The integral you've uploaded is:

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

To solve this, we'll use trigonometric substitution. The key is recognizing that the expression inside the square root, $9 - x^2$, resembles the Pythagorean identity. Here's the step-by-step breakdown:

Step 1: Trigonometric Substitution

We substitute $x = 3\sin(\theta)$, because $9 - x^2 = 9(1 - \sin^2(\theta)) = 9\cos^2(\theta)$. The differential $dx$ becomes:

$dx = 3\cos(\theta) \, d\theta$

Step 2: Substitution into the Integral

Now substitute $x = 3\sin(\theta)$ and $dx = 3\cos(\theta) \, d\theta$ into the original integral:

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

This simplifies the square root:

$\sqrt{9 - 9\sin^2(\theta)} = \sqrt{9\cos^2(\theta)} = 3\cos(\theta)$

Substitute this into the integral:

$\int \frac{3\cos(\theta)}{9\sin^2(\theta)} \cdot 3\cos(\theta) \, d\theta$

Simplify the constants:

$\int \frac{9\cos^2(\theta)}{9\sin^2(\theta)} \, d\theta = \int \frac{\cos^2(\theta)}{\sin^2(\theta)} \, d\theta = \int \cot^2(\theta) \, d\theta$

Step 3: Simplifying the Integral

Now, we integrate $\cot^2(\theta)$:

Recall the identity:

$\cot^2(\theta) = \csc^2(\theta) - 1$

So the integral becomes:

$\int \cot^2(\theta) \, d\theta = \int (\csc^2(\theta) - 1) \, d\theta$

We know the integrals of $\csc^2(\theta)$ and $1$:

$\int \csc^2(\theta) \, d\theta = -\cot(\theta)$ $\int -1 \, d\theta = -\theta$

Thus, the integral is:

$-\cot(\theta) - \theta + C$

Step 4: Substitute Back in Terms of $x$

We need to convert $\theta$ back to $x$. From the substitution $x = 3\sin(\theta)$, we have:

$\sin(\theta) = \frac{x}{3} \quad \Rightarrow \quad \theta = \arcsin\left(\frac{x}{3}\right)$

Also, $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$. Since $\cos(\theta) = \sqrt{1 - \sin^2(\theta)}$, we get:

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

Therefore:

$\cot(\theta) = \frac{\frac{\sqrt{9 - x^2}}{3}}{\frac{x}{3}} = \frac{\sqrt{9 - x^2}}{x}$

Final Answer:

Thus, the solution to the integral is:

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

Let me know if you'd like further clarification or if you have any other questions!

Further Questions:

1. How does trigonometric substitution help in solving integrals involving square roots?
2. Can you explain why $x = 3\sin(\theta)$ was chosen for this substitution?
3. What is the significance of the Pythagorean identity in integrals like these?
4. How do you convert back from trigonometric functions to the original variable after substitution?
5. Can this method be applied to other integrals with similar square root forms?

Tip:

When using trigonometric substitution, always identify the pattern that matches a Pythagorean identity to choose the correct substitution.