Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 14

$\frac{4\pi -3\sqrt{3}}{12}$

\begin{aligned} \int_{-\pi / 3}^{\pi / 3} \sin ^{2} t d t&=\frac{1}{2}\int_{-\pi / 3}^{\pi / 3} (1-\cos(2 t) )d t\\ &= \frac{1}{2}\left(t-\frac{\sin (2t)}{2}\right)_{-\pi/3}^{\pi/3}\\ &= \frac{1}{2}\left(\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\right)\\ &=\frac{4\pi -3\sqrt{3}}{12}\end{aligned}