STEP II 1992 question 3 solution

y = OP\sin\phi + PB\sin (\phi +\theta ) = 3\sin\phi + \sin (-3\phi ) = 3\sin\phi - \sin 3\phi \\
\therefore (x, y) = (3\cos\phi + \cos 3\phi , 3\sin\phi - \sin 3\phi ).\\
A = \int_0^{2\pi} x \frac{\text{d}y}{\text{d}\phi} \text{d}\phi = \int_0^{2\pi} (3\cos\phi + \cos 3\phi )(3\cos\phi + 3\cos 3\phi ) \text{d}\phi \\
= \int_0^{2\pi} (9\cos^2 \phi - 6\cos\phi\cos 3\phi - 3\cos^2 3\phi) \text{d}\phi \\
= \frac{1}{2} \int_0^{2\pi} (9\cos 2\phi + 9 - 6\cos 2\phi - 6\cos 4\phi - 3\cos 6\phi - 3) \text{d}\phi \\
= \frac{1}{2} \left[6\phi + \tfrac{9}{2} \sin 2\phi - 3\sin 2\phi - \tfrac{3}{4} \sin 4\phi - \tfrac{1}{2} \sin 6\phi \right]_0^{2\pi} \\