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The integral you have provided is:

\[
\int_0^2 \sqrt{16 - x^2} \, dx.
\]

This is a standard integral that represents part of the area of a circle. We can solve it using trigonometric substitution, specifically recognizing that \( \sqrt{16 - x^2} \) resembles the equation for the upper half of a circle with radius 4. Here's the step-by-step solution.

### Step 1: Trigonometric Substitution
Let:
\[
x = 4 \sin\theta.
\]
Then:
\[
dx = 4 \cos\theta \, d\theta.
\]

Now, substituting into the integral, we get:
\[
\sqrt{16 - x^2} = \sqrt{16 - 16\sin^2\theta} = 4\cos\theta.
\]

The limits of integration will change accordingly:
- When \(x = 0\), \( \sin\theta = 0 \), so \( \theta = 0 \).
- When \(x = 2\), \( \sin\theta = \frac{1}{2} \), so \( \theta = \frac{\pi}{6} \).

Thus, the integral becomes:
\[
\int_0^{\frac{\pi}{6}} 4\cos^2\theta \cdot 4\cos\theta \, d\theta = 16 \int_0^{\frac{\pi}{6}} \cos^2\theta \, d\theta.
\]

### Step 2: Simplify the Integral
Use the half-angle identity for \( \cos^2\theta \):
\[
\cos^2\theta = \frac{1 + \cos(2\theta)}{2}.
\]

Now, substitute this into the integral:
\[
16 \int_0^{\frac{\pi}{6}} \frac{1 + \cos(2\theta)}{2} \, d\theta = 8 \int_0^{\frac{\pi}{6}} (1 + \cos(2\theta)) \, d\theta.
\]

### Step 3: Solve the Integral
Now, integrate each term:
\[
8 \left[ \theta + \frac{\sin(2\theta)}{2} \right]_0^{\frac{\pi}{6}}.
\]

Evaluate this expression:
\[
8 \left[ \frac{\pi}{6} + \frac{\sin\left(\frac{\pi}{3}\right)}{2} - 0 \right] = 8 \left[ \frac{\pi}{6} + \frac{\sqrt{3}}{4} \right].
\]

Thus, the final result is:
\[
8 \left( \frac{\pi}{6} + \frac{\sqrt{3}}{4} \right) = \frac{4\pi}{3} + 2\sqrt{3}.
\]

### Final Answer:
The value of the integral is:
\[
\boxed{\frac{4\pi}{3} + 2\sqrt{3}}.
\]

Would you like more details or further clarification?

Here are 5 related questions to deepen your understanding:
1. How is trigonometric substitution chosen for integrals involving radicals like \( \sqrt{a^2 - x^2} \)?
2. Why is \( x = 4\sin\theta \) a suitable substitution for this integral?
3. Can this integral be interpreted geometrically in terms of the area of a circle?
4. What is the significance of using the half-angle identity in this type of integral?
5. How would the integral change if the limits were different?

**Tip:** When you encounter integrals with square roots of quadratic expressions, trigonometric substitution is often an effective method.

Evaluate the definite integral of the square root of 16 minus x squared, from 0 to 2.

Learn how to evaluate the integral ∫(√(16 - x²) dx) from 0 to 2 using trigonometric substitution. This problem covers key calculus concepts such as trigonometric identities and the geometry of circles. The solution is suitable for undergraduate calculus students.

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