Evaluate the indefinite integral of y=sec x?

First, we'll substitute the expression of the function y =
secx by y = 1/cos x.

We'll write the
integral:

Int dx/cos x = Int cos xdx/(cos
x)^2

From the fundamental formula of trigonometry, we'll
get:

(cos x)^2 = 1 - (sin
x)^2

Int cos xdx/(cos x)^2 = Int cos xdx/[1 - (sin
x)^2]

We'll note sin x = t

cos
x*dx = dt

We'll re-write the integral in
t:

Int cos xdx/[1 - (sin x)^2] = Int dt/(1 -
t^2)

We'll analyze the
integrand:

1/(1 - t^2) =
1/(1-t)(1+t)

We'll separate the integrand into partial
fractions:

1/(1-t)(1+t) = A/(1-t) +
B/(1+t)

1 = A(1+t) + B(1-t)

1
= A + At + B - Bt

We'll factorize by
t:

1 = t(A-B) + A+B

The
coefficient of t from the left side has to be equal to the coefficient of t from the
right side:

A-B = 0

A =
B

A+B = 1

2B =
1

B = A = 1/2

1/(1-t)(1+t) =
1/2(1-t) + 1/2(1+t)

Int dt/(1-t)(1+t) = Intdt/2(1-t) + Int
dt/2(1+t)

Int dt/(1-t)(1+t) = (1/2)ln |1-t| + (1/2)ln|1+t|
+ C