Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises - Page 238: 60

$$\tan t$$

$$\eqalign{ & f\left( t \right) = {\sec ^2}t \cr & {\text{find an antiderivative of }}f\left( t \right) \cr & F\left( t \right) = \tan t + C \cr & {\text{using the initial condition }}F\left( {\pi /4} \right) = 1 \cr & 1 = \tan \left( {\frac{\pi }{4}} \right) + C \cr & 1 = 1 + C \cr & {\text{then}} \cr & C = 0 \cr & {\text{so}}{\text{,}} \cr & = \tan t + 0 \cr & = \tan t \cr} $$