Question

cos y log (sec x + tan x) dx = cos x log (sec y + tan y) dy

Answer 1

We have,
cos y log (sec x + tan x) dx = cos x log (sec y + tan y) dy
$$\begin{gathered} \Rightarrow \frac{\log \left(\sec y+\tan y\right)}{\cos y}dy=\frac{\log \left(\sec x+\tan x\right)}{\cos x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{\log \left(\sec y+\tan y\right)}{\cos y}dy=\int \frac{\log \left(\sec x+\tan x\right)}{\cos x}dx.....\left(1\right) \\ \mathrm{Putting}\log \left(\sec y+\tan y\right)=t\mathrm{and}\log \left(\sec x+\tan x\right)=u \\ \Rightarrow \frac{{\sec }^{2}y+\sec y\tan y}{\sec y+\tan y}dy=dt\mathrm{and}\frac{{\sec }^{2}x+\sec x\tan x}{\sec x+\tan x}dx=du \\ \Rightarrow \sec ydy=dt\mathrm{and}\sec xdx=du \\ \mathrm{Therefore},\left(1\right)\mathrm{becomes} \\ \int tdt=\int udu \\ \Rightarrow \frac{t^2}{2}=\frac{u^2}{2}+C \\ \Rightarrow \frac{{\left[\log \left(\sec y+\tan y\right)\right]}^2}{2}=\frac{{\left[\log \left(\sec x+\tan x\right)\right]}^2}{2}+C \\ \Rightarrow {\left[\log \left(\sec y+\tan y\right)\right]}^2={\left[\log \left(\sec x+\tan x\right)\right]}^2+2C \\ \Rightarrow {\left[\log \left(\sec y+\tan y\right)\right]}^2={\left[\log \left(\sec x+\tan x\right)\right]}^2+k,\mathrm{where}k=2C \end{gathered}$$