Answer
$4\ln\left(\frac{2}{3}\right)+2$
Work Step by Step
If $t$ = $\tan(\frac{x}{2})$
then $\frac{x}{2}$ = $\tan^{-1}t$
$\cos\frac{x}{2}$ = $\frac{1}{\sqrt {1+t^2}}$
$\sin\frac{x}{2}$ = $\frac{t}{\sqrt {1+t^2}}$
$\cos x=\frac{1-t^2}{1+t^2}$
$\sin x=\frac{2t}{1+t^2}$
$dx=\frac{2}{1+t^2}dt$
$\int_0^{\frac{\pi}{2}}\frac{\sin 2x}{2+\cos x}dx$ = $\int_0^{\frac{\pi}{2}}\frac{2\sin x\cos x}{2+\cos x}dx$
= $2\int_0^1\frac{2\cdot\frac{2t}{1+t^2}\frac{1-t^2}{1+t^2}}{2+\frac{1-t^2}{1+t^2}}\cdot\frac{2}{1+t^2}dt$
= $\int_0^1\frac{8t(1-t^2)}{(t^2+3)(t^2+1)^2}dt$
Let
$u$ = $t^2$
$du$ = $2tdt$
$\frac{1-t^2}{(t^2+3)(t^2+1)^2}$ = $\frac{1-u}{(u+3)(u+1)}$ = $\frac{A}{u+3}+\frac{B}{u+1}+\frac{C}{(u+1)^2}$
$1-u$ = $A(u+1)^2+B(u+3)(u+1)+C(u+3)$
$A$ = $1$, $B$ = $-1$, $C$ = $1$
So
$\int_0^1\left[8t\frac{1-t^2}{(t^2+3)(t^2+1)^2}\right]dt$ = $\int_0^1\left[\frac{8t}{t^2+3}-\frac{8t}{t^2+1}+\frac{8t}{(t^2+1)^2}\right]dt$ = $\left[4\ln|t^2+3|-4\ln|t^2+1|-\frac{4}{t^2+1}\right]_0^1$ = $4\ln\left(\frac{2}{3}\right)+2$
