Math, asked by monsoonchandapbshqx, 1 year ago

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

Answers

Answered by Shivaya1
2

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



Similar questions