Answer
$$\frac{1}{3}x\sin 3x + \frac{1}{9}\cos 3x + C$$
Work Step by Step
$$\eqalign{
& \int {x\cos 3x} dx \cr
& \cr
& {\text{by using the integration by parts method}} \cr
& u = x,\,\,\,\,\,\,\,\,\,\,\,\,dv = \cos 3xdx \cr
& du = dx,\,\,\,\,\,\,\,\,\,v = \frac{1}{3}\sin 3x \cr
& \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {x\cos 3x} dx = \left( x \right)\left( {\frac{1}{3}\sin 3x} \right) - \int {\left( {\frac{1}{3}\sin 3x} \right)\left( {dx} \right)} \cr
& \int {x\cos 3x} dx = \frac{1}{3}x\sin 3x - \frac{1}{3}\int {\sin 3xdx} \cr
& = \frac{1}{3}x\sin 3x - \frac{1}{3}\left( { - \frac{1}{3}\cos 3x} \right) + C \cr
& {\text{simplifying}} \cr
& = \frac{1}{3}x\sin 3x + \frac{1}{9}\cos 3x + C \cr} $$
