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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


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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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