## Thomas' Calculus 13th Edition

$$\frac{\pi }{3} - \frac{{\sqrt 3 }}{4}$$
\eqalign{ & \int_{ - \pi /3}^{\pi /3} {{{\sin }^2}t} dt \cr & {\text{use identity si}}{{\text{n}}^2}t = \frac{{1 - \cos 2t}}{2} \cr & = \int_{ - \pi /3}^{\pi /3} {\left( {\frac{{1 - \cos 2t}}{2}} \right)} dt \cr & = \frac{1}{2}\int_{ - \pi /3}^{\pi /3} {\left( {1 - \cos 2t} \right)} dt \cr & {\text{solve the integral}} \cr & = \frac{1}{2}\left( {t - \frac{1}{2}\sin 2t} \right)_{ - \pi /3}^{\pi /3} \cr & = \left( {\frac{1}{2}t - \frac{1}{4}\sin 2t} \right)_{ - \pi /3}^{\pi /3} \cr & {\text{evaluating, we get:}} \cr & = \left( {\frac{1}{2}\left( {\frac{\pi }{3}} \right) - \frac{1}{4}\sin 2\left( {\frac{\pi }{3}} \right)} \right) - \left( {\frac{1}{2}\left( { - \frac{\pi }{3}} \right) - \frac{1}{4}\sin 2\left( { - \frac{\pi }{3}} \right)} \right) \cr & = \left( {\frac{\pi }{6} - \frac{1}{4}\left( {\frac{{\sqrt 3 }}{2}} \right)} \right) - \left( { - \frac{\pi }{6} - \frac{1}{4}\left( { - \frac{{\sqrt 3 }}{2}} \right)} \right) \cr & = \frac{\pi }{6} - \frac{{\sqrt 3 }}{8} + \frac{\pi }{6} - \frac{{\sqrt 3 }}{8} \cr & = \frac{\pi }{3} - \frac{{\sqrt 3 }}{4} \cr}