ln(sec x+yan x)/cos x dx=ln(sec y+tan y)/cosy dy

Dear upendra

ln(sec x+yan x)/cos x dx=ln(sec y+tan y)/cosy dy

integrate both side

∫ln(sec x+yan x)/cos x dx=∫ln(sec y+tan y)/cosy dy

let secx +tan x =u                               and sexy +tan y =t

  (secx tanx +sec²x ) dx =du

or secx (secx +tanx) dx=du

or dx/cosx  =du/u

so intigral become

 ∫1/u logu   du  =∫1/t logt   dt

let I=∫1/u logu   du

intigrate using first and second finction

I=logu*logu -∫1/u logu   du

2I=(logu)²

I=1/2 (logu)²

similerly for right hand side

intigral become

1/2 (logu)²     = 1/2 (logt)²     +C

 (log(secx+tanx))²     =  (log(secy+tany))²     +C