It seems you want to integrate the function \(\sqrt{25 - 4x^2}\) with respect to \(x\). This integral can be solved using a trigonometric substitution.

The integral is:

\[
\int \sqrt{25 - 4x^2} \, dx
\]

### Step-by-Step Solution:

1. **Trigonometric Substitution**:
   We use the substitution:

   \[
   x = \frac{5}{2} \sin \theta
   \]

   Then, \(dx = \frac{5}{2} \cos \theta \, d\theta\).

2. **Rewrite the Integral**:
   Substitute \(x\) and \(dx\) in the integral:

   \[
   \int \sqrt{25 - 4\left(\frac{5}{2} \sin \theta\right)^2} \cdot \frac{5}{2} \cos \theta \, d\theta
   \]

   Simplify the expression inside the square root:

   \[
   25 - 4 \left(\frac{25}{4} \sin^2 \theta\right) = 25 - 25 \sin^2 \theta = 25 (1 - \sin^2 \theta) = 25 \cos^2 \theta
   \]

   So, the integral becomes:

   \[
   \int \sqrt{25 \cos^2 \theta} \cdot \frac{5}{2} \cos \theta \, d\theta = \int 5 \cos \theta \cdot \frac{5}{2} \cos \theta \, d\theta
   \]

   This simplifies to:

   \[
   \frac{25}{2} \int \cos^2 \theta \, d\theta
   \]

3. **Integrate**:
   Use the half-angle identity for cosine:

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

   Thus, the integral becomes:

   \[
   \frac{25}{2} \int \frac{1 + \cos(2\theta)}{2} \, d\theta = \frac{25}{4} \int (1 + \cos(2\theta)) \, d\theta
   \]

   Integrating term-by-term:

   \[
   \frac{25}{4} \left( \theta + \frac{\sin(2\theta)}{2} \right) + C
   \]

4. **Back-Substitute \(\theta\) with \(x\)**:
   Recall that \(x = \frac{5}{2} \sin \theta\), so \(\sin \theta = \frac{2x}{5}\). Then, \(\theta = \arcsin\left(\frac{2x}{5}\right)\).

   Also, using the double-angle identity, \(\sin(2\theta) = 2 \sin \theta \cos \theta\). Since \(\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{2x}{5}\right)^2}\), we have:

   \[
   \sin(2\theta) = 2 \cdot \frac{2x}{5} \cdot \sqrt{1 - \left(\frac{2x}{5}\right)^2} = \frac{4x}{5} \sqrt{1 - \frac{4x^2}{25}}
   \]

5. **Final Answer**:

   \[
   \frac{25}{4} \left( \arcsin\left(\frac{2x}{5}\right) + \frac{1}{2} \cdot \frac{4x}{5} \sqrt{1 - \frac{4x^2}{25}} \right) + C
   \]

   Simplify the expression:

   \[
   \frac{25}{4} \arcsin\left(\frac{2x}{5}\right) + \frac{5x}{2} \sqrt{1 - \frac{4x^2}{25}} + C
   \]

### Conclusion:

The integral of \(\sqrt{25 - 4x^2}\) with respect to \(x\) is:

\[
\int \sqrt{25 - 4x^2} \, dx = \frac{25}{4} \arcsin\left(\frac{2x}{5}\right) + \frac{5x}{2} \sqrt{1 - \frac{4x^2}{25}} + C
\]

Would you like more details or have any questions?

### Related Questions:

1. How do you perform trigonometric substitutions for integrals?
2. What are some common integrals involving square roots of quadratic expressions?
3. How do you derive the half-angle identities in trigonometry?
4. What is the method for back-substitution after solving an integral using substitution?
5. Can this method be applied to definite integrals? How?

### Tip:

When using trigonometric substitution, always consider the domain of the function and the appropriate range for the inverse trigonometric functions.

Explore the step-by-step solution for integrating √(25 - 4x^2) using trigonometric substitution. This guide covers key calculus concepts and techniques, including trigonometric identities and back-substitution. Ideal for advanced high school or college-level students.

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