Solve the integral: $ \iint_D (x^2 \tan(x) + y^3 + 4) dxdy$ Solve the following integral
$$ \iint_{D} (x^2 \tan(x) + y^3 + 4)dxdy$$
where $D = \{ (x,y): x^2 + y^2 \leq 2\}$
I thought that a polar coordinates transformation would work but it takes an horrible form. Any suggestion to solve it?
Thanks!

Being odd function, 
$$\int_{-\sqrt{2}}^{\sqrt 2 } \int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}} x^2 \tan(x) \, dydx = \int_{-\sqrt{2}}^{\sqrt 2 } \int_{-\sqrt{2-y^2}}^{\sqrt{2-y^2}} x^2 \tan(x) \, dxdy = 0$$
rest is easy.
EDIT:: the integral consist of three parts. The first part which I showed above is zero by symmetry of bounds for odd function.
$$\iint_{D} (x^2 \tan(x) + y^3 + 4)dxdy$$
The second part is also zero since it is odd function of $y$
$$\int_D y^3 \, dx dy = \int_{-\sqrt{2}}^{\sqrt 2 } \int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}} y^3 \, dy dx = 0$$
The third part is 
$$\int_D 4 \, dx dy = 4 \int_{-\sqrt{2}}^{\sqrt 2 } \int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}} 1 \, dx dy = 4  \cdot \pi 2 = 8 \pi $$
The answer is $8 \pi $ you can check it here.
