# Evaluate the indefinite integral of y=sec x?

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To find the integral of y = secx.

solution:

We know that sec x= secx(secx +tanx)/ (secx+tanx).

Also d/dx (sec x +tanx ) = d/dx (secx ) +d/dx(tanx)

d/dx(sec x + tanx) = secxtanx + sec^2x = secx(tanx+secx).

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

Int secx = log (secx+tanx) + C, where C is a constant.

Therefore Int y = Int secx = log(secx+tanx) + C.

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