1. **State the problem:** We want to evaluate the integral $$\int \sec^3\left(\frac{1}{2}x + 3\right) \, dx.$$\n\n2. **Recall the formula and approach:** The integral of $$\sec^3 u \, du$$ is a standard integral. We use the reduction formula or integration by parts: $$\int \sec^3 u \, du = \frac{1}{2} \sec u \tan u + \frac{1}{2} \ln|\sec u + \tan u| + C.$$\n\n3. **Substitution:** Let $$u = \frac{1}{2}x + 3,$$ then $$du = \frac{1}{2} dx \implies dx = 2 du.$$\n\n4. **Rewrite the integral:** $$\int \sec^3\left(\frac{1}{2}x + 3\right) dx = \int \sec^3 u \cdot 2 \, du = 2 \int \sec^3 u \, du.$$\n\n5. **Apply the formula:** $$2 \int \sec^3 u \, du = 2 \left( \frac{1}{2} \sec u \tan u + \frac{1}{2} \ln|\sec u + \tan u| \right) + C = \sec u \tan u + \ln|\sec u + \tan u| + C.$$\n\n6. **Back-substitute:** Replace $$u$$ with $$\frac{1}{2}x + 3$$ to get the final answer: $$\sec\left(\frac{1}{2}x + 3\right) \tan\left(\frac{1}{2}x + 3\right) + \ln\left|\sec\left(\frac{1}{2}x + 3\right) + \tan\left(\frac{1}{2}x + 3\right)\right| + C.$$