`f'(t) = 2cos(t) + sec^2 (t), -pi/2<t<pi/2, f(pi/3) = 4` Find `f`.

Textbook Question

Chapter 4, 4.9 - Problem 35 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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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).`

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