1. We are asked to evaluate the integral $$\int \frac{\cos(3x)}{8 + \sin^2(3x)} \, dx$$ 2. The problem involves a trigonometric integral with a quadratic expression in the denominator. 3. Let us use the substitution method. Set $$u = \sin(3x)$$ Then, $$\frac{du}{dx} = 3 \cos(3x) \implies du = 3 \cos(3x) \, dx$$ 4. Rearranging for $\cos(3x) \, dx$: $$\cos(3x) \, dx = \frac{du}{3}$$ 5. Substitute into the integral: $$\int \frac{\cos(3x)}{8 + \sin^2(3x)} \, dx = \int \frac{1}{8 + u^2} \cdot \frac{du}{3} = \frac{1}{3} \int \frac{1}{8 + u^2} \, du$$ 6. Recall the integral formula: $$\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$$ Here, $a^2 = 8$, so $a = \sqrt{8} = 2\sqrt{2}$. 7. Applying the formula: $$\frac{1}{3} \int \frac{1}{8 + u^2} \, du = \frac{1}{3} \cdot \frac{1}{2\sqrt{2}} \arctan\left(\frac{u}{2\sqrt{2}}\right) + C = \frac{1}{6\sqrt{2}} \arctan\left(\frac{\sin(3x)}{2\sqrt{2}}\right) + C$$ 8. Therefore, the final answer is: $$\boxed{\frac{1}{6\sqrt{2}} \arctan\left(\frac{\sin(3x)}{2\sqrt{2}}\right) + C}$$