Answer
$$\ln 3$$
Work Step by Step
$$\eqalign{
& \int_0^\pi {\frac{{\sin t}}{{2 - \cos t}}} dt \cr
& {\text{Use substitution:}}\cr
& {\text{Let }}u = 2 - \cos t,{\text{ so that }}du = \sin tdt \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}t = \pi,{\text{ }}u = 2 - \cos \left( \pi \right) = 3 \cr
& \,\,\,\,\,\,{\text{If }}t = 0,{\text{ }}u = 2 - \cos \left( 0 \right) = 1 \cr
& {\text{write the integral in terms of }}u \cr
& \int_0^\pi {\frac{{\sin t}}{{2 - \cos t}}} dt = \int_1^3 {\frac{{du}}{u}} \cr
& {\text{integrate}} \cr
& = \left( {\ln \left| u \right|} \right)_1^3 \cr
& {\text{use the fundamental theorem of calculus: }}\cr
& \int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \ln \left| 3 \right| - \ln \left| 1 \right| \cr
& {\text{simplifying}} \cr
& = \ln 3 \cr} $$
