Answer
$$\int_{0}^{2\pi}sin^{2}\frac{\theta}{3}d\theta=\pi+\frac{3\sqrt{3}}{8}$$
Work Step by Step
$$\int_{0}^{2\pi}sin^{2}\frac{\theta}{3}d\theta=\frac{1}{2}\int_{0}^{2\pi}(1-cos\frac{2\theta}{3})d\theta$$$$=\frac{1}{2}\left [ \theta -\frac{3}{2}sin\frac{2\theta }{3}\right ]_{0}^{2\pi}=\pi+\frac{3\sqrt{3}}{8}$$