Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 240: 78

$s=\sin t -\cos t$

The anti-derivative for $\dfrac{ds}{dx}=\cos t+\sin t$ is $s=\sin t -\cos t+C$ Apply the initial conditions $s(\pi)=1$ to get the value of $C$. This implies $1=\sin \pi -\cos \pi+C$ or, $C=0$ Hence, $s=\sin t -\cos t$