Let $g:[0,1] \to \mathbb{R}$ be a smooth, simple, closed curve. Prove that: $$\int_g x^2 dx = \int_g \exp (-\cos(y^2)) dy = 0$$
I am thinking there is some application of Green's Theorem here, however, I am not sure how to do it. I would appreciate it if anyone had any hints/ideas for how to solve this.
Edit: Here is my attempt, although it seems entirely too simple for this question. Green's theorem tells us that:
$$\int_g x^2 dx + \exp(-\cos(y^2)) dy = {\int \int}_D \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} = {\int \int}_D 0 - 0 = 0 (???)$$ question marks added to express my uncertainty with the answer. Even then, if this integral is equal to $0$, this does not guarantee that both integrals are $0$, but instead that their sums are both $0$. Perhaps this has to do with the fact that both integrals are necessarily positive, as each integrand is necessarily positive over the domain? I am not so sure.
Your approach works very simply if instead of adding the integrals you just use $\int_g{c^2dx + 0y}$ and $\int_g{0dx +\exp(-\cos y^2)dy}$