Answer
$$2\sqrt {1 - \cos x} + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {1 + \cos x} dx} \cr
& {\text{Rationalizing}} \cr
& = \int {\sqrt {1 + \cos x} \left( {\frac{{\sqrt {1 - \cos x} }}{{\sqrt {1 - \cos x} }}} \right)dx} \cr
& = \int {\frac{{\sqrt {\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)} }}{{\sqrt {1 - \cos x} }}dx} \cr
& = \int {\frac{{\sqrt {1 - {{\cos }^2}x} }}{{\sqrt {1 - \cos x} }}dx} \cr
& {\text{Recall that si}}{{\text{n}}^2}x + {\cos ^2}x = 1 \cr
& = \int {\frac{{\sqrt {{{\sin }^2}x} }}{{\sqrt {1 - \cos x} }}dx} \cr
& = \int {\frac{{\sin x}}{{\sqrt {1 - \cos x} }}dx} \cr
& {\text{Let }}u = 1 - \cos x,{\text{ }}du = \sin xdx \cr
& {\text{Substituting}} \cr
& = \int {\frac{{du}}{{\sqrt u }}} = \int {{u^{ - 1/2}}du} \cr
& = \frac{{{u^{1/2}}}}{{1/2}} + C \cr
& = 2\sqrt u + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}1 - \cos x{\text{ for }}u \cr
& = 2\sqrt {1 - \cos x} + C \cr} $$
