help plz !!!
integration of sec x log(sec x + tan x) dx
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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