Answer
$$ - \ln \left| {1 + \cos x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin x}}{{1 + \cos x}}} dx \cr
& {\text{set }}u = 1 + \cos x{\text{ then }}\frac{{du}}{{dx}} = - \sin x,\,\,\,\,\,\,\,\,\frac{{ - du}}{{\sin x}} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{\sin x}}{{1 + \cos x}}} dx = \int {\frac{{\sin x}}{u}} \left( {\frac{{ - du}}{{\sin x}}} \right) \cr
& {\text{cancel common factor}} \cr
& = \int {\frac{1}{u}} \left( { - du} \right) \cr
& = - \int {\frac{1}{u}} du \cr
& {\text{integrate by using the integration rule }}\int {\frac{1}{x}} dx = \ln \left| x \right| + C \cr
& {\text{then}} \cr
& = - \ln \left| u \right| + C \cr
& {\text{write in terms of }}x \cr
& = - \ln \left| {1 + \cos x} \right| + C \cr} $$
