$\begin{align}  & \int_{0}^{\tfrac{\pi }{2}}{\sin 2xdx}=\left[ -\tfrac{1}{2}\cos 2x \right]_{0}^{\tfrac{\pi }{2}}=-\tfrac{1}{2}\left[ \cos 2x \right]_{0}^{\tfrac{\pi }{2}} \\ & =-\tfrac{1}{2}\left[ \cos 2(\tfrac{\pi }{2})-\cos 2(0) \right] \\ & =-\tfrac{1}{2}\left[ \cos \pi -\cos 0 \right]=-\tfrac{1}{2}\left[ -1-1 \right]=1 \\\end{align}$