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

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

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

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

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