We are tasked with simplifying or analyzing the given expression:

$\frac{1}{x^2 \sqrt{4 + x^2}}, \quad x = 2 \tan \theta.$

Step 1: Substituting $x = 2 \tan \theta$

Given $x = 2 \tan \theta$, we substitute it into the expression: $x^2 = (2 \tan \theta)^2 = 4 \tan^2 \theta.$

The term $\sqrt{4 + x^2}$ becomes: $\sqrt{4 + x^2} = \sqrt{4 + 4 \tan^2 \theta}.$

Factoring $4$ out of the square root: $\sqrt{4 + 4 \tan^2 \theta} = \sqrt{4(1 + \tan^2 \theta)} = 2 \sqrt{1 + \tan^2 \theta}.$

Using the trigonometric identity $1 + \tan^2 \theta = \sec^2 \theta$: $\sqrt{4 + x^2} = 2 \sec \theta.$

Step 2: Substituting into the expression

Now substitute $x^2 = 4 \tan^2 \theta$ and $\sqrt{4 + x^2} = 2 \sec \theta$ into the original expression: $\frac{1}{x^2 \sqrt{4 + x^2}} = \frac{1}{(4 \tan^2 \theta)(2 \sec \theta)}.$

Simplify the denominator: $\frac{1}{4 \tan^2 \theta \cdot 2 \sec \theta} = \frac{1}{8 \tan^2 \theta \sec \theta}.$

Step 3: Simplify further

Recall that $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$. Substitute these: $8 \tan^2 \theta \sec \theta = 8 \left(\frac{\sin^2 \theta}{\cos^2 \theta}\right) \left(\frac{1}{\cos \theta}\right).$

Combine terms: $8 \tan^2 \theta \sec \theta = 8 \frac{\sin^2 \theta}{\cos^3 \theta}.$

Thus, the original expression becomes: $\frac{1}{8 \tan^2 \theta \sec \theta} = \frac{\cos^3 \theta}{8 \sin^2 \theta}.$

Final Answer:

$\boxed{\frac{\cos^3 \theta}{8 \sin^2 \theta}}$

Would you like further explanation on any step or to explore applications of this result?

Related Questions:

1. How does this relate to trigonometric substitutions in integrals?
2. Can this expression be simplified further for specific values of $\theta$?
3. What is the geometric significance of the substitution $x = 2 \tan \theta$?
4. How would the solution differ if $x$ were expressed in terms of $\sin \theta$ instead?
5. What are common scenarios where $1 + \tan^2 \theta = \sec^2 \theta$ is used?

Tip:

Always double-check trigonometric identities and substitutions to simplify complex expressions efficiently.