help plz !!!

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

## CERTIFIED ANSWER

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**)+(**sec ^{2}x**)}

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

du = **secx dx**

Therefore: I = ⌡u du

I = (u^{2}/2) + C

I = (* [* * log(secx+tanx)] *

^{2}/2) + C

If you are satified do give thumbs up.

*This conversation is already closed by Expert*