1. **State the problem:** Evaluate the integral $$\int 6 \cos \sqrt{x} \, dx$$. 2. **Substitution:** Let $$t = \sqrt{x} = x^{1/2}$$, then $$x = t^2$$. 3. **Find differential:** $$dx = 2t \, dt$$. 4. **Rewrite the integral:** $$\int 6 \cos \sqrt{x} \, dx = \int 6 \cos t \cdot 2t \, dt = \int 12 t \cos t \, dt$$. 5. **Integration by parts:** Use formula $$\int u \, dv = uv - \int v \, du$$. Choose $$u = t$$ so $$du = dt$$, and $$dv = \cos t \, dt$$ so $$v = \sin t$$. 6. **Apply integration by parts:** $$\int 12 t \cos t \, dt = 12 \int t \cos t \, dt = 12 (t \sin t - \int \sin t \, dt)$$. 7. **Integrate $$\int \sin t \, dt$$:** $$\int \sin t \, dt = -\cos t + C$$. 8. **Substitute back:** $$12 (t \sin t + \cos t) + C$$. 9. **Replace $$t$$ with $$\sqrt{x}$$:** $$\boxed{12 \sqrt{x} \sin \sqrt{x} + 12 \cos \sqrt{x} + C}$$. This is the evaluated integral.