\begin{aligned} \sin A \cdot \sin B = \frac{1}{2}\sin(A+B)+\frac{1}{2}\sin(A-B) \end{aligned}

\begin{aligned} \int_0^{ \frac{\pi}{6} } \sin \left( x + \frac{\pi}{3} \right) \cos \left( x + \frac{\pi}{3} \right) \ dx &= \int_0^{30^\circ} \sin(x+60^\circ) \cos(x+60^\circ) \ dx \\[8pt] &= \int_0^{30^\circ} \left( \frac{1}{2} \sin(2x+120^\circ) + \frac{1}{2} \sin (0) \right) \ dx \\[8pt] &= \int_0^{30^\circ} \frac{1}{2} \sin(2x+120^\circ) \ dx = \frac{1}{2} \int_0^{30^\circ} \sin(2x+120^\circ) \ dx \\[8pt] &= \frac{1}{2} \left[ -\frac{1}{2}\cos(2x+120^\circ) \right]_0^{30^\circ} \\[8pt] &= \frac{1}{2} \left[ -\frac{1}{2}\cos(60^\circ+120^\circ) + \frac{1}{2} \cos(0^\circ+120^\circ) \right] \\[8pt] &= \frac{1}{2} \left[ -\frac{1}{2} \cos (180^\circ) + \frac{1}{2} \cos(120^\circ) \right] \\[8pt] &= \frac{1}{2} \left[ -\frac{1}{2}(-1)+\frac{1}{2} \left( -\frac{1}{2} \right) \right] \\[8pt] &= \frac{1}{2} \left[ \frac{1}{2}-\frac{1}{4} \right] \\[8pt] &= \frac{1}{8} \end{aligned}