Calculate the double integral. double integral xsin(x+y)dA, R=[0, pi/6]x[0, pi/3]

$\text{Firstly, integrate with respect to y}$.

$$\begin{aligned} &\int_{0}^{\pi/6} \int_{0}^{\pi/3} x\sin{(x+y)}dydx \\ &=\int_{0}^{\pi/6}\left( (-x\cos{(x+\frac{\pi}{3})})+x\cos{x}\right) dx \\ &=-\int_{0}^{\pi/6} (x\cos{(x+\frac{\pi}{3})})dx + \int_{0}^{\pi/6}x\cos{x} dx \\ &=- \biggl[ x\sin{\left(x+ \frac{\pi}{3}\right)}+\cos{\left( x+ \frac{\pi}{3}\right)}\biggr]_{x=0}^{x=\pi/6}+ \biggl[x\sin{x+\cos{x}} \biggr]_{x=0}^{x=\pi/6} \\ &=-\frac{\pi}{6}\sin{\frac{\pi}{2}}-\cos{\frac{\pi}{2}}+\cos{\frac{\pi}{3}}+\frac{\pi}{6}\sin{\frac{\pi}{6}}+\cos{\frac{\pi}{6}}-\cos{0} \\ &=-\frac{\pi}{6}+\frac{1}{2}+\frac{\pi}{12}+ \frac{\sqrt{3}}{2}-1 \\ &=-\frac{\pi}{12} + \frac{\sqrt{3}-1}{2} \end{aligned}$$

We calculate the iterated integral:

$$\begin{align*} \int_0^{\frac{\pi}{6}}\int_0^{\frac{\pi}{3}}x\sin(x+y)\,dydx &=\int_0^{\frac{\pi}{6}}-x\cos(x+y)\Big|_{y=0}^{y=\frac{\pi}{3}}\,dx \\ &=\int_{0}^{\frac{\pi}{6}}x\cos x-x\cos(x+\dfrac{\pi}{3})\,dx\\ &=\dfrac{-1+\sqrt{3}}{2}-\dfrac{\pi}{12} \end{align*}$$

$$\iint_R x\sin(x+y)\hspace{1mm}dA = \int_0^{\pi/3}\left[\int_0^{\pi/6}x\sin(x+y)\hspace{1mm}dx\right]\hspace{1mm}dy$$