Exact constants from a trigonometric integral

1. Show that \(\dfrac{\mathrm{d}}{\mathrm{d}x}(\sin3x-3x)=-6\sin^2\!\left(\dfrac{3x}{2}\right)\).[3]
2. Without using a calculator, find the values of the constants \(a\), \(b\) and \(c\) for which \(\displaystyle\int_0^{\pi/9}\left[\sin^2\!\left(\dfrac{3x}{2}\right)-\sin3x\right]\,\mathrm{d}x=a\sqrt3+b\pi+c\).[5]

The coefficients in the requested exact form are rational numbers.