Evaluate $\int_{0}^{\frac{\pi}{3}} \left( \frac{1}{2} \sin 3x - \cos^2 x \right) dx$, correct to 2 decimal places.

1. Use the double angle identity to rewrite $\cos^2 x$ in an integrable form.

$\cos^2 x = \frac{1}{2}(\cos 2x + 1)$

$\int_{0}^{\frac{\pi}{3}} \left( \frac{1}{2} \sin 3x - \cos^2 x \right) dx = \int_{0}^{\frac{\pi}{3}} \left( \frac{1}{2} \sin 3x - \frac{1}{2}\cos 2x - \frac{1}{2} \right) dx$

2. Integrate the expression with respect to $x$.

$= \left[ -\frac{1}{6} \cos 3x - \frac{1}{4} \sin 2x - \frac{1}{2} x \right]_0^{\frac{\pi}{3}}$

3. Substitute the upper and lower limits to evaluate the definite integral.

Substitute $x = \frac{\pi}{3}$:
$\left( -\frac{1}{6} \cos \pi - \frac{1}{4} \sin \frac{2\pi}{3} - \frac{1}{2} \left(\frac{\pi}{3}\right) \right) = \frac{1}{6} - \frac{\sqrt{3}}{8} - \frac{\pi}{6}$

Substitute $x = 0$:
$\left( -\frac{1}{6} \cos 0 - \frac{1}{4} \sin 0 - 0 \right) = -\frac{1}{6}$

Evaluate the difference:
$\left( \frac{1}{6} - \frac{\sqrt{3}}{8} - \frac{\pi}{6} \right) - \left( -\frac{1}{6} \right) = \frac{1}{3} - \frac{\sqrt{3}}{8} - \frac{\pi}{6}$

4. Calculate the final numerical value correct to 2 decimal places.

$\frac{1}{3} - \frac{\sqrt{3}}{8} - \frac{\pi}{6} \approx -0.40677$

$\approx -0.41$ (to 2 d.p.)