Answer
$$\iint \limits_{R} x \sin (x+y) d A= \frac{\sqrt 3-1}{2} -\frac{\pi}{12}$$
Work Step by Step
Given $$\iint \limits_{R} x \sin (x+y) d A, \quad R=[0, \pi / 6] \times[0, \pi / 3]$$
So, we have
\begin{aligned}
I&=\iint \limits_{R} x \sin (x+y) d A\\
&=\int_{0}^{\pi / 3}\left[\int_{0}^{\pi / 6} x \sin (x+y) d x\right] d y\\
&=
\end{aligned}
Let
\begin{aligned}
I_1&= \int_{0}^{\pi / 6} x \sin (x+y) d x\\
\end{aligned}
So by partition technique Let $$u=x \Rightarrow du= dx$$ $$dv=\sin (x+y) \ dx \Rightarrow v= -\cos (x+y) $$ So, we get
\begin{aligned}
I_1&=uv|_{0}^{\pi / 6}- \int_{0}^{\pi / 6} v du\\
&= -x \cos (x+y)|_{0}^{\pi / 6}+\int_{0}^{\pi / 6} \cos (x+y) d x\\
&= -x \cos (x+y)|_{0}^{\pi / 6}+\sin(x+y)|_{0}^{\pi / 6}\\
&= -\frac{\pi}{6} \cos (\frac{\pi}{6}+y) +0+\sin(\frac{\pi}{6}+y) -\sin( y)\\
&= -\frac{\pi}{6} \cos (\frac{\pi}{6}+y) +\sin(\frac{\pi}{6}+y) -\sin( y)\\
\end{aligned}
so, we get
\begin{aligned}
I&
=\int_{0}^{\pi / 3}\left[ -\frac{\pi}{6} \cos (\frac{\pi}{6}+y) +\sin(\frac{\pi}{6}+y) -\sin( y) \right] d y\\
&= \left[ -\frac{\pi}{6} \sin (\frac{\pi}{6}+y) -\cos(\frac{\pi}{6}+y)+\cos( y) \right] _{0}^{\pi / 3}\\
&= \left[ -\frac{\pi}{6} \sin (\frac{\pi}{6}+\frac{\pi}{3}) -\cos(\frac{\pi}{6}+\frac{\pi}{3})+\cos( \frac{\pi}{3}) \right] - \left[ -\frac{\pi}{6} \sin (\frac{\pi}{6}+0) -\cos(\frac{\pi}{6}+0)+\cos( 0) \right]\\
&= \left[ -\frac{\pi}{6} \sin (\frac{\pi}{2}) -\cos(\frac{\pi}{2})+\cos( \frac{\pi}{3}) \right] - \left[ -\frac{\pi}{6} \sin (\frac{\pi}{6} ) -\cos(\frac{\pi}{6} )+1 \right]\\
&= \left[ -\frac{\pi}{6} + \frac{1}{2}\right] - \left[ -\frac{\pi}{6} \frac{1}{2} - \frac{\sqrt 3}{2} +1 \right]\\
&= \frac{\sqrt 3-1}{2} -\frac{\pi}{12} \\
\end{aligned}
