Answer
$\frac{2\pi}{3}-\frac{\pi\sqrt{3}}{4}$
Work Step by Step
We begin with the iterated integral
$$\int_0^{\pi/6}\int_0^{\pi/2}(\sin x + \sin y)\,dy\,dx$$
Solving the inner integral, we are left with:
$$\int_0^{\pi/6}\bigg[y\sin x -\cos y\bigg]_{y=0}^{y=\pi/2}dx\\
=\int_0^{\pi/6}\bigg(\frac{\pi}{2}\sin x-0-0+1\bigg)dx\\
=\int_0^{\pi/6}\bigg(\frac{\pi}{2}\sin x+1\bigg)dx$$
Solving this integral, we get:
$$\bigg[-\frac{\pi}{2}\cos x + x\bigg]_{0}^{\pi/6}\\
=-\frac{\pi}{2}\frac{\sqrt{3}}{2}+\frac{\pi}{6}+\frac{\pi}{2}-0\\
=\frac{2\pi}{3}-\frac{\pi\sqrt{3}}{4}$$
