Calculus: Early Transcendentals 8th Edition

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