You may also use the following approach, such that:

`int sec x dx = int 1/cos x dx => int 1/cos x dx = int (sin^2 x + cos^2 x)/cos x dx`

Using the property of linearity of integral yields:

`int (sin^2 x + cos^2 x)/cos x dx = int (sin^2 x)/ (cos x) dx + int (cos^2 x)/cos x dx`

`int (sin^2 x + cos^2 x)/cos x dx = int (sin^2 x)/ (cos x) dx + int (cos^2 x)/cos x dx`

`int (sin^2 x + cos^2 x)/cos x dx = int sin x*tan x dx + int cos x dx`

You may use the following substitution, such that:

`tan (x/2) = t => (1 + tan^2(x/2))/2 dx = dt => dx = (2dt)/(1 + t^2)`

`sin x = (2t)/(1 + t^2)`

`tan x = (2t)/(1 - t^2)`

Changing the variable yields:

`int sin x*tan x dx = int (4t^2)/(1 - t^4)(2dt)/(1 + t^2)`

Using paratial fraction decomposition yields:

`t^2/((1 - t^2)(1 + t^2)^2) = A/(1 - t) + B/(1 + t) + (Ct + D)/(1 + t^2) + (Et + F)/(1 + t^2)^2`

`t^2 = A(1 + t)(1 + 2t^2 + t^4) + B(1 - t)(1 + 2t^2 + t^4) + (Ct + D)(1 - t^4) + (Et + F)(1 - t^2)`

`t^2 = A + 2At^2 + At^4 + At + 2At^3 + At^5 + B + 2Bt^2 + Bt^4 - Bt - 2Bt^3 - Bt^5 + Ct - Ct^5 + D - Dt^4 + Et - Et^3 + F - Ft^2`

`A - B - C = 0`

`A + B - D = 0`

`2A - 2B - E = 0`

`2A +2B - F = 1 => 2D + 2D = 1 => D = 1/4`

`A - B + C + E = 0 => A - B + A - B + 2(A - B) = 0 => A - B = 0 => A = B => E = C = 0`

`A + B + D + F = 0=> 2D + F = 0 => -F = 2D => F = -1/2`

`A + B = 1/2 - 1/4 => A + B = 1/4 => 2A = 1/4 => A = B = 1/8`

`t^2/((1 - t^2)(1 + t^2)^2) = 1/(8(1 - t)) + 1/(8(1 + t)) + 1/(4(1 + t^2)) -1/(2(1 + t^2)^2)`

`int t^2/((1 - t^2)(1 + t^2)^2) dt = int 1/(8(1 - t)) dt + int 1/(8(1 + t)) dt + int 1/(4(1 + t^2))dt -int 1/(2(1 + t^2)^2) dt`

`int t^2/((1 - t^2)(1 + t^2)^2) dt = (1/8)ln|1 - t| + (1/8)ln|1 + t| + (1/4) tan^(-1) t - tan^(-1) t + c`

Substituting back `tan (x/2)` for t yields:

`int sin x*tan x dx = (1/8)ln|1 - tan (x/2)| + (1/8)ln|1 + tan (x/2)| -3 (1/4)(x/2) + c`

`int sec x dx = (1/8)ln|1 - tan (x/2)| + (1/8)ln|1 + tan (x/2)| -3 (1/4)(x/2) + sin x + c`

**Hence, evaluating the integral yields **`int sec x dx = (1/8)ln|1 - tan (x/2)| + (1/8)ln|1 + tan (x/2)| -3 (1/4)(x/2) + sin x + c. `

To find the antiderivative of the given function, we'll have to calculate the indefinite integral of the function.

Int f(x)dx = Int sec x dx

We'll substitute the expression of the function f(x)= secx by f(x) = 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