Let $K$ be an oriented curve starting at $a = (\pi, -1)$ and ending at $b = (3\pi, 1)$. $$ I = \int_K (e^y+e^{-y})\cos xdx + (e^y-e^{-y})\sin xdy $$ Does the value of the integral depend on the choice of the path between $a$ and $b$? Calculate the value of $I$.
Let $H = (e^y+e^{-y})\sin x$ then $grad H = [(e^y+e^{-y})\cos x, (e^y-e^{-y})\sin x]^T$ so I suppose that the value of integral does not depend on the choice of the path between $a$ and $b$. Is it
But how to calculate the value? I cannot use Green Theorem because it is not the closed path and therefore is not the edge of any area.
You are correct that the integral is path-independent, because the 1-form you are integrating is exact (by which I mean, it is the gradient of a function).
This hugely simplifies the evaluation of the integral, because we can simply use the fundamental theorem of calculus!
$$ \int_a^b \,\nabla H \cdot \mathrm{d}x = \int_a^b \,\mathrm{d}H = H|_b - H|_a \,.$$