Answer
$${x^2}\cos \left( {1 - x} \right) + 2x\sin \left( {1 - x} \right) - 2\cos \left( {1 - x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}\sin \left( {1 - x} \right)} dx \cr
& {\text{Using the integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}u = {x^2},\,\,\,\,du = 2xdx\,\,\,\,\, \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = \sin \left( {1 - x} \right)dx,\,\,\,\,v = \cos \left( {1 - x} \right) \cr
& \cr
& {\text{Integration by parts then gives}} \cr
& \int {{x^2}\sin \left( {1 - x} \right)} dx = {x^2}\cos \left( {1 - x} \right) - \int {\cos \left( {1 - x} \right)\left( {2xdx} \right)} \cr
& \int {{x^2}\sin \left( {1 - x} \right)} dx = {x^2}\cos \left( {1 - x} \right) - \int {2x\cos \left( {1 - x} \right)dx} \cr
& \cr
& {\text{Integrate by parts again to get }}\int {2x\cos \left( {1 - x} \right)dx} \cr
& \,\,\,{\text{Let }}u = 2x,\,\,\,\,du = 2dx\,\,\,\,\, \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = \cos \left( {1 - x} \right)dx,\,\,\,\,v = - \sin \left( {1 - x} \right) \cr
& \cr
& \int {{x^2}\sin \left( {1 - x} \right)} dx = {x^2}\cos \left( {1 - x} \right) - \left( { - 2x\sin \left( {1 - x} \right) - \int {\left( { - \sin \left( {1 - x} \right)} \right)\left( {2dx} \right)} } \right) \cr
& \int {{x^2}\sin \left( {1 - x} \right)} dx = {x^2}\cos \left( {1 - x} \right) - \left( { - 2x\sin \left( {1 - x} \right) + \int {2\sin \left( {1 - x} \right)dx} } \right) \cr
& \int {{x^2}\sin \left( {1 - x} \right)} dx = {x^2}\cos \left( {1 - x} \right) + 2x\sin \left( {1 - x} \right) - \int {2\sin \left( {1 - x} \right)dx} \cr
& \int {{x^2}\sin \left( {1 - x} \right)} dx = {x^2}\cos \left( {1 - x} \right) + 2x\sin \left( {1 - x} \right) - 2\cos \left( {1 - x} \right) + C \cr} $$
