Find the line integral using the fundamental theorem of line integrals

Question

$$\int_C (1+\cosh(y),x\sinh(y))d\vec{s}$$ Where C is a curve that goes from $(0,0)$ to $(1,1)$

I am not sure on how to proceed, I can find

$\vec{F}=(1+\cosh(y),x\sinh(y))$

$\vec{f}:\nabla f=\vec{F}$

$\vec f=(x+x\cosh(y), x\cosh(y))$

Answer 1

Fundamental theorem of line integrals: if $f = \nabla F$ and $c$ is a contour from $a$ to $b$

$\int_c f \cdot dr = F(b) - F(a)$

The line integral does not depend on the path. It only depends on the endpoints.

$f = \nabla (x+x \cosh y)$

$\int_c f\cdot dr = 1 + \cosh 1$

Answer 2

Hint: You are in search of a function $f$ with the property that $$ \frac{\partial f}{\partial x}=1+\cosh y\implies f(x,y)=x+x\cosh y+g(y) $$ furthermore, you would like $$ \frac{\partial f}{\partial y}=x\sinh y+g'(y)=x\sinh y\implies g(y)=c $$ and $$ f(x,y)=x+x\cosh y+c $$ try evaluating the potential $f$ at the end points.