Answer
-1
Work Step by Step
$\int^{\pi}_{\frac{\pi}{2}}\frac{sin2x}{2sinx}dx$
=$\int^{\pi}_{\frac{\pi}{2}}\frac{2 \sin x \cos x}{2sin x}dx$
=$\int^{\pi}_{\frac{\pi}{2}}\cos x dx$
=$[\sin x]^{\pi}_{\frac{\pi}{2}}$
=$(sin(\pi)-(sin\frac{\pi}{2}))$
=-1
