Use sin² 3x = (1 - cos 6x)/2 to integrate sin² 3x…

Step 1. Rewrite. $\sin^2 3x = \tfrac{1}{2} - \tfrac{1}{2}\cos 6x$.

Step 2. Integrate. $\displaystyle\int \tfrac{1}{2}\,dx = \dfrac{x}{2}$. Next, $\displaystyle\int -\tfrac{1}{2}\cos 6x\,dx = -\tfrac{1}{2} \times \dfrac{\sin 6x}{6} = -\dfrac{\sin 6x}{12}$.

Step 3. Read off. $a = \dfrac{1}{2} = 0.5$ and $b = -\dfrac{1}{12} = -0.0833$.