Answer
$2 - 4\ln \left( {\frac{3}{2}} \right)$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {\frac{{\sin 2x}}{{2 + \cos x}}} dx \cr
& {\text{Use the double angle identity }}\sin 2x = 2\sin x\cos x \cr
& = \int_0^{\pi /2} {\frac{{2\sin x\cos x}}{{2 + \cos x}}} dx \cr
& {\text{Let }}u = \cos x,{\text{ }}du = - \sin xdx \cr
& {\text{The new limits of integration are:}} \cr
& x = 0 \to u = 1 \cr
& x = \pi /2 \to u = 0 \cr
& = - 2\int_1^0 {\frac{u}{{2 + u}}} du \cr
& {\text{By the long division}} \cr
& = - 2\int_1^0 {\left( {1 - \frac{2}{{2 + u}}} \right)} du \cr
& {\text{Integrating}} \cr
& = 2\left[ {u - 2\ln \left| {2 + u} \right|} \right]_0^1 \cr
& = 2\left[ {1 - 2\ln \left| {2 + 1} \right|} \right] - 2\left[ {0 - 2\ln \left| {2 + 0} \right|} \right] \cr
& = 2\left( {1 - 2\ln 3} \right) - 2\left( { - 2\ln 2} \right) \cr
& = 2 - 4\ln 3 + 4\ln 2 \cr
& = 2 - 4\left( {\ln 3 - \ln 2} \right) \cr
& = 2 - 4\ln \left( {\frac{3}{2}} \right) \cr} $$
