[tex]\displaystyle\int_{y=-8}^{y=8}\int_{x=0}^{x=\pi/2}(y+y^2\cos x)\,\mathrm dx\,\mathrm dy=\int_{y=-8}^{y=8}(xy+y^2\sin x)\bigg|_{x=0}^{x=\pi/2}\,\mathrm dy[/tex]
[tex]=\displaystyle\int_{y=-8}^{y=8}\left(\dfrac\pi2y+y^2\right)\,\mathrm dy[/tex]
[tex]=\left(\dfrac\pi4y^2+\dfrac13y^3\right)\bigg|_{y=-8}^{y=8}[/tex]
[tex]=\dfrac{1024}3[/tex]
