The given integral is:

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

Step 1: Use Trigonometric Substitution

We recognize the expression $\sqrt{16 - x^2}$, which suggests the substitution:

$x = 4\sin\theta$

Then, we differentiate:

$dx = 4\cos\theta \, d\theta$

Substituting into the square root term:

$\sqrt{16 - x^2} = \sqrt{16 - 16\sin^2\theta} = \sqrt{16(1 - \sin^2\theta)} = \sqrt{16\cos^2\theta} = 4\cos\theta.$

Thus, our integral transforms into:

$I = \int \frac{4\cos\theta}{4\sin\theta} \cdot 4\cos\theta \, d\theta$

$I = \int \frac{16\cos^2\theta}{4\sin\theta} \, d\theta = \int \frac{4\cos^2\theta}{\sin\theta} \, d\theta$

$I = 4 \int \frac{\cos^2\theta}{\sin\theta} \, d\theta.$

Step 2: Express in Terms of $u$

Let $u = \sin\theta$, so that $du = \cos\theta \, d\theta$.

Since $\cos^2\theta = 1 - \sin^2\theta = 1 - u^2$, we rewrite:

$I = 4 \int \frac{(1 - u^2)}{u} \, du$

$I = 4 \int \left( \frac{1}{u} - u \right) du$

Step 3: Compute the Integral

$I = 4 \left( \ln |u| - \frac{u^2}{2} \right) + C$

$I = 4 \left( \ln |\sin\theta| - \frac{\sin^2\theta}{2} \right) + C.$

Step 4: Convert Back to $x$

Since $\sin\theta = \frac{x}{4}$, we substitute:

$I = 4 \left( \ln \left| \frac{x}{4} \right| - \frac{1}{2} \left( \frac{x}{4} \right)^2 \right) + C.$

$I = 4 \ln |x| - 4 \ln 4 - \frac{x^2}{2 \cdot 16} + C.$

$I = 4 \ln |x| - 4 \ln 4 - \frac{x^2}{32} + C.$

Thus, the final result is:

$I = 4 \ln |x| - 4 \ln 4 - \frac{x^2}{32} + C.$

Questions for Further Understanding:

1. Why is trigonometric substitution useful in solving this integral?
2. How can we generalize this method for similar integrals?
3. What happens if we use a different trigonometric identity for substitution?
4. How does this integral relate to arc lengths or areas in geometry?
5. Can we solve this integral using a different method, such as hyperbolic substitution?

Tip:

Trigonometric substitutions are particularly useful when dealing with square roots of quadratic expressions, making integrals more manageable.