The issue is that you're still using FTC on the interval [2, t] for large t values, where the function is discontinuous at infinitely many discontinuities as you mentioned in a comment at the end. Note by the way that the discontinuities are not quite at n pi/2 for all natural numbers n (for instance n = 2 gives x = pi on which sec(pi) = -1 is well defined and continuous). The discontinuities of sec^2(x) = 1/cos^2(x) occur at all x of the form n pi/2 where n is any odd integer (even negative), since at those values of x the denominator cos^2(x) is equal to 0. While you understand that the integral of sec^2(x) from 1 to 2 is infinite due to limit of tan(x) for x approaching pi/2 from the left is infinity, you're not showing the steps as to how exactly that fact is being used in the computation.sec-(x)dx - we write the integral as a limit as t approaches t DO lim sec ( x ) du to0- 2 Since the derivative of ton (" ) is sec ( a ), the integral of sec(x ) is tan ( n) lim ton ( x)] , - Evaluate tan (t ) at t & at 2 lim tan(t ) - tan ( 2 ) = OO There are infinite discontinuities in -" of tan(t) for n EN