Answer
$$4\left( {\sin x - x\cos x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {4x\sin x} dx \cr
& {\text{setting }}\,\,\,\,\,\,u = 4x{\text{ then }}du = 4dx\,\,\,\,\,\,\,\,\,\, \cr
& {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = \sin xdx{\text{ then }}v = - \cos x \cr
& {\text{Substituting these values into the formula for integration by parts}} \cr
& \int u dv = uv - \int {vdu} \cr
& \int {4x\sin x} dx = \left( {4x} \right)\left( { - \cos x} \right) - \int {\left( { - \cos x} \right)\left( {4dx} \right)} \cr
& {\text{simplifying}} \cr
& = - 4x\cos x + 4\int {\cos x} dx \cr
& {\text{integrate by using the Basic Trigonometric integral }}\int {\cos x} dx = \sin x + C \cr
& = - 4x\cos x + 4\sin x + C \cr
& {\text{factor}} \cr
& = 4\left( {\sin x - x\cos x} \right) + C \cr} $$
