Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 357: 59

$s(t) = -cos(t) - sin(t) + 1$

$v(t) = -1(-sin(t)) - cos(t)$ $s(t) = -cos(t) - sin(t) + C$ Given: $s(0) = 0$ $s(0) = -cos(0) - sin(0) + C$ $0 = -1 - 0 + C$ $0 = -1 + C$ $C = 1$ Replace $C$ in the original function with the newly found value: $s(t) = -cos(t) - sin(t) + 1$