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

6 years ago


Answers : (1)


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


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

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6 years ago

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