To solve the indefinite integral $\int \frac{1}{\sqrt{9 - 36x^2}} \, dx$ using $u$-substitution, we'll follow the steps systematically.

Step 1: Identify a substitution

We can recognize that $9 - 36x^2$ looks like a form that might benefit from a substitution involving a trigonometric identity. In particular, this resembles the form $a^2 - u^2$, which suggests we could use a trigonometric substitution.

Let's start by simplifying the expression inside the square root: $9 - 36x^2 = 9(1 - 4x^2)$ Now, let $u = 2x$, so that $du = 2dx$, and the integral becomes simpler.

Step 2: Substitution

Let $u = 2x$, implying $du = 2dx$, or $\frac{du}{2} = dx$.

Substitute into the integral: $\int \frac{1}{\sqrt{9 - 36x^2}} \, dx = \int \frac{1}{\sqrt{9(1 - u^2)}} \cdot \frac{du}{2}$ Simplify the constant factor: $= \frac{1}{2} \int \frac{1}{\sqrt{9(1 - u^2)}} \, du$ Factor out the constant 9 from the square root: $= \frac{1}{2} \int \frac{1}{3\sqrt{1 - u^2}} \, du$ $= \frac{1}{6} \int \frac{1}{\sqrt{1 - u^2}} \, du$

Step 3: Recognize the standard integral

The integral $\int \frac{1}{\sqrt{1 - u^2}} \, du$ is a standard trigonometric integral, which is the inverse sine function: $\int \frac{1}{\sqrt{1 - u^2}} \, du = \arcsin(u) + C$

Step 4: Substitute back in terms of $x$

Recall that $u = 2x$, so substitute back: $\frac{1}{6} \arcsin(u) + C = \frac{1}{6} \arcsin(2x) + C$

Thus, the solution to the integral is: $\int \frac{1}{\sqrt{9 - 36x^2}} \, dx = \frac{1}{6} \arcsin(2x) + C$

Recap:

• Final Answer: $\frac{1}{6} \arcsin(2x) + C$

Would you like a deeper explanation of any of the steps, or do you have any further questions?

Here are a few related questions you might explore:

1. What are some other common forms of integrals that involve trigonometric substitution?
2. How would you handle an integral with a similar form, but with different constants?
3. What is the geometric interpretation of the inverse sine function?
4. How does the method of trigonometric substitution apply to integrals of rational functions?
5. What other integration methods can be used for integrals involving square roots?

Tip: For integrals involving square roots of expressions like $1 - u^2$, trigonometric substitution is often the go-to method. Identifying the right substitution early is key!