The solution of 2cos $\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+4\mathrm{y} \sin \mathrm{x}=\sin 2\mathrm{x}$

$\begin{array}{l}Given2\cos x.\frac{dy}{dx}+4y\sin x=2\sin x\cos x\\ \Rightarrow \frac{dy}{dx}+\left(2Tanx\right)y=\sin x\\ I.F=e^{\int 2Tanx}=e^{2\log \left|\sec x\right|}\\ =e^{\log \left|\sec {}^{2}x\right|}={\sec }^{2}x\\ Solutionisy(I.F)=\int \left(\sin x\right)(I.F)dx\\ \Rightarrow y.{\sec }^{2}x=\int \sin x.{\sec }^{2}xdx\\ =\int \sec xTanx.dx\\ y{\sec }^{2}x=\sec x+c\end{array}$