Answer
$$0$$
Work Step by Step
First we need to set up the integration then we will solve the integral as follows:
$\int_0^{\pi/2} \int_{-\pi}^{\pi} \sin x \sin y \ dx \ dy= \int_0^{\pi/2} \sin y \ dy \int_{-\pi}^{\pi} \sin x dx$
We know that for an odd function $f(-x)=-f(x)$, we have $\int_{-a}^{a} f(x) \ dx=0$
Thus, our integral becomes:
$\int_0^{\pi/2} \int_{-\pi}^{\pi} \sin x \sin y \ dx \ dy= \int_0^{\pi/2} \sin y \ dy \times (0)=0$
