(a)[3]
Express $3\cos 2x - \sqrt{3}\sin 2x$ in the form $R\cos(2x + \alpha)$, where $R > 0$ and $0 < \alpha < \frac{1}{2}\pi$. Give the exact values of $R$ and $\alpha$.
(b)[5]
Hence find the exact value of $\displaystyle \int_{0}^{\tfrac{1}{2}\pi} \frac{3}{\left(3\cos 2x - \sqrt{3}\sin 2x\right)^2}\,dx$, and simplify your answer.