Answer
$$ - \frac{1}{5}\cos {x^2} + C$$
Work Step by Step
$$\eqalign{
& \int {x\sin {x^2}} dx \cr
& {\text{set }}u = {x^2}{\text{ then }}\frac{{du}}{{dx}} = 2x,\,\,\,\,\,\,\,\,\frac{{du}}{{2x}} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {x\sin {x^2}} dx = \int {x\sin u} \left( {\frac{{du}}{{2x}}} \right) \cr
& = \int {\sin u} \left( {\frac{{du}}{2}} \right) \cr
& {\text{factor out the constant}} \cr
& = \frac{1}{2}\int {\sin u} du \cr
& {\text{integrate by using the Basic Trigonometric integral }}\int {\sin x} dx = - \cos x + C \cr
& = \frac{1}{5}\left( { - \cos u} \right) + C \cr
& = - \frac{1}{2}\cos u + C \cr
& {\text{write in terms of }}x,{\text{ replacing }}{x^2}{\text{ for }}u{\text{ gives}} \cr
& = - \frac{1}{5}\cos {x^2} + C \cr} $$
