Answer
$$\tan x - \sec x + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{1 + \sin \theta }}} d\theta \cr
& {\text{Multiply the numerator and denominator by }}1 - \sin \theta \cr
& = \int {\frac{1}{{1 + \sin \theta }}} \left( {\frac{{1 - \sin \theta }}{{1 - \sin \theta }}} \right)d\theta \cr
& = \int {\frac{{1 - \sin \theta }}{{1 - {{\sin }^2}\theta }}} d\theta \cr
& {\text{Recall that si}}{{\text{n}}^2}x + {\cos ^2}x = 1 \cr
& = \int {\frac{{1 - \sin \theta }}{{{{\cos }^2}x}}} d\theta \cr
& = \int {\left( {\frac{1}{{{{\cos }^2}x}} - \frac{{\sin \theta }}{{{{\cos }^2}x}}} \right)} d\theta \cr
& = \int {\left( {{{\sec }^2}x - \sec x\tan x} \right)} d\theta \cr
& {\text{Integrating}} \cr
& = \tan x - \sec x + C \cr} $$
