To solve \(\int_{0}^{\frac{1}{6}\pi} x \cos 3x \, dx\), we use integration by parts. Let \(u = x\) and \(dv = \cos 3x \, dx\). Then \(du = dx\) and \(v = \frac{1}{3} \sin 3x\).
Applying integration by parts:
\(\int u \, dv = uv - \int v \, du\)
\(\int x \cos 3x \, dx = x \cdot \frac{1}{3} \sin 3x - \int \frac{1}{3} \sin 3x \, dx\)
\(= \frac{1}{3} x \sin 3x - \frac{1}{3} \int \sin 3x \, dx\)
Integrate \(\int \sin 3x \, dx\):
\(\int \sin 3x \, dx = -\frac{1}{3} \cos 3x\)
Substitute back:
\(\frac{1}{3} x \sin 3x + \frac{1}{9} \cos 3x\)
Evaluate from \(0\) to \(\frac{1}{6}\pi\):
\(\left[ \frac{1}{3} x \sin 3x + \frac{1}{9} \cos 3x \right]_{0}^{\frac{1}{6}\pi}\)
At \(x = \frac{1}{6}\pi\):
\(\frac{1}{3} \cdot \frac{1}{6}\pi \cdot \sin \frac{1}{2}\pi + \frac{1}{9} \cdot \cos \frac{1}{2}\pi = \frac{1}{18}\pi \cdot 1 + 0 = \frac{1}{18}\pi\)
At \(x = 0\):
\(\frac{1}{3} \cdot 0 \cdot \sin 0 + \frac{1}{9} \cdot \cos 0 = 0 + \frac{1}{9} \cdot 1 = \frac{1}{9}\)
Subtract the results:
\(\frac{1}{18}\pi - \frac{1}{9} = \frac{1}{18}\pi - \frac{2}{18} = \frac{1}{18}(\pi - 2)\)
