Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 287: 59

$y=\int_{2}^{x} \sec t dt+3$

Given: $\dfrac{dy}{dx}=\sec x$ $\implies \dfrac{d}{dx}[\int_{2}^{x} (\sec t) dt]+3=\sec x$ $\implies y(2)=\int_{(-2)}^{2} (\sec t) dt+3$ $\implies y(2)=0+3=3$ Hence, $y=\int_{2}^{x} \sec t dt+3$