1. **Problem statement:** (i) Find $$\int \sin^2 3x \, dx$$. (ii) Find $$\int x (x^2 + 4)^{3/2} \, dx$$. 2. **Formulas and rules:** - Use the power-reduction identity for sine: $$\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$$. - For substitution, use $$u = x^2 + 4$$ for the second integral. 3. **Solution for (i):** $$\int \sin^2 3x \, dx = \int \frac{1 - \cos 6x}{2} \, dx = \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos 6x \, dx$$ 4. Integrate each term: $$\frac{1}{2} x - \frac{1}{2} \cdot \frac{\sin 6x}{6} + C = \frac{x}{2} - \frac{\sin 6x}{12} + C$$ 5. **Solution for (ii):** Let $$u = x^2 + 4$$, then $$du = 2x \, dx$$ or $$x \, dx = \frac{du}{2}$$. 6. Rewrite the integral: $$\int x (x^2 + 4)^{3/2} \, dx = \int (u)^{3/2} \cdot \frac{du}{2} = \frac{1}{2} \int u^{3/2} \, du$$ 7. Integrate: $$\frac{1}{2} \cdot \frac{u^{5/2}}{\frac{5}{2}} + C = \frac{1}{2} \cdot \frac{2}{5} u^{5/2} + C = \frac{1}{5} (x^2 + 4)^{5/2} + C$$ **Final answers:** (i) $$\int \sin^2 3x \, dx = \frac{x}{2} - \frac{\sin 6x}{12} + C$$ (ii) $$\int x (x^2 + 4)^{3/2} \, dx = \frac{1}{5} (x^2 + 4)^{5/2} + C$$