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.