Let $\mathcal{C}$ the set of all regular, simple and closed curves in $\mathbb{R}^2$. Consider the function $F: \mathcal{C} \to \mathbb{R}$ defined by $$F(\gamma) := \int_\gamma(y^3 - y + \sin(x^5)) \, \text{d}x + (\log(y^2 + 1) - 2x^3) \, \text{d}y.$$ Show that $F$ has a maximum in $\mathcal{C}$.
Let $D_\gamma$ the set such that $\gamma$ is border of $D_\gamma$. So,
\begin{align} F(\gamma) & := \int_\gamma(y^3 - y + \sin(x^5)) \, \text{d}x + (\log(y^2 + 1) - 2x^3) \, \text{d}y \\[10pt] & =\int_\gamma(y^3 - y + \sin(x^5),\log(y^2 + 1) - 2x^5) \, \text{d}x \, \text{d}y \int_{D_\gamma}(-6x^2 - 3y^2 + 1) \, \text{d}x \, \text{d}y \end{align}
So, Green theorem simplifies the integral, but I don't know how to show that there is a maximum. Can someone help me? Is possible explicitly calculate this maximum?
We have $$\begin{align} F(\gamma) &= \int_{\gamma}(y^{3} - y + \sin(x^{5})) \, \mathrm{d}x + (\log(y^{2} + 1) - 2x^{3}) \, \mathrm{d}y \\ &= \int_{D_\gamma} \left( -6x^2 - 3y^2 + 1 \right) \, \mathrm{d}x \, \mathrm{d}y \\ \end{align}$$
This expression attains its maximal value when $-6x^2 - 3y^2 + 1 > 0$ on all of $D_\gamma,$ i.e. when $\gamma$ is the ellipse $6x^2 + 3y^2 = 1.$