Answer
$$2\ln 2 - \ln 3$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /3} {\frac{{\sin x}}{{1 + \cos x}}dx} \cr
& {\text{Set }}u = 1 + \cos x \to du = - \sin xdx \cr
& u = 1 + \cos x,{\text{ for }}x = \pi /3,\,\,u = 3/2 \cr
& u = 1 + \cos x,{\text{ for }}x = 0,\,\,u = 2 \cr
& {\text{using the substitution}} \cr
& \int_0^{\pi /3} {\frac{{\sin x}}{{1 + \cos x}}dx} = - \int_2^{3/2} {\frac{1}{u}} du \cr
& {\text{integrate}} \cr
& = - \left[ {\ln \left| u \right|} \right]_2^{3/2} \cr
& {\text{evaluate}} \cr
& = - \left[ {\ln \left| {\frac{3}{2}} \right| - \ln \left| 2 \right|} \right] \cr
& = 2\ln 2 - \ln 3 \cr} $$
