We are given the integral:

$\int \frac{1}{9 + x^2} \, dx$

and the substitution $x = 3\tan t$, with $-\frac{\pi}{2} \leq t \leq \frac{\pi}{2}$.

Step-by-step solution:

1. Differentiate the substitution $x = 3\tan t$: $dx = 3\sec^2 t \, dt$

2. Substitute $x = 3\tan t$ and $dx = 3\sec^2 t \, dt$ into the integral: $\int \frac{1}{9 + x^2} \, dx = \int \frac{1}{9 + (3\tan t)^2} \cdot 3\sec^2 t \, dt$

3. Simplify the expression $9 + (3\tan t)^2$: $9 + (3\tan t)^2 = 9 + 9\tan^2 t = 9(1 + \tan^2 t)$ Using the trigonometric identity $1 + \tan^2 t = \sec^2 t$, we get: $9(1 + \tan^2 t) = 9\sec^2 t$

4. Substitute this back into the integral: $\int \frac{1}{9\sec^2 t} \cdot 3\sec^2 t \, dt = \int \frac{3\sec^2 t}{9\sec^2 t} \, dt = \int \frac{1}{3} \, dt$

5. Integrate with respect to $t$: $\int \frac{1}{3} \, dt = \frac{t}{3} + C$

6. Substitute back $t$ in terms of $x$ using the inverse of $x = 3\tan t$, which gives $t = \tan^{-1}\left(\frac{x}{3}\right)$: $\frac{t}{3} = \frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right)$

Thus, the simplified integral is:

$\int \frac{1}{9 + x^2} \, dx = \frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right) + C$

Would you like further details or clarification?

Here are 5 questions related to this process:

1. How does substitution simplify integrals involving squares of variables?
2. What is the significance of using $\tan^{-1}$ in integrals?
3. Why do we restrict the bounds for $t$ to $-\frac{\pi}{2} \leq t \leq \frac{\pi}{2}$?
4. Can we use other trigonometric substitutions for similar integrals?
5. How do we handle definite integrals when performing substitutions?

Tip: Always check for trigonometric identities to simplify expressions when using substitutions.