Answer
$$\frac{{2{{\left( {\sin x} \right)}^{5/2}}}}{5} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\left( {\sin x} \right)}^{3/2}}} \cos xdx \cr
& {\text{set }}u = \sin x{\text{ then }}\frac{{du}}{{dx}} = \cos x,\,\,\,\,\,\,\,\,\frac{{du}}{{\cos x}} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{{\left( {\sin x} \right)}^{3/2}}\cos x} dx = \int {{u^{3/2}}\cos x} \left( {\frac{{du}}{{\cos x}}} \right) \cr
& = \int {{u^{3/2}}} du \cr
& {\text{integrate by using the power rule for integration}} \cr
& = \frac{{{u^{5/2}}}}{{5/2}} + C \cr
& = \frac{{2{u^{5/2}}}}{5} + C \cr
& {\text{write in terms of }}x \cr
& = \frac{{2{{\left( {\sin x} \right)}^{5/2}}}}{5} + C \cr} $$
