1. **State the problem:** Evaluate the integral $$\int_0^2 \frac{dx}{\sqrt{16 - x^2}}.$$\n\n2. **Choose substitution:** We recognize the integrand involves $$\sqrt{a^2 - x^2}$$ with $$a=4$$. A common substitution is $$x = 4 \sin \theta$$ because $$\sin^2 \theta + \cos^2 \theta = 1$$ helps simplify the square root.\n\n3. **Find dx:** Differentiating, $$dx = 4 \cos \theta \, d\theta.$$\n\n4. **Rewrite the integral:** Substitute $$x = 4 \sin \theta$$ and $$dx = 4 \cos \theta \, d\theta$$ into the integral. The limits change as follows: when $$x=0$$, $$\sin \theta = 0 \Rightarrow \theta=0$$; when $$x=2$$, $$2 = 4 \sin \theta \Rightarrow \sin \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6}.$$\n\nThe integral becomes:\n$$\int_0^{\pi/6} \frac{4 \cos \theta \, d\theta}{\sqrt{16 - (4 \sin \theta)^2}} = \int_0^{\pi/6} \frac{4 \cos \theta \, d\theta}{\sqrt{16 - 16 \sin^2 \theta}}.$$\n\n5. **Simplify the square root:**\n$$\sqrt{16 - 16 \sin^2 \theta} = \sqrt{16 (1 - \sin^2 \theta)} = \sqrt{16 \cos^2 \theta} = 4 \cos \theta.$$\n\n6. **Simplify the integrand:**\n$$\frac{4 \cos \theta}{4 \cos \theta} = 1.$$\n\n7. **Evaluate the integral:**\n$$\int_0^{\pi/6} 1 \, d\theta = \left. \theta \right|_0^{\pi/6} = \frac{\pi}{6} - 0 = \frac{\pi}{6}.$$\n\n**Final answer:** $$\boxed{\frac{\pi}{6}}.$$