`(ln(secx+tanx)/(cosx)dx=(ln(secy+tany)/(cosy)dy`

Explanation

`frac{ln (sec x+tan x)}{cos x} d x=frac{ln (sec y+tan y)}{cos y} d y`


Integrating on both side of equation, we get


`int frac{ln (sec x+tan x)}{cos x} d x=int frac{ln (sec y+tan y)}{cos y} d y`


let assume `z=ln (sec x+tan x)` in LHS


`Rightarrow frac{{dz}}{{dx}}=frac{sec{x} cdot tan{x}+sec ^{2}{x}}{sec{x}+tan{x}}`


`Rightarrow frac{d z}{d x}=frac{sec x(tan x+sec x)}{sec x+tan x}`


`Rightarrow d z=frac{d x}{cos x}`


Let assume `t=ln (sec y+tan y)` in RHS


...