Answer
$${x^2}\sin x + 2x\cos x - 2\sin x + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}\cos x} dx \cr
& u = {x^2},{\text{ }}du = 2xdx \cr
& dx = \cos xdx,{\text{ }}v = \sin x \cr
& {\text{integration by parts }} = uv - \int {vdu} \cr
& = {x^2}\sin x - \int {\sin x\left( {2x} \right)dx} \cr
& = {x^2}\sin x - 2\int {x\sin xdx} \cr
& u = x,{\text{ }}du = dx \cr
& dx = \sin xdx,{\text{ }}v = - \cos xdx \cr
& = {x^2}\sin x - 2\left( { - x\cos x - \int {\left( { - \cos x} \right)\left( {dx} \right)} } \right) \cr
& = {x^2}\sin x - 2\left( { - x\cos x + \int {\cos xdx} } \right) \cr
& = {x^2}\sin x + 2x\cos x - 2\int {\cos xdx} \cr
& = {x^2}\sin x + 2x\cos x - 2\sin x + C \cr} $$
