To solve \(\int_{\frac{1}{3}\pi}^{\pi} x \sin \frac{1}{2}x \, dx\), we use integration by parts.

Let \(u = x\) and \(dv = \sin \frac{1}{2}x \, dx\). Then \(du = dx\) and \(v = -2 \cos \frac{1}{2}x\).

Applying integration by parts:

\(\int u \, dv = uv - \int v \, du\)

\(= -2x \cos \frac{1}{2}x + 2 \int \cos \frac{1}{2}x \, dx\)

Integrate \(\cos \frac{1}{2}x\):

\(\int \cos \frac{1}{2}x \, dx = 4 \sin \frac{1}{2}x\)

Thus, the integral becomes:

\(-2x \cos \frac{1}{2}x + 4 \sin \frac{1}{2}x\)

Evaluate from \(\frac{1}{3}\pi\) to \(\pi\):

\(\left[ -2x \cos \frac{1}{2}x + 4 \sin \frac{1}{2}x \right]_{\frac{1}{3}\pi}^{\pi}\)

Calculate at \(x = \pi\):

\(-2\pi \cos \frac{1}{2}\pi + 4 \sin \frac{1}{2}\pi = 4\)

Calculate at \(x = \frac{1}{3}\pi\):

\(-2 \left( \frac{1}{3}\pi \right) \cos \frac{1}{6}\pi + 4 \sin \frac{1}{6}\pi\)

\(= -\frac{2}{3}\pi \cdot \frac{\sqrt{3}}{2} + 4 \cdot \frac{1}{2}\)

\(= -\frac{\sqrt{3}}{3}\pi + 2\)

Subtract the two results:

\(4 - \left( -\frac{\sqrt{3}}{3}\pi + 2 \right)\)

\(= 2 + \frac{\sqrt{3}}{3}\pi\)