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

147 Points
11 years ago

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

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