Answer
$$y = - 2\sqrt {1 + \cos \theta } + C$$
Work Step by Step
$$\eqalign{
& y' = \sqrt {1 - \cos \theta } \cr
& \frac{{dy}}{{d\theta }} = \sqrt {1 - \cos \theta } \cr
& {\text{Separate the variables}} \cr
& dy = \sqrt {1 - \cos \theta } d\theta \cr
& {\text{Integrate both sides}} \cr
& y = \int {\sqrt {1 - \cos \theta } } d\theta \cr
& {\text{Rationalizing}} \cr
& y = \int {\sqrt {1 - \cos \theta } \left( {\frac{{\sqrt {1 + \cos \theta } }}{{\sqrt {1 + \cos \theta } }}} \right)d\theta } \cr
& y = \int {\frac{{\sqrt {\left( {1 - \cos \theta } \right)\left( {1 + \cos \theta } \right)} }}{{\sqrt {1 + \cos \theta } }}d\theta } \cr
& y = \int {\frac{{\sqrt {1 - {{\cos }^2}\theta } }}{{\sqrt {1 + \cos \theta } }}d\theta } \cr
& {\text{Recall that si}}{{\text{n}}^2}x + {\cos ^2}x = 1 \cr
& y = \int {\frac{{\sqrt {{{\sin }^2}\theta } }}{{\sqrt {1 + \cos \theta } }}d} \theta \cr
& y = \int {\frac{{\sin \theta }}{{\sqrt {1 + \cos \theta } }}d} \theta \cr
& {\text{Let }}u = 1 + \cos \theta ,{\text{ }}du = - \sin \theta d\theta \cr
& {\text{Substituting}} \cr
& y = - \int {\frac{{du}}{{\sqrt u }}} = - \int {{u^{ - 1/2}}du} \cr
& y = - \frac{{{u^{1/2}}}}{{1/2}} + C \cr
& y = - 2\sqrt u + C \cr
& {\text{Write in terms of }}\theta ,{\text{ substitute }}1 + \cos \theta {\text{ for }}u \cr
& y = - 2\sqrt {1 + \cos \theta } + C \cr} $$
