$\begin{align}  & \text{ }\int\limits_{{{e}^{4}}}^{{{e}^{2}}}{\frac{1}{x}\sin (\log x)dx} \\ & \text{    Let }u=\log x \\ & \text{   }\frac{du}{dx}=\frac{1}{x},\text{ }dx=xdu \\ & \int{\frac{1}{x}\sin (\log x)dx}=\int{\frac{1}{x}(\sin u)(xdu)}=\int{\sin udu} \\ & \int{\frac{1}{x}\sin (\log x)dx}=-\cos u=-\cos (\log x)+C \\ & \therefore \int\limits_{{{e}^{4}}}^{{{e}^{2}}}{\frac{1}{x}\sin (\log x)dx}=\left[ -\cos (\log x) \right]_{{{e}^{4}}}^{{{e}^{2}}} \\ & \int\limits_{{{e}^{4}}}^{{{e}^{2}}}{\frac{1}{x}\sin (\log x)dx}=[-\cos (\log {{e}^{2}})]-[-\cos (\log {{e}^{4}})] \\ & \int\limits_{{{e}^{4}}}^{{{e}^{2}}}{\frac{1}{x}\sin (\log x)dx}=-\cos 2+\cos 4=0.00183 \\\end{align}$