Answer
$$\frac{{{{\sin }^5}{x^2}}}{{10}} + C$$
Work Step by Step
$$\eqalign{
& \int {x{{\sin }^4}{x^2}\cos {x^2}} dx \cr
& or \cr
& \int {\left( {{{\sin }^4}{x^2}} \right)\left( {\cos {x^2}} \right)} xdx \cr
& {\text{set }}u = {x^2},\,\,\,\,\,du = 2xdx,\,\,\,\,\,xdx = \frac{1}{2}du \cr
& {\text{use the substitution}} \cr
& \int {\left( {{{\sin }^4}{x^2}} \right)\left( {\cos {x^2}} \right)} xdx = \int {\left( {{{\sin }^4}u} \right)\left( {\cos u} \right)\left( {\frac{1}{2}du} \right)} \cr
& = \frac{1}{2}\int {\left( {{{\sin }^4}u} \right)\left( {\cos u} \right)du} \cr
& \cr
& {\text{use the substitution }}v = \sin u,\,\,\,\,\,\,\,\,dv = \cos udu \cr
& = \frac{1}{2}\int {{v^4}dv} \cr
& {\text{integrate}} \cr
& = \frac{1}{2}\left( {\frac{{{v^5}}}{5}} \right) + C \cr
& = \frac{{{v^5}}}{{10}} + C \cr
& {\text{replace }}v = \sin u \cr
& = \frac{{{{\sin }^5}u}}{{10}} + C \cr
& {\text{replace }}u = {x^2} \cr
& = \frac{{{{\sin }^5}{x^2}}}{{10}} + C \cr} $$
