Chapter 7 - Techniques of Integration - 7.2 Trigonometric Integrals - 7.2 Exercises - Page 524: 8

$$\int_{0}^{2\pi}sin^{2}(\frac{1}{3}\theta)d\theta=\pi+\frac{3\sqrt{3}}{8}$$

$$\int_{0}^{2\pi}sin^{2}(\frac{1}{3}\theta)d\theta=\int_{0}^{2\pi}\frac{1-cos(\frac{2}{3}\theta)}{2}d\theta$$ $$=\left |\frac{\theta}{2}-\frac{3}{4}sin(\frac{2}{3}\theta) \right |_{0}^{2\pi}$$ $$=\pi-\frac{3}{4}(-\frac{\sqrt{3}}{2})$$ $$=\pi+\frac{3\sqrt{3}}{8}$$