Answer
$$\frac{{2{{\left( {\sin x} \right)}^{3/2}}}}{3} - \frac{{2{{\left( {\sin x} \right)}^{7/2}}}}{7} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\cos }^3}x} \sqrt {\sin x} dx \cr
& {\text{split off }}\cos \theta \cr
& = \int {{{\cos }^2}x} \sqrt {\sin x} \cos xdx \cr
& {\text{pythagorean identity}} \cr
& = \int {\left( {1 - {{\sin }^2}x} \right)} {\left( {\sin x} \right)^{1/2}}\cos xdx \cr
& u = \sin x,{\text{ }}du = \cos xdx \cr
& = \int {\left( {1 - {{\sin }^2}x} \right)} {\left( {\sin x} \right)^{1/2}}\cos xdx \cr
& = \int {\left( {1 - {u^2}} \right){u^{1/2}}du} \cr
& = \int {\left( {{u^{1/2}} - {u^{5/2}}} \right)du} \cr
& {\text{evaluate the integral}} \cr
& = \frac{{{u^{3/2}}}}{{3/2}} - \frac{{{u^{7/2}}}}{{7/2}} + C \cr
& {\text{replace }}u = \sin x \cr
& = \frac{{2{{\left( {\sin x} \right)}^{3/2}}}}{3} - \frac{{2{{\left( {\sin x} \right)}^{7/2}}}}{7} + C \cr} $$
