$I=\int_{0}^{\frac{\pi }{2}}{\sin x\left( x+\cos 2x \right)dx}=\int_{0}^{\frac{\pi }{2}}{x.\sin xdx}+\int_{0}^{\frac{\pi }{2}}{\sin x.\cos 2xdx}={{I}_{1}}+{{I}_{2}}$

${{I}_{1}}=\int_{0}^{\frac{\pi }{2}}{x.\sin xdx}=-\left. x\cos x \right|_{0}^{\frac{\pi }{2}}+\int_{0}^{\frac{\pi }{2}}{\cos xdx}=0+\left. \sin x \right|_{0}^{\frac{\pi }{2}}=1$

$\begin{align}  & {{I}_{2}}=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}{\left( \sin 3x-\sin x \right)dx}=\frac{1}{6}\int_{0}^{\frac{\pi }{2}}{\sin 3x.d\left( 3x \right)}-\frac{1}{2}\int_{0}^{\frac{\pi }{2}}{\sin x.dx} \\  & =-\frac{1}{6}\left. \cos 3x \right|_{0}^{\frac{\pi }{2}}+\frac{1}{2}\left. \cos x \right|_{0}^{\frac{\pi }{2}}=\frac{1}{6}-\frac{1}{2}=-\frac{1}{3} \\ \end{align}$

$I={{I}_{1}}+{{I}_{2}}=\frac{2}{3}$