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 help plz !!!

integration of sec x log(sec x + tan x) dx 

Asked by Rima Bareh(student) , on 13/9/12


BEST ANSWER Certified by MeritNation Expert  

Ans : I = ⌡ secx*log(secx+tanx) dx

Let u=log(secx+tanx)

Differentiate wrt "x"

du/dx = {1/(secx+tanx)} *{ d(secx)/dx + d(tanx)/dx}

du/dx = {1/(secx+tanx)} *{ (secx*tanx)+(sec2x)}

du/dx ={1/(secx+tanx)} *{ (tanx)+(secx)} * secx

du =secx dx

Therefore: I =⌡u du

I = (u2/2) + C

I = ([ log(secx+tanx)] 2/2) + C

If you are satified do give thumbs up.

Posted by Mohd Imad(student)on 13/9/12

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