Answer
$$ - \frac{{2{{\left( {\cos x} \right)}^{3/2}}}}{3} + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {\cos x} \sin x} dx \cr
& {\text{substitute }}u = \cos x,{\text{ }}du = - \sin \pi xdx \cr
& = \int {\sqrt u } \left( { - du} \right) \cr
& = - \int {\sqrt u } du \cr
& = - \int {{u^{1/2}}} du \cr
& {\text{find antiderivative}} \cr
& = - \frac{{{u^{3/2}}}}{{3/2}} + C \cr
& = - \frac{{2{u^{3/2}}}}{3} + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{{2{{\left( {\cos x} \right)}^{3/2}}}}{3} + C \cr} $$
