Answer
$${\sin ^{ - 1}}\left( {\frac{{\sin \theta }}{2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\cos \theta }}{{\sqrt {2 - {{\sin }^2}\theta } }}} d\theta \cr
& \cr
& {\text{substitute }}t = \sin \theta ,{\text{ }}dt = \sin \theta d\theta \cr
& \int {\frac{{\cos \theta }}{{\sqrt {2 - {{\sin }^2}\theta } }}} d\theta = \int {\frac{{dt}}{{\sqrt {2 - {t^2}} }}} \cr
& \cr
& {\text{Let sin}}\alpha = \frac{t}{{\sqrt 2 }},{\text{ }}dt = \sqrt 2 \cos \alpha d\alpha \cr
& \int {\frac{{dt}}{{\sqrt {2 - {t^2}} }}} = \int {\frac{{\sqrt 2 \cos \alpha d\alpha }}{{\sqrt {2 - 2{{\sin }^2}\alpha } }}} \cr
& = \int {\frac{{\sqrt 2 \cos \alpha d\alpha }}{{\sqrt 2 \sqrt {1 - {{\sin }^2}\alpha } }}} \cr
& \cr
& {\text{pythagorean identity}} \cr
& = \int {\frac{{\sqrt 2 \cos \alpha d\alpha }}{{\sqrt 2 \sqrt {{{\cos }^2}\alpha } }}} \cr
& = \int {d\alpha } \cr
& = \alpha + C \cr
& \cr
& {\text{Where }}\alpha = {\sin ^{ - 1}}\left( {\frac{t}{2}} \right) \cr
& = {\sin ^{ - 1}}\left( {\frac{t}{2}} \right) + C \cr
& {\text{ replace }}t = \sin \theta \cr
& = {\sin ^{ - 1}}\left( {\frac{{\sin \theta }}{2}} \right) + C \cr} $$
