Double integral $\int\int x \sin(x+y)dA$ with $R=[0,\pi/6]\times[0,\pi/3]$ The closest of my tries was:
$$\int^{\pi/6}_0 x \sin(x+y)dx=-x \cos(x+y)+ \sin(x+y) \bigg \lvert ^{\frac{\pi}{6}}_{0}$$ 
having used the formula $\int uv' =uv-\int u'v$. Integrating this with respect to $y$, we find
$$\int^{\pi/3}_0 -\frac{\pi}{6} \cos(\frac{\pi}{6}+y)+ \sin(\frac{\pi}{6}+y)- \sin(y)dy = -\frac{\pi}{6} \sin(\frac{\pi}{6}+y)- \cos(\frac{\pi}{6}+y)+ \cos(y) \bigg \lvert^{\pi/3}_0$$
which got me something around $-0.0094011004$, when the answer should be $0.5(\sqrt{3}-1) \approx 0.3660$.

Your calculations are perfectly correct, up until the numerical evaluation, which is wrong. You should get
$$
\frac{\sqrt3-1}{2} - \frac{\pi}{12} \approx 0.1042
,
$$
which is the correct answer. (The solution that you quote is wrong, it's lacking the term $-\pi/12$.)
