Question

$\frac{dy}{dx}=\frac{1-\cos 2y}{1+\cos 2y}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}=\frac{1-\cos 2y}{1+\cos 2y} \\ \Rightarrow \frac{dx}{dy}=\frac{1+\cos 2y}{1-\cos 2y} \\ \Rightarrow dx=\frac{1+\cos 2y}{1-\cos 2y}dy \\ \Rightarrow dx=\frac{2{\cos }^{2}y}{2{\sin }^{2}y}dy \\ \Rightarrow dx={\cot }^{2}ydy \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow \int dx=\int {\cot }^{2}ydy \\ \Rightarrow x=\int \left({\mathrm{cosec}}^{2}y-1\right)dy \\ \Rightarrow x=\int {\mathrm{cosec}}^{2}ydy-\int dy \\ \Rightarrow x=-\cot y-y+C \\ \Rightarrow x+\cot y+y=C \\ \mathrm{Hence},x+\cot y+y=C\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$