Answer
$\displaystyle\int\limits_\frac{\pi}{6}^\pi \sin\theta d\theta=\dfrac{2+\sqrt{3}}{2}$
Work Step by Step
$\displaystyle\int\limits_\frac{\pi}{6}^\pi \sin\theta d\theta$
Evaluate the integral and apply the second part of the fundamental theorem of calculus:
$\displaystyle\int\limits_\frac{\pi}{6}^\pi \sin\theta d\theta=-\cos\theta\Big|_\frac{\pi}{6}^\pi=-(\cos\pi-\cos\dfrac{\pi}{6})=-(-1-\dfrac{\sqrt{3}}{2})=...$
$...=\dfrac{\sqrt{3}}{2}+1=\dfrac{2+\sqrt{3}}{2}$