https://r2.erweima.ai/mathbot/image/1726095856355b8w3pt86.jpg

The integral in the image is:

\[
\int \frac{dx}{\sqrt{25 - 9x^2}}
\]

This is a standard trigonometric substitution integral, where we recognize the expression inside the square root as similar to the form \(a^2 - (bx)^2\). To solve it, we'll use the substitution:

\[
x = \frac{5}{3} \sin(\theta)
\]

Now, let me solve the integral step by step.

### Step 1: Substitution
We start by using the substitution \( x = \frac{5}{3} \sin(\theta) \). The differential \(dx\) becomes:

\[
dx = \frac{5}{3} \cos(\theta) d\theta
\]

Now, substitute \(x = \frac{5}{3} \sin(\theta)\) into the square root and the differential:

\[
\sqrt{25 - 9x^2} = \sqrt{25 - 9\left(\frac{5}{3} \sin(\theta)\right)^2} = \sqrt{25 - 25 \sin^2(\theta)} = 5 \cos(\theta)
\]

So the integral becomes:

\[
\int \frac{\frac{5}{3} \cos(\theta) d\theta}{5 \cos(\theta)} = \int \frac{\frac{5}{3} \cos(\theta)}{5 \cos(\theta)} d\theta = \int \frac{1}{3} d\theta
\]

### Step 2: Integrate
Now, integrate:

\[
\int \frac{1}{3} d\theta = \frac{\theta}{3} + C
\]

### Step 3: Back-substitution
Since \( x = \frac{5}{3} \sin(\theta) \), we have \( \sin(\theta) = \frac{3x}{5} \), so \( \theta = \arcsin\left(\frac{3x}{5}\right) \).

Thus, the final solution is:

\[
\frac{\arcsin\left(\frac{3x}{5}\right)}{3} + C
\]

### Final Answer:
\[
\int \frac{dx}{\sqrt{25 - 9x^2}} = \frac{\arcsin\left(\frac{3x}{5}\right)}{3} + C
\]

Would you like further details or have any questions?

Here are 5 related questions for deeper understanding:
1. What are the general forms of integrals involving \( \sqrt{a^2 - x^2} \)?
2. How does trigonometric substitution simplify certain integrals?
3. What other substitution techniques are useful in integration?
4. How can you handle integrals that involve hyperbolic functions?
5. What are other integral forms of inverse trigonometric functions?

**Tip:** Always verify the substitution step by differentiating your final result to ensure it matches the original integrand.

This guide solves the integral ∫dx/√(25 - 9x^2) using trigonometric substitution. Follow step-by-step instructions, with key concepts such as substitution and inverse trigonometric functions explained. Ideal for students in Grade 12 or early university courses.

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