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