Answer
$4\pi^2$
Work Step by Step
$\iint_R(1+x^2\sin y+y^2\sin x)dA=\iint_R 1dA+\iint_R x^2\sin ydA +\iint_Ry^2\sin x dA$
Firstly, find each double integral:
(i) $\iint_R 1dA=\iint_RdA=$ Area of $R$ $=(\pi-(-\pi))\cdot (\pi-(-\pi))=2\pi\cdot 2\pi =4\pi^2$
(ii) We know that $\sin y$ is an odd function, so that $\iint_Rx^2\sin ydA=\int_{-\pi}^\pi x^2(\int_{-\pi}^\pi\sin y dy)dx=\int_{-\pi}^\pi x^2\cdot 0dx=\int_{-\pi}^\pi 0dx=0$
(iii) $\iint_Ry^2\sin xdA=\iint_{-\pi}^\pi y^2\int_{-\pi}^\pi\sin xdx)dy=\int_{-\pi}^\pi y^2\cdot 0dy=0$
Thus,
$\iint_R(1+x^2\sin y+y^2\sin x)dA=4\pi^2+0+0=4\pi^2$
