Answer
$$2\sqrt {1 + \sin x} + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {1 - \sin x} } dx \cr
& {\text{Rationalizing}} \cr
& = \int {\sqrt {\left( {1 - \sin x} \right)\left( {\frac{{1 + \sin x}}{{1 + \sin x}}} \right)} } dx \cr
& = \int {\sqrt {\frac{{1 - {{\sin }^2}x}}{{1 + \sin x}}} } dx \cr
& {\text{Where }}1 - {\sin ^2}x = {\cos ^2}x \cr
& = \int {\sqrt {\frac{{{{\cos }^2}x}}{{1 + \sin x}}} } dx \cr
& = \int {\frac{{\cos x}}{{\sqrt {1 + \sin x} }}} dx \cr
& {\text{Rewrite}} \cr
& = \int {{{\left( {1 + \sin x} \right)}^{ - 1/2}}\left( {\cos x} \right)} dx \cr
& {\text{Inregrating}} \cr
& = \frac{{{{\left( {1 + \sin x} \right)}^{1/2}}}}{{1/2}} + C \cr
& = 2\sqrt {1 + \sin x} + C \cr} $$
