Calculus: Early Transcendentals, Chapter 4, 4.9, Section 4.9, Problem 35

You need to evaluate the antiderivative of the function f'(t), such that:
int f'(t)dt = f(t) + c
int (2cos t + sec^2 t) dt = int 2cos t dt + int sec^2 t dt
int (2cos t + sec^2 t) dt = 2sin t + tan t + c
The function is indeterminate, because of the constant c, but the problem provides the information that f(pi/3) =4, hence, you may evaluate the constant c, such that:
f(pi/3) = 2sin (pi/3) + tan (pi/3) + c => 4 = 2sqrt3/2+ sqrt 3 + c => c = 4 - 2sqrt 3
Hence, evaluating the function yields f(t) = 2sin t + tan t + 4 - 2sqrt 3, for t in (-pi/2, pi/2).