What is the anti derivative of 1/cosx? (with working please)?

Answer 1

Solve the integral with the substitution
u = sin(x)
→
du/dx = cos(x)
→
dx = 1/cos(x) du
with the Pythagorean identity
sin²(x) + cos²(x)
→
cos²(x) = sqrt(1 - sin²(x)) = (1 - u²)


∫ 1/cos(x) dx
= ∫ 1/cos²(x) du
= ∫ 1/(1 - u²) du
which is an elementary integral: the integral of the inverse hyperbolic tangent
= artanh( u ) + c
= artanh( sin(x) ) + c
with the definition of the inverse hyperbolic tangent:
(artanh(a) = 1/2 · ln[(1 + a) / (1 - a)]
= 1/2 · ln[(1 + sin(x)) / (1 - sin(x))] + c
multiply nominator and denominator of the fraction with (1 + sin(x))
= 1/2 · ln[(1 + sin(x))² / (1 - sin²(x))] + c
= 1/2 · ln[(1 + sin(x))² / cos²(x)] + c
because 1/2 · ln(a) = ln(sqrt(a))
= ln[(1 + sin(x)) / cos(x)] + c
= ln[1/cos(x) + sin(x)/cos(x)] + c
= ln[sec(x) + tan(x)] + c

Answer 2

∫ (1/ cos x).dx
= ∫ sec x .dx
= log (sec x + tan x) + C
NB this integral is given in text books as a standard integral.