Answer
$$\ln 2$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {\frac{{\sin x}}{{1 + \cos x}}} dx \cr
& {\text{substitute }}u = 1 + \cos x,{\text{ }}du = - \sin xdx \cr
& {\text{express the limits in terms of }}u \cr
& x = 0{\text{ implies }}u = 1 + \cos 0 = 2 \cr
& x = \pi /2{\text{ implies }}u = 1 + \cos \left( {\pi /2} \right) = 1 \cr
& {\text{the entire integration is carried out as follows}} \cr
& \int_0^{\pi /2} {\frac{{\sin x}}{{1 + \cos x}}} dx = - \int_2^1 {\frac{{du}}{u}} \cr
& {\text{find the antiderivative}} \cr
& = - \left. {\left( {\ln \left| u \right|} \right)} \right|_2^1 \cr
& {\text{use the fundamental theorem}} \cr
& = - \left( {\ln \left| 1 \right| - \ln \left| 2 \right|} \right) \cr
& {\text{simplify}} \cr
& = - \left( {0 - \ln 2} \right) \cr
& = \ln 2 \cr} $$
