Answer
$-2$
Work Step by Step
When the vector field $F(x,y)=Pi+Qj$ is a conservative field, then throughout the domain $D$, we have
$\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$
$P$ and $Q$ are the first-order partial derivatives on the domain $D$.
Here, we have $\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}=\cos y$
This implies that the vector field $F$ is conservative.
Now, we have $f(x,y)=x \sin y+\cos y+ g(y)$
$f_y(x,y)=x \cos y-\sin y$
Here, we get $g(y)=C$; $C$ is a constant
Thus, we get $F(x,y)=x \sin y+\cos y+ C$
Now, $\int_C F \cdot dr =F(1,\pi)-F(2,0)=(1 \sin \pi+\cos \pi+ C)-(2 \sin 0+\cos 0+ C)$
or, $\int_C F \cdot dr=(0-1+C)-(0+1+C)=-2$
