The integral you're looking to solve is:

$\int \sqrt{16 - x^2} \, dx$

This is a standard integral that can be solved using a trigonometric substitution. Here's the step-by-step approach:

Step 1: Use Trigonometric Substitution

We will use the substitution $x = 4 \sin(\theta)$, since $\sqrt{16 - x^2}$ resembles the identity $\sqrt{1 - \sin^2(\theta)} = \cos(\theta)$.

Thus, let: $x = 4 \sin(\theta)$ Then, $dx = 4 \cos(\theta) \, d\theta$

Step 2: Simplify the Integrand

Substitute $x = 4 \sin(\theta)$ into the integrand:

$\sqrt{16 - x^2} = \sqrt{16 - 16\sin^2(\theta)} = 4\cos(\theta)$

So the integral becomes:

$\int \sqrt{16 - x^2} \, dx = \int 4\cos(\theta) \cdot 4\cos(\theta) \, d\theta = \int 16\cos^2(\theta) \, d\theta$

Step 3: Use a Trigonometric Identity

We can simplify $\cos^2(\theta)$ using the half-angle identity:

$\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$

So, the integral becomes:

$\int 16\cos^2(\theta) \, d\theta = \int 16 \cdot \frac{1 + \cos(2\theta)}{2} \, d\theta = 8 \int (1 + \cos(2\theta)) \, d\theta$

Step 4: Integrate

Now, integrate:

$8 \int (1 + \cos(2\theta)) \, d\theta = 8 \left[ \theta + \frac{\sin(2\theta)}{2} \right] + C$

Step 5: Reverse the Substitution

Recall that $x = 4 \sin(\theta)$, so $\sin(\theta) = \frac{x}{4}$. Now, solve for $\theta$ using $\theta = \arcsin\left(\frac{x}{4}\right)$.

Also, $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$, and from the Pythagorean identity $\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{x}{4}\right)^2} = \frac{\sqrt{16 - x^2}}{4}$.

Thus, $\sin(2\theta) = 2 \cdot \frac{x}{4} \cdot \frac{\sqrt{16 - x^2}}{4} = \frac{x \sqrt{16 - x^2}}{8}$.

Now substitute these into the integral result:

[ 8 \left[ \theta + \frac{\sin(2\theta)}{2} \right] = 8 \left[ \arcsin\left( \frac{x}{4} \right) + \frac{x \sqrt{16 - x^2}}{16} \right] + C ]

Final Answer:

$\int \sqrt{16 - x^2} \, dx = 8 \arcsin\left( \frac{x}{4} \right) + \frac{x \sqrt{16 - x^2}}{2} + C$

Would you like further details or have any questions about this solution?

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Related Questions:

1. How does the trigonometric substitution method help simplify integrals involving square roots of quadratic expressions?
2. Can you derive the integral of $\sqrt{a^2 - x^2}$ using a similar substitution method?
3. What is the significance of the half-angle identity in reducing trigonometric integrals?
4. How do you handle definite integrals with square roots of quadratic functions using trigonometric substitution?
5. Can the result be applied to other integrals, like $\int \sqrt{9 - x^2} \, dx$?

Tip: Always remember that recognizing a standard form like $\sqrt{a^2 - x^2}$ can greatly simplify the process of integration. Trigonometric substitution is a powerful tool for this!