Solution
= Solution
Use <integration by parts>, taking $u=x$ and $dv=\sin3x\,dx$:
$$
\int_0^{\pi/6}x\sin3x\,dx=\left[-\frac x3\cos3x+\frac19\sin3x\right]_0^{\pi/6}
=\boxed{\frac19.}
$$
= Solution
Use <integration by parts>, taking $u=x$ and $dv=\sin3x\,dx$:
$$
\int_0^{\pi/6}x\sin3x\,dx=\left[-\frac x3\cos3x+\frac19\sin3x\right]_0^{\pi/6}
=\boxed{\frac19.}
$$