Consider the differential equation $ \left(\frac{\mathrm{d}}{\mathrm{d}x}+a\right)y=f(x)\tag{1} $ This can be solved using an integrating factor. Suppose $ g'(x)=ag(x)\tag{2} $ then $ \frac{\mathrm{d}}{\mathrm{d}x}(g(x)y)=g(x)y'+ag(x)y=g(x)f(x)\tag{3} $ Equation $(3)$ is simply $(1)$ multiplied by $g(x)$. Note that $(2)$ is the same as $ \frac{\mathrm{d}}{\mathrm{d}x}\log(g(x))=a\tag{4} $ which is satisfied by $ g(x)=e^{ax}\tag{5} $ plugging $(5)$ back into $(3)$ yields $ \frac{\mathrm{d}}{\mathrm{d}x}\left(e^{ax}y\right)=e^{ax}f(x)\tag{6} $ which becomes $ y=e^{-ax}\int e^{ax}f(x)\,\mathrm{d}x\tag{7} $
$y''+y=\sin(x)+x\cos(x)$ is simply $ \left(\frac{\mathrm{d}}{\mathrm{d}x}+i\right)\left(\frac{\mathrm{d}}{\mathrm{d}x}-i\right)y=\sin(x)+x\cos(x)\tag{8} $ Using $(7)$ once with $a=i$ gives $ \begin{align} \left(\frac{\mathrm{d}}{\mathrm{d}x}-i\right)y &=e^{-ix}\int e^{ix}(\sin(x)+x\cos(x))\,\mathrm{d}x\\ &=e^{-ix}\left(\frac{x^2}4+\frac{ix}2+c_1\right)-e^{ix}\left(\frac{ix}4+\frac18\right)\tag{9} \end{align} $ Using $(7)$ again with $a=-i$ gives $ \begin{align} y &=e^{ix}\int e^{-ix}\left(e^{-ix}\left(\frac{x^2}4+\frac{ix}2+c_1\right)-e^{ix}\left(\frac{ix}4+\frac18\right)\right)\,\mathrm{d}x\\ &=e^{-ix}\left(\frac{ix^2}8-\frac x8\right)-e^{ix}\left(\frac{ix^2}8+\frac x8\right)+c_1^\prime e^{-ix}+c_2^\prime e^{ix}\\ &=\frac{x^2}4\sin(x)-\frac x4\cos(x)+c_1^{\prime\prime}\cos(x)+c_2^{\prime\prime}\sin(x)\tag{10} \end{align} $