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