To find the anti derivative of secx.

Or to find the In secx dx.

We know that d/dx( se cx) = secx tanx.....(1)

We know that d/dx tanx = sec^2 x...(2)

Therefore Integral secx dx = Integral secx {(secx+tanx)/(secx+tanx)} dx.

Integral secx dx = Integral {secx*tanx +sec^2x}/(secx+tantax) dx.

Integral secx dx = { dt/t} , where t = (secx+tanx) , as from (1) and (2) d/dx { secx +tanx ) = (secx*tanx +sec^2x) .

Therefore Integral secx dx= log (secx +tanx) +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

anti derivative of sec(x)= inte ( sec(x).(sec(x)+tan(x))/(sec(x)+tan(x))) dx= log(sex(x)+tan(x))+c