Answer
$\ln \left| {1 + \sin \theta } \right| + C$
Work Step by Step
$$\eqalign{
& \text{Let }I=\int {\frac{{\cos \theta }}{{1 + \sin \theta }}} d\theta \cr
& {\text{Let }}u = 1 + \sin \theta ,{\text{ then }}du = \cos \theta d\theta \cr
& {\text{Apply the substitution}} \cr
& I = \int {\frac{{du}}{u}} \cr
& {\text{Integrating}} \cr
& I= \ln \left| u \right| + C \cr
& {\text{Write in terms of }}\theta ,{\text{ substitute }}u = 1 + \sin\theta \cr
& I = \ln \left| {1 + \sin\theta } \right| + C \cr} $$
