Answer
$$\csc \theta - \cot \theta + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{d\theta }}{{1 + \cos \theta }}} \cr
& = \int {\frac{1}{{1 + \cos \theta }}} \left( {\frac{{1 - \cos \theta }}{{1 - \cos \theta }}} \right)d\theta \cr
& = \int {\frac{{1 - \cos \theta }}{{1 - {{\cos }^2}\theta }}} d\theta \cr
& = \int {\left( {\frac{1}{{1 - {{\cos }^2}\theta }} - \frac{{\cos \theta }}{{1 - {{\cos }^2}\theta }}} \right)} d\theta \cr
& = \int {\left( {\frac{1}{{{{\sin }^2}\theta }} - \frac{{\cos \theta }}{{{{\sin }^2}\theta }}} \right)} d\theta \cr
& = \int {\left( {{{\csc }^2}\theta - \cot \theta \csc \theta } \right)} d\theta \cr
& = \int {{{\csc }^2}\theta } d\theta - \int {\cot \theta \csc \theta } d\theta \cr
& {\text{find antiderivatives}} \cr
& = - \cot \theta - \left( { - \csc \theta } \right) + C \cr
& = \csc \theta - \cot \theta + C \cr} $$
