The area of the region between y = cosx and $\mathrm{y}=\sin 2\mathrm{x} \mathrm{in} 0\le \mathrm{x}\le \frac{\mathrm{\pi }}{2} \mathrm{is} \mathrm{\_\_\_\_\_}$

A$Area={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\left(\cos x -\sin 2x\right)dx+{\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\sin 2x-\cos x \right)dx$

$Area={\left(\sin x+\frac{\cos 2x}{2}\right)}_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}+{\left(\frac{-\cos 2x}{2}-\sin x\right)}_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$=\left(\frac{1}{2}+\frac{1}{4}\right)-\left(0+\frac{1}{2}\right)+\left(\frac{1}{2}-1\right)-\left(\frac{-1}{4}-\frac{1}{2}\right)=\frac{1}{2}$