Answer
$$ - \frac{1}{3}\cos {x^3} + C $$
Work Step by Step
$$\eqalign{
& \int {{x^2}\sin {x^3}} dx \cr
& {\text{Integrate by substitution method}} \cr
& {\text{Let }}u = {x^3},\,\,\,\,du = 3{x^2}dx,\,\,\,\,dx = \frac{{du}}{{3{x^2}}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {{x^2}\sin {x^3}} dx = \int {{x^2}\sin u} \left( {\frac{{du}}{{3{x^2}}}} \right) \cr
& {\text{Cancel common factor }}{x^2} \cr
& = \int {\sin u} \left( {\frac{{du}}{3}} \right) \cr
& = \frac{1}{3}\int {\sin u} du \cr
& {\text{integrating}} \cr
& = - \frac{1}{3}\cos u + C \cr
& {\text{replacing }}u = {x^3} \cr
& = - \frac{1}{3}\cos {x^3} + C \cr} $$
