Answer
$\frac{1}{3}{\sin ^{3/2}}\left( {{x^2}} \right) - \frac{1}{7}{\sin ^{7/2}}\left( {{x^2}} \right) + C$
Work Step by Step
$$\eqalign{
& \int x {\cos ^3}\left( {{x^2}} \right)\sqrt {\sin \left( {{x^2}} \right)} dx \cr
& = \int x {\cos ^{2 + 1}}\left( {{x^2}} \right)\sqrt {\sin \left( {{x^2}} \right)} dx \cr
& {\text{Use the law of exponents }}{a^{m + n}} = {a^m}{a^n} \cr
& = \int x {\cos ^2}\left( {{x^2}} \right)\cos \left( {{x^2}} \right)\sqrt {\sin \left( {{x^2}} \right)} dx \cr
& = \int {{{\cos }^2}\left( {{x^2}} \right)\sqrt {\sin \left( {{x^2}} \right)} } \left( x \right)\cos \left( {{x^2}} \right)dx \cr
& {\text{Use the pythagorean identity }}{\cos ^2}\theta = 1 - {\sin ^2}\theta \cr
& = \int {\left( {1 - {{\sin }^2}\left( {{x^2}} \right)} \right)\sqrt {\sin \left( {{x^2}} \right)} } \left( x \right)\cos \left( {{x^2}} \right)dx \cr
& \cr
& {\text{Let }}u = \sin \left( {{x^2}} \right),{\text{ }}du = 2x\cos \left( {{x^2}} \right)dx \cr
& {\text{Substituting}} \cr
& = \int {\left( {1 - {u^2}} \right)\sqrt u } \left( {\frac{1}{2}} \right)du \cr
& = \frac{1}{2}\int {\left( {{u^{1/2}} - {u^{5/2}}} \right)} du \cr
& {\text{Integrating}} \cr
& = \frac{1}{2}\left( {\frac{{{u^{3/2}}}}{{3/2}} - \frac{{{u^{7/2}}}}{{7/2}}} \right) + C \cr
& = \frac{1}{3}{u^{3/2}} - \frac{1}{7}{u^{7/2}} + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}\sin \left( {{x^2}} \right){\text{ for }}u \cr
& = \frac{1}{3}{\sin ^{3/2}}\left( {{x^2}} \right) - \frac{1}{7}{\sin ^{7/2}}\left( {{x^2}} \right) + C \cr} $$
