In this question you must show detailed reasoning.
\includegraphics{figure_9}
The diagram shows the curve \(y = \frac{4\cos 2x}{3 - \sin 2x}\) for \(x > 0\), and the normal to the curve at the point \((\frac{1}{4}\pi, 0)\).
Show that the exact area of the shaded region enclosed by the curve, the normal to the curve and the \(y\)-axis is \(\ln \frac{2}{3} + \frac{1}{128}\pi^2\). [10]