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