Question

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

Answer 1

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

Answer 2

Answer:

Step-by-step explanation: