(i)[4]
Prove that $\frac{2 \sin x - \sin 2x}{1 - \cos 2x} = \frac{\sin x}{1 + \cos x}$.
(ii)[4]
Hence, with full working shown, evaluate $\int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \frac{2 \sin x - \sin 2x}{1 - \cos 2x}\, dx$, and present your answer in the form $\ln k$.