Answer
$$ - \ln 3$$
Work Step by Step
$$\eqalign{
& \int_{ - \pi }^0 {\frac{{\sin x}}{{2 + \cos x}}dx} \cr
& {\text{substitute }}u = 2 + \cos x,{\text{ }}du = - \sin xdx \cr
& {\text{express the limits in terms of }}u \cr
& x = - \pi {\text{ implies }}u = 2 + \cos \left( { - \pi } \right) = 1 \cr
& x = 0{\text{ implies }}u = 2 + \cos \left( 0 \right) = 3 \cr
& {\text{the entire integration is carried out as follows}} \cr
& \int_{ - \pi }^0 {\frac{{\sin x}}{{2 + \cos x}}dx} = - \int_1^3 {\frac{{du}}{u}} \cr
& {\text{find the antiderivative}} \cr
& = - \left. {\left( {\ln \left| u \right|} \right)} \right|_1^3 \cr
& {\text{use the fundamental theorem}} \cr
& = - \left( {\ln \left| 3 \right| - \ln \left| 1 \right|} \right) \cr
& {\text{simplify}} \cr
& = - \ln 3 \cr} $$
