1. **State the problem:** We want to find the integral $$\int \sqrt{\cos x (3 + 2 \cos x)} \, dx$$. 2. **Rewrite the integrand:** The expression inside the square root is $$\cos x (3 + 2 \cos x) = 3 \cos x + 2 \cos^2 x$$. 3. **Use substitution:** Let $$u = \cos x$$, then $$du = -\sin x \, dx$$. However, the integral does not contain a clear $$\sin x$$ term to substitute directly, so we try to express the integral in terms of $$u$$ and $$dx$$. 4. **Express dx in terms of du:** Since $$du = -\sin x \, dx$$, we have $$dx = \frac{-du}{\sin x}$$. 5. **Rewrite the integral:** $$\int \sqrt{u (3 + 2u)} \, dx = \int \sqrt{u (3 + 2u)} \cdot \frac{-du}{\sin x}$$. 6. **Express $$\sin x$$ in terms of $$u$$:** Since $$u = \cos x$$, $$\sin x = \sqrt{1 - u^2}$$. 7. **Substitute:** $$\int \sqrt{u (3 + 2u)} \cdot \frac{-du}{\sqrt{1 - u^2}} = - \int \frac{\sqrt{u (3 + 2u)}}{\sqrt{1 - u^2}} \, du$$. 8. **Simplify the integrand:** $$\frac{\sqrt{u (3 + 2u)}}{\sqrt{1 - u^2}} = \sqrt{\frac{u (3 + 2u)}{1 - u^2}}$$. 9. **Final integral:** $$- \int \sqrt{\frac{u (3 + 2u)}{1 - u^2}} \, du$$. This integral is non-elementary and requires advanced techniques or numerical methods. **Answer:** The integral $$\int \sqrt{\cos x (3 + 2 \cos x)} \, dx$$ can be expressed as $$- \int \sqrt{\frac{u (3 + 2u)}{1 - u^2}} \, du$$ with $$u = \cos x$$, which is not elementary.