$\begin{align}  & \int_{0}^{\tfrac{\pi }{2}}{{{x}^{2}}\sin x\cos xdx}=\int_{0}^{\tfrac{\pi }{2}}{{{x}^{2}}(\tfrac{1}{2}\sin 2x)dx}=\frac{1}{2}\int_{0}^{\tfrac{\pi }{2}}{{{x}^{2}}\sin 2xdx} \\ & \text{Using integration by part for }\int{{{x}^{2}}\sin 2xdx} \\ & \text{Let }u={{x}^{2}},\text{ }du=2xdx \\ & dv=\sin 2xdx,\text{ }v=-\frac{\cos 2x}{2} \\ & \frac{1}{2}\int_{0}^{\tfrac{\pi }{2}}{{{x}^{2}}\sin 2xdx}=\frac{1}{2}\left[ \frac{-{{x}^{2}}\cos 2x}{2}-\int{\left( \frac{-\cos 2x}{2} \right)(2xdx)} \right] \\ & \frac{1}{2}\int_{0}^{\tfrac{\pi }{2}}{{{x}^{2}}\sin 2xdx}=\frac{1}{2}\left[ \frac{-{{x}^{2}}\cos 2x}{2}+x\cos 2x \right]_{0}^{\tfrac{\pi }{2}} \\ & =\left[ \frac{-{{x}^{2}}\cos 2x}{2}+(x\int \cos 2xdx)-\int{[(\int \cos 2x)\tfrac{d}{dx}(x)]dx} \right] \\ & =\frac{1}{2}\left[ \frac{-{{x}^{2}}\cos 2x}{2}+\left( \frac{x\sin 2x}{2}-\frac{1}{2}\int{\sin 2xdx} \right) \right]_{0}^{\tfrac{\pi }{2}} \\ & =\frac{1}{2}\left[ \frac{-{{x}^{2}}\cos 2x}{2}+\frac{x\sin 2x}{2}+\frac{\cos 2x}{4} \right]_{0}^{\tfrac{\pi }{2}} \\ & =\frac{1}{2}\left[ \left( \frac{{{\pi }^{2}}}{8}-\frac{1}{4} \right)-\frac{1}{4} \right]=\frac{1}{2}\left( \frac{\pi }{8}-\frac{1}{2} \right)=\frac{\pi }{16}-\frac{1}{4} \\\end{align}$