$$ \int_8^{16}\!x\sqrt{16x-x^2}\,\mathrm{d}x $$

To solve the integral $ \int_8^{16} x \sqrt{16x - x^2} \, dx $, we'll proceed by simplifying the integrand using a trigonometric substitution. Here's the detailed process:

### Step 1: Simplifying the Expression
First, recognize that the expression $ 16x - x^2 $ can be rewritten as:
\[
16x - x^2 = - (x^2 - 16x) = -(x - 8)^2 + 64.
\]
Thus, we have:
\[
\sqrt{16x - x^2} = \sqrt{64 - (x - 8)^2}.
\]
This is the equation of a semicircle with radius 8, centered at $ x = 8 $. Therefore, trigonometric substitution is a useful approach.

### Step 2: Substitution
Let’s make the substitution:
\[
x - 8 = 8 \sin \theta \quad \text{so that} \quad dx = 8 \cos \theta \, d\theta.
\]
Thus, $ x = 8(1 + \sin \theta) $, and the limits of integration need to be adjusted. When $ x = 8 $, $ \theta = 0 $; and when $ x = 16 $, $ \theta = \frac{\pi}{2} $.

### Step 3: Substituting into the Integral
The square root term becomes:
\[
\sqrt{64 - (x - 8)^2} = \sqrt{64 - 64\sin^2\theta} = 8\cos\theta.
\]
Substitute into the original integral:
\[
\int_8^{16} x \sqrt{16x - x^2} \, dx = \int_0^{\frac{\pi}{2}} 8(1 + \sin\theta) \cdot 8 \cos\theta \cdot 8 \cos\theta \, d\theta.
\]
Simplifying this expression gives:
\[
512 \int_0^{\frac{\pi}{2}} (1 + \sin\theta) \cos^2\theta \, d\theta.
\]

### Step 4: Solving the Integral
Expand the integrand:
\[
512 \int_0^{\frac{\pi}{2}} (\cos^2\theta + \sin\theta \cos^2\theta) \, d\theta.
\]
Now solve these two integrals separately.

#### Integral 1: $ \int_0^{\frac{\pi}{2}} \cos^2\theta \, d\theta $
Use the identity $ \cos^2\theta = \frac{1 + \cos 2\theta}{2} $:
\[
\int_0^{\frac{\pi}{2}} \cos^2\theta \, d\theta = \frac{1}{2} \int_0^{\frac{\pi}{2}} (1 + \cos 2\theta) \, d\theta = \frac{1}{2} \left[ \frac{\pi}{2} + 0 \right] = \frac{\pi}{4}.
\]

#### Integral 2: $ \int_0^{\frac{\pi}{2}} \sin\theta \cos^2\theta \, d\theta $
Use the substitution $ u = \cos\theta $, $ du = -\sin\theta \, d\theta $, so the integral becomes:
\[
\int_0^{\frac{\pi}{2}} \sin\theta \cos^2\theta \, d\theta = \int_1^0 -u^2 \, du = \int_0^1 u^2 \, du = \frac{1}{3}.
\]

### Step 5: Final Answer
Now, combine the results:
\[
512 \left( \frac{\pi}{4} + \frac{1}{3} \right) = 512 \left( \frac{3\pi + 4}{12} \right) = \frac{512(3\pi + 4)}{12} = \frac{128(3\pi + 4)}{3}.
\]

Thus, the value of the integral is:
\[
\boxed{\frac{128(3\pi + 4)}{3}}.
\]

---

Would you like further details on any of the steps, or have any questions?

Here are some related questions:
1. How does trigonometric substitution help in evaluating integrals?
2. What are the common forms of integrals involving square roots of quadratic expressions?
3. What is the geometric interpretation of integrals involving circles or semicircles?
4. How can the cosine double-angle identity simplify integrals?
5. What other types of integrals can be solved using trigonometric substitution?

**Tip:** Trigonometric substitution is particularly useful when dealing with integrals involving $ \sqrt{a^2 - x^2} $, $ \sqrt{a^2 + x^2} $, or $ \sqrt{x^2 - a^2} $.

Learn how to solve the integral $$ \int_8^{16} x \sqrt{16x - x^2} \, dx $$ using trigonometric substitution. This problem involves concepts like trigonometric identities, integration techniques, and the geometry of circles. The detailed solution simplifies the integral into a solvable form using substitution, resulting in $$ \frac{128(3\pi + 4)}{3} $$.

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