Answer
$$ = - {\sinh ^{ - 1}}\left( {\cos \theta } \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin \theta }}{{\sqrt {1 + {{\cos }^2}\theta } }}} d\theta \cr
& {\text{substitute }}u = \cos \theta ,{\text{ }}du = - \sin \theta d\theta \cr
& = \int {\frac{{\sin \theta }}{{\sqrt {1 + {{\cos }^2}\theta } }}} d\theta = \int {\frac{{ - du}}{{\sqrt {1 + {u^2}} }}} \cr
& = - \int {\frac{{du}}{{\sqrt {1 + {u^2}} }}} \cr
& {\text{find the antiderivarive using the theorem 6}}{\text{.8}}{\text{.6 }}\left( {{\text{see page 480}}} \right) \cr
& = - {\sinh ^{ - 1}}u + C \cr
& {\text{write in terms of }}x \cr
& = - {\sinh ^{ - 1}}\left( {\cos \theta } \right) + C \cr} $$
