Yes, this integral can be solved using u-substitution. Let’s break it down.

The given integral is:

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

Step-by-Step Solution Using $u$-Substitution:

1. Identify a substitution: Let $u = 3x$, so that $du = 3 \, dx$, or $dx = \frac{du}{3}$.

2. Rewrite the integral: Substitute $x = \frac{u}{3}$, which gives $9x^2 = u^2$. The integral becomes:

   $\int \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{3}$

   Factor out $\frac{1}{3}$:

   $\frac{1}{3} \int \frac{1}{\sqrt{1 - u^2}} \, du$

3. Recognize the standard integral: The integral $\int \frac{1}{\sqrt{1 - u^2}} \, du$ is the standard form for $\arcsin(u) + C$.

   So, the result is:

   $\frac{1}{3} \arcsin(u) + C$

4. Back-substitute $u = 3x$: Replace $u$ with $3x$:

   $\frac{1}{3} \arcsin(3x) + C$

Final Answer:

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

Would you like further clarification or additional details?

Expanding Ideas with Questions:

1. How do you identify when $u$-substitution is needed in an integral?
2. What is the difference between this method and trigonometric substitution for similar integrals?
3. Can you explain why $\arcsin(x)$ is the antiderivative of $\frac{1}{\sqrt{1 - x^2}}$?
4. How would the integral change if the expression was $\sqrt{1 - 9x^2}$ in the numerator?
5. Can you verify the result by differentiating $\frac{1}{3} \arcsin(3x) + C$?

Tip:

For integrals involving $\sqrt{1 - kx^2}$, try factoring $k$ out of the square root and look for substitutions involving $kx$.