Answer
$$ - \frac{2}{3}\cos x\sqrt {\cos x} + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {\cos x} } \sin xdx \cr
& {\text{or we can write }}\sqrt {\cos x} {\text{ as}} \cr
& = \int {{{\left( {\cos x} \right)}^{1/2}}\left( {\sin x} \right)} dx \cr
& {\text{set }}u = \cos x{\text{ then }}\frac{{du}}{{dx}} = - \sin x,\,\,\,\,\,\,\,\,\frac{{ - du}}{{\sin x}} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{{\left( {\cos x} \right)}^{1/2}}\left( {\sin x} \right)} dx = \int {{u^{1/2}}\left( {\sin x} \right)} \left( {\frac{{ - du}}{{\sin x}}} \right) \cr
& = - \int {{u^{1/2}}du} \cr
& {\text{integrate by using the power rule for integration}} \cr
& = - \frac{{{u^{3/2}}}}{{3/2}} + C \cr
& {\text{simplifying}} \cr
& = - \frac{2}{3}{u^{3/2}} + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{2}{3}{\left( {\cos x} \right)^{3/2}} + C \cr
& or \cr
& = - \frac{2}{3}\cos x{\left( {\cos x} \right)^{1/2}} + C \cr
& = - \frac{2}{3}\cos x\sqrt {\cos x} + C \cr} $$
