What is the anti derivative of tan(x)?

Question

also, how do you get to it.

Answer 1

What is the anti derivative of tan(x)?

∫tan(x) dx

= ∫sin(x)/cos(x) dx

Let u = cos(x) Thus du = -sin(x)dx so -du = sin(x) dx

= ∫-du/u

= - ln|u| + c

= -ln|cos(x)| + c

= ln(1/|cos(x)| ) + c

= ln|sec(x)| + c

Answer 2

Use a trig substitution and then integrate.

∫(tan x) dx = ∫(sin x/cos x) dx

Let u = cos x
du = -(sin x) dx

∫(sin x/cos x) dx = ∫(1/u)(-du) = -ln|u| + C = -ln|cos x| + C

Answer 3

f(x) = tan(x)
f'(x) = sec^2(x)

Explained in the link below...