Answer
$$ - \frac{1}{{12}}\cos 4{x^3} + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}\sin 4{x^3}} dx \cr
& {\text{set }}u = 4{x^3}{\text{ then }}\frac{{du}}{{dx}} = 12{x^2},\,\,\,\,\,\,\,\,\frac{{du}}{{12{x^2}}} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{x^2}\sin 4{x^3}} dx = \int {{x^2}\sin u} \left( {\frac{{du}}{{12{x^2}}}} \right) \cr
& {\text{cancel the common factors}} \cr
& = \int {\sin u} \left( {\frac{{du}}{{12}}} \right) \cr
& = \frac{1}{{12}}\int {\sin u} du \cr
& {\text{integrate by using the Basic Trigonometric integral }}\int {\sin x} dx = - \cos x + C \cr
& = - \frac{1}{{12}}\cos u + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{1}{{12}}\cos 4{x^3} + C \cr} $$
