Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 273: 98

$s=\sin t -\cos t$

We need to find the anti-derivative for $\dfrac{ds}{dx}=\cos t+\sin t$ Thus, we have: $s=\sin t -\cos t+C$ Apply the initial condition $s(\pi)=1$ in the above equation to solve for $C$. we get: $1=\sin \pi -\cos \pi+C \implies C=0$ Hence, $s=\sin t -\cos t$