integral of [ ( secx ) / ( 1+ tanx ) ] ^2?

Question

i tried to solve it but i end up with ln(cos^2 x) + ln(sin^2 x) + 1/2(integral of 1/(sinx.cosx) and i can't continue. Please help me solve it

Answer 1

This first thing that you gotta realise is that sec^2(x) is the derivative of tan(x) and therefore 1 + tan(x), once this is known the integration becomes a lot easier. make the substitution tan(x) + 1 = u
sec^2(x) dx = du
Sub this in to get
int 1/(u^2) du which is nice and easy
= -1/u + C = -1/(1 + tan(x)) + C

Answer 2

Let u = tan x + 1
du/dx = sec ² x
du = sec ² x.dx
I = ∫ (1 / u²).du
I = - 1 / u + C
I = - 1 / (tan x + 1) + C