Question

cos (x + y) dy = dx

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \cos \left(x+y\right)dy=dx \\ \Rightarrow \frac{dy}{dx}=\frac{1}{\cos \left(x+y\right)}.....\left(1\right) \\ \mathrm{Let}x+y=v \\ \Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx} \\ \Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1 \\ \mathrm{Therefore},\left(1\right)\mathrm{becomes} \\ \therefore \frac{dv}{dx}-1=\frac{1}{\cos v} \\ \Rightarrow \frac{dv}{dx}=\frac{\cos v+1}{\cos v} \\ \Rightarrow \frac{\cos v}{\cos v+1}dv=dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{\cos v}{\cos v+1}dv=\int dx \\ \Rightarrow \int \frac{\cos v\left(1-\cos v\right)}{1-{\cos }^{2}v}dv=\int dx \\ \Rightarrow \int \frac{\cos v\left(1-\cos v\right)}{{\sin }^{2}v}dv=\int dx \\ \Rightarrow \int \left(\cot v\mathrm{cosec}v-\mathrm{co}{t}^{2}v\right)dv=\int dx \\ \Rightarrow \int \left(\cot v\mathrm{cosec}v-{\mathrm{cosec}}^{2}v+1\right)dv=\int dx \\ \Rightarrow -\mathrm{cosec}v+\cot v+v=x+C \\ \Rightarrow -\mathrm{cosec}\left(x+y\right)+\cot \left(x+y\right)+x+y=x+C \\ \Rightarrow -\mathrm{cosec}\left(x+y\right)+\cot \left(x+y\right)+y=C \\ \Rightarrow \mathrm{cosec}\left(x+y\right)-\cot \left(x+y\right)=y-C \\ \Rightarrow \frac{1-\cos \left(x+y\right)}{\sin \left(x+y\right)}=y-C \\ \Rightarrow \frac{2{\sin }^2\left({\displaystyle \frac{x+y}{2}}\right)}{2\sin \left({\displaystyle \frac{x+y}{2}}\right)\cos \left({\displaystyle \frac{x+y}{2}}\right)}=y-C \\ \Rightarrow \frac{\sin \left({\displaystyle \frac{x+y}{2}}\right)}{\cos \left(x+y\right)}=y-C \\ \Rightarrow \tan \left(\frac{x+y}{2}\right)=y-C \end{gathered}$$