Question: (b) Use an appropriate trigonometric substitution to evaluate $$\int \frac{\sqrt{x^2 + 16}}{x^4} \, dx$$.
1. **State the problem:** We want to evaluate the integral $$\int \frac{\sqrt{x^2 + 16}}{x^4} \, dx$$ using a trigonometric substitution.
2. **Choose the substitution:** Since the integrand contains $$\sqrt{x^2 + 16}$$, which is of the form $$\sqrt{x^2 + a^2}$$ with $$a=4$$, we use the substitution $$x = 4 \tan \theta$$.
3. **Compute differentials and expressions:**
- $$dx = 4 \sec^2 \theta \, d\theta$$
- $$\sqrt{x^2 + 16} = \sqrt{16 \tan^2 \theta + 16} = \sqrt{16(\tan^2 \theta + 1)} = 4 \sec \theta$$
- $$x^4 = (4 \tan \theta)^4 = 4^4 \tan^4 \theta = 256 \tan^4 \theta$$
4. **Rewrite the integral in terms of $$\theta$$:**
$$\int \frac{\sqrt{x^2 + 16}}{x^4} \, dx = \int \frac{4 \sec \theta}{256 \tan^4 \theta} \cdot 4 \sec^2 \theta \, d\theta = \int \frac{16 \sec^3 \theta}{256 \tan^4 \theta} \, d\theta = \int \frac{\sec^3 \theta}{16 \tan^4 \theta} \, d\theta$$
5. **Simplify the integral:**
$$\int \frac{\sec^3 \theta}{16 \tan^4 \theta} \, d\theta = \frac{1}{16} \int \frac{\sec^3 \theta}{\tan^4 \theta} \, d\theta$$
6. **Express in sine and cosine:**
Recall $$\sec \theta = \frac{1}{\cos \theta}$$ and $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$, so
$$\frac{\sec^3 \theta}{\tan^4 \theta} = \frac{\frac{1}{\cos^3 \theta}}{\left(\frac{\sin \theta}{\cos \theta}\right)^4} = \frac{1}{\cos^3 \theta} \cdot \frac{\cos^4 \theta}{\sin^4 \theta} = \frac{\cos \theta}{\sin^4 \theta}$$
7. **Rewrite the integral:**
$$\frac{1}{16} \int \frac{\cos \theta}{\sin^4 \theta} \, d\theta = \frac{1}{16} \int \cos \theta \sin^{-4} \theta \, d\theta$$
8. **Use substitution:** Let $$u = \sin \theta$$, then $$du = \cos \theta \, d\theta$$.
9. **Integral in terms of $$u$$:**
$$\frac{1}{16} \int u^{-4} \, du = \frac{1}{16} \int u^{-4} \, du$$
10. **Integrate:**
$$\int u^{-4} \, du = \int u^{-4} \, du = \frac{u^{-3}}{-3} + C = -\frac{1}{3 u^3} + C$$
11. **Substitute back:**
$$= \frac{1}{16} \left(-\frac{1}{3 \sin^3 \theta}\right) + C = -\frac{1}{48 \sin^3 \theta} + C$$
12. **Rewrite $$\sin \theta$$ in terms of $$x$$:**
From $$x = 4 \tan \theta$$, we have $$\tan \theta = \frac{x}{4}$$.
Using the identity $$\sin \theta = \frac{\tan \theta}{\sec \theta}$$ and $$\sec \theta = \sqrt{1 + \tan^2 \theta}$$,
$$\sin \theta = \frac{\frac{x}{4}}{\sqrt{1 + \left(\frac{x}{4}\right)^2}} = \frac{\frac{x}{4}}{\sqrt{1 + \frac{x^2}{16}}} = \frac{\frac{x}{4}}{\frac{\sqrt{x^2 + 16}}{4}} = \frac{x}{\sqrt{x^2 + 16}}$$
13. **Final answer:**
$$-\frac{1}{48 \sin^3 \theta} + C = -\frac{1}{48} \left(\frac{\sqrt{x^2 + 16}}{x}\right)^3 + C = -\frac{(x^2 + 16)^{3/2}}{48 x^3} + C$$