1. **State the problem:** Evaluate the integral $$\int \sin(5x) \cdot \cos(2x) \, dx.$$\n\n2. **Use product-to-sum formula:** Recall the identity $$\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)].$$\n\n3. **Apply the formula:** Here, $A=5x$ and $B=2x$, so\n$$\sin(5x) \cos(2x) = \frac{1}{2} [\sin(5x+2x) + \sin(5x-2x)] = \frac{1}{2} [\sin(7x) + \sin(3x)].$$\n\n4. **Rewrite the integral:**\n$$\int \sin(5x) \cos(2x) \, dx = \int \frac{1}{2} [\sin(7x) + \sin(3x)] \, dx = \frac{1}{2} \int \sin(7x) \, dx + \frac{1}{2} \int \sin(3x) \, dx.$$\n\n5. **Integrate each term:**\n- $$\int \sin(7x) \, dx = -\frac{1}{7} \cos(7x) + C_1,$$\n- $$\int \sin(3x) \, dx = -\frac{1}{3} \cos(3x) + C_2.$$\n\n6. **Combine results:**\n$$\frac{1}{2} \left(-\frac{1}{7} \cos(7x) - \frac{1}{3} \cos(3x) \right) + C = -\frac{1}{14} \cos(7x) - \frac{1}{6} \cos(3x) + C,$$\nwhere $C$ is the constant of integration.\n\n**Final answer:**\n$$\int \sin(5x) \cos(2x) \, dx = -\frac{1}{14} \cos(7x) - \frac{1}{6} \cos(3x) + C.$$