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`        ln(sec x+yan x)/cos x dx=ln(sec y+tan y)/cosy dy`
8 years ago

147 Points
```										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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```
8 years ago
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