To find the antiderivative of $\sqrt{1 - 9x^2}$, we can use a trigonometric substitution method.

Step-by-step solution:

We start with the expression:

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

Notice that $1 - 9x^2$ suggests a substitution involving sine or cosine. Let's use the following trigonometric substitution:

$x = \frac{1}{3} \sin(\theta)$

Thus, $dx = \frac{1}{3} \cos(\theta) \, d\theta$, and we substitute into the integrand.

Step 1: Substitute into the expression

$1 - 9x^2 = 1 - 9\left(\frac{1}{3} \sin(\theta)\right)^2 = 1 - \sin^2(\theta) = \cos^2(\theta)$

So, $\sqrt{1 - 9x^2} = \sqrt{\cos^2(\theta)} = \cos(\theta)$.

Step 2: Update the integral

Now substitute all these into the original integral:

$\int \sqrt{1 - 9x^2} \, dx = \int \cos(\theta) \cdot \frac{1}{3} \cos(\theta) \, d\theta = \frac{1}{3} \int \cos^2(\theta) \, d\theta$

Step 3: Simplify the integral

We can use the double angle identity for cosine to simplify $\cos^2(\theta)$:

$\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$

Thus, the integral becomes:

$\frac{1}{3} \int \frac{1 + \cos(2\theta)}{2} \, d\theta$

Step 4: Integrate

Now, integrate the expression:

$\frac{1}{3} \cdot \frac{1}{2} \left( \int 1 \, d\theta + \int \cos(2\theta) \, d\theta \right)$

$= \frac{1}{6} \left( \theta + \frac{\sin(2\theta)}{2} \right) + C$

Step 5: Back-substitute

Now, we back-substitute $\theta$ in terms of $x$. Recall that:

$x = \frac{1}{3} \sin(\theta) \quad \Rightarrow \quad \sin(\theta) = 3x$

Thus, $\theta = \arcsin(3x)$.

Also, $\sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2(3x) \sqrt{1 - 9x^2} = 6x \sqrt{1 - 9x^2}$.

So, the final result is:

$\frac{1}{6} \left( \arcsin(3x) + \frac{6x \sqrt{1 - 9x^2}}{2} \right) + C$

Simplifying the constants:

$= \frac{1}{6} \arcsin(3x) + \frac{x \sqrt{1 - 9x^2}}{2} + C$

Final answer:

$\int \sqrt{1 - 9x^2} \, dx = \frac{1}{6} \arcsin(3x) + \frac{x \sqrt{1 - 9x^2}}{2} + C$

Would you like any further clarifications or details on this process?

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Further questions for you:

1. How does trigonometric substitution help in simplifying integrals involving square roots?
2. What is the general form for using trigonometric substitution in integrals of the type $\sqrt{a^2 - x^2}$?
3. Can you explain the significance of the double-angle identity in this solution?
4. What other methods can be used to approach integrals involving square roots, aside from trigonometric substitution?
5. How do you back-substitute after performing a trigonometric substitution?

Tip:

When using trigonometric substitution, always look for a structure in the integrand that matches known trigonometric identities.