Calculus of a Single Variable, Chapter 4, 4.1, Section 4.1, Problem 30

You need to evaluate the indefinite integral, such that:
int sec y(tan y - sec y) dy= int (sin y - 1)/(cos^2 y) dy = int (sin y)/(cos^2 y) dy - int 1/(cos^2 y) dy
You need to solve int (sin y)/(cos^2 y) dy , using substitution cos y = t => -sin ydy = dt
int (sin y)/(cos^2 y) dy = int (-dt)/(t^2) = 1/t + c = 1/(cos y) + c
You need to remember that 1/(cos^2 y) = (tan y)'
int sec y(tan y - sec y) dy= 1/(cos y) - tan y + c
Hence, evaluating the indefinite integral yields int sec y(tan y - sec y) dy= 1/(cos y) - tan y + c.