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.