The value of definite integral ${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }} \frac{{\mathrm{x}}^2}{1+\sin \mathrm{x}+\sqrt{1+{\sin }^2\mathrm{x}}}\mathrm{dx}$ is

$\mathrm{I}={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }} \frac{{\mathrm{x}}^2}{(1+\sin \mathrm{x})+\sqrt{1+{\sin }^2\mathrm{x}}}\mathrm{dx}$

$={\int }_0^{\mathrm{\pi }} \left(\frac{{\mathrm{x}}^2}{(1+\sin \mathrm{x})+\sqrt{1+{\sin }^2\mathrm{x}}}+\frac{{\mathrm{x}}^2}{1-\sin \mathrm{x}+\sqrt{1+{\sin }^2\mathrm{x}}}\right)\mathrm{dx}$

$={\int }_0^{\mathrm{\pi }} {\mathrm{x}}^2\left[\frac{1-\sin \mathrm{x}+\sqrt{1+{\sin }^2\mathrm{x}}+1+\sin \mathrm{x}+\sqrt{1+{\sin }^2\mathrm{x}}}{\left[1+\sin \mathrm{x}+\sqrt{1+{\sin }^2\mathrm{x}}\right]\left[1-\sin \mathrm{x}+\sqrt{1+{\sin }^2\mathrm{x}}\right]}\mathrm{dx}\right.$

$={\int }_0^{\mathrm{\pi }} {\mathrm{x}}^2\left[\frac{2\left(1+\sqrt{1+{\sin }^2\mathrm{x}}\right)}{{\left(1+\sqrt{1+{\sin }^2\mathrm{x}}\right)}^2-{\sin }^2\mathrm{x}}\right]\mathrm{dx}$

$={\int }_0^{\mathrm{\pi }} {\mathrm{x}}^2\left[\frac{2\left(1+\sqrt{1+{\sin }^2\mathrm{x}}\right)}{1+1+{\sin }^2\mathrm{x}+2\sqrt{1+{\sin }^2\mathrm{x}}-{\sin }^2\mathrm{x}}\right]\mathrm{dx}={\int }_0^{\mathrm{\pi }} {\mathrm{x}}^2\mathrm{dx}={\left[\frac{{\mathrm{x}}^3}{3}\right]}_0^{\mathrm{\pi }}=\frac{{\mathrm{\pi }}^3}{3}$