$\begin{array}{ll} & \mathrm{e}^{y-x} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y(\sin x+\cos x)}{(1+y \log y)} \\ \therefore \quad & \mathrm{e}^y \frac{(1+y \log y)}{y} \mathrm{~d} y=\mathrm{e}^x(\sin x+\cos x) \mathrm{d} x \\ \therefore \quad & \int \mathrm{e}^y\left(\log y+\frac{1}{y}\right) \mathrm{d} y=\int \mathrm{e}^x(\sin x+\cos x) \mathrm{d} x \\ & \Rightarrow \mathrm{e}^y \log y=\mathrm{e}^x \sin x+\mathrm{c} \\ & \ldots\left[\because \int \mathrm{e}^x\left(\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right) \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right]\end{array}$