

Evaluate from \(\frac{1}{4}\pi\) to \(\frac{1}{2}\pi\):

\(\left[ -x \cot x + \ln |\sin x| \right]_{\frac{1}{4}\pi}^{\frac{1}{2}\pi}\)

At \(x = \frac{1}{2}\pi\), \(-x \cot x = 0\) and \(\ln |\sin x| = 0\).

At \(x = \frac{1}{4}\pi\), \(-x \cot x = -\frac{1}{4}\pi\) and \(\ln |\sin x| = \ln \frac{1}{\sqrt{2}} = -\frac{1}{2} \ln 2\).

\(= 0 - \left( -\frac{1}{4}\pi - \frac{1}{2} \ln 2 \right)\)

\(= \frac{1}{4}\pi + \frac{1}{2} \ln 2\)

\(= \frac{1}{4}(\pi + \ln 4)\)