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Rima Bareh from St.Anthony's Higher Secondary School, asked a question
Subject: Math , asked on 13/9/12

 help plz !!!

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

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Mohd Imad From Gems Our Own English High School, added an answer, on 14/9/12
757 helpful votes in Math

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

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