Aloha :)
$$I=\int\limits_{u=-1}^1\;\int\limits_{v=0}^{2\pi}\left[-u(2-u^2)\cos^2(v)-u(2-u^2)\sin^2(v)+u(2-u^2)^3\sin(v)\right]\,du\,dv$$$$\phantom I=\int\limits_{u=-1}^1\;\int\limits_{v=0}^{2\pi}\left[-u(2-u^2)\cdot\underbrace{(\cos^2(v)+\sin^2(v))}_{=1}+u(2-u^2)^3\sin(v)\right]\,du\,dv$$$$\phantom I=\int\limits_{v=0}^{2\pi}dv\int\limits_{u=-1}^1\left[-u(2-u^2)+u(2-u^2)^3\sin(v)\right]\,du$$Now let \(\,f(u)\coloneqq-u(2-u^2)+u(2-u^2)^3\sin(v)\,\) and use the symmetrie
$$\pink{f(-u)}=-(-u)(2-(-u)^2)+(-u)(2-(-u)^2)^3\sin(v)$$$$\phantom{f(-u)}=u(2-u^2)-u(2-u^2)^3\sin(v)=\pink{-f(u)}$$to rewrite the integral:$$\small I=\int\limits_{v=0}^{2\pi}dv\int\limits_{u=-1}^1f(u)\,du=\int\limits_{v=0}^{2\pi}dv\int\limits_{u=\pink0}^1\left[f(u)+\pink{f(-u)}\right]\,du=\int\limits_{v=0}^{2\pi}dv\int\limits_{u=0}^1\left[f(u)\pink{-f(u)}\right]\,du=0$$
