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upendra hembram Grade: 12
        

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

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
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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Badiuddin



7 years ago
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