how do you calculate the integral of the function f(x) = sqrt(5-x^2)

To calculate the integral of the function \( f(x) = \sqrt{5 - x^2} \), we proceed as follows:

\[
\int \sqrt{5 - x^2} \, dx
\]

1. **Recognize the Structure**: This integral resembles the form \( \sqrt{a^2 - x^2} \), which often suggests using a trigonometric substitution.

2. **Choose an Appropriate Substitution**: Set \( x = \sqrt{5} \sin \theta \). This choice works because \( \sqrt{a^2 - x^2} \) simplifies under such substitutions.
   - Then, \( dx = \sqrt{5} \cos \theta \, d\theta \).

3. **Rewrite the Integral**: Substitute \( x = \sqrt{5} \sin \theta \) and \( dx = \sqrt{5} \cos \theta \, d\theta \) into the integral:
   \[
   \int \sqrt{5 - (\sqrt{5} \sin \theta)^2} \cdot \sqrt{5} \cos \theta \, d\theta
   \]
   
4. **Simplify Inside the Square Root**:
   \[
   \sqrt{5 - 5 \sin^2 \theta} = \sqrt{5 (1 - \sin^2 \theta)} = \sqrt{5 \cos^2 \theta} = \sqrt{5} \cos \theta
   \]
   
   Substitute this back in:
   \[
   \int \sqrt{5} \cos \theta \cdot \sqrt{5} \cos \theta \, d\theta = \int 5 \cos^2 \theta \, d\theta
   \]

5. **Use a Trigonometric Identity**: Rewrite \( \cos^2 \theta \) using the identity \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \):
   \[
   \int 5 \cos^2 \theta \, d\theta = \int 5 \cdot \frac{1 + \cos 2\theta}{2} \, d\theta = \frac{5}{2} \int (1 + \cos 2\theta) \, d\theta
   \]

6. **Integrate**:
   \[
   \frac{5}{2} \left( \int 1 \, d\theta + \int \cos 2\theta \, d\theta \right)
   \]
   \[
   = \frac{5}{2} \left( \theta + \frac{\sin 2\theta}{2} \right) = \frac{5}{2} \theta + \frac{5}{4} \sin 2\theta
   \]

7. **Return to \( x \)-Terms**: Since \( x = \sqrt{5} \sin \theta \), we have \( \sin \theta = \frac{x}{\sqrt{5}} \) and \( \theta = \arcsin \left( \frac{x}{\sqrt{5}} \right) \). Also, \( \sin 2\theta = 2 \sin \theta \cos \theta = 2 \cdot \frac{x}{\sqrt{5}} \cdot \frac{\sqrt{5 - x^2}}{\sqrt{5}} = \frac{2x \sqrt{5 - x^2}}{5} \).

   Substitute these back:
   \[
   \frac{5}{2} \arcsin \left( \frac{x}{\sqrt{5}} \right) + \frac{5}{4} \cdot \frac{2x \sqrt{5 - x^2}}{5}
   \]
   Simplify:
   \[
   = \frac{5}{2} \arcsin \left( \frac{x}{\sqrt{5}} \right) + \frac{x \sqrt{5 - x^2}}{2} + C
   \]

The final answer is:
\[
\int \sqrt{5 - x^2} \, dx = \frac{5}{2} \arcsin \left( \frac{x}{\sqrt{5}} \right) + \frac{x \sqrt{5 - x^2}}{2} + C
\]

Would you like any further details, or have questions?

### Related Questions
1. How can trigonometric substitution be applied to different forms of integrals?
2. What are alternative methods to evaluate integrals involving square roots?
3. How do we interpret the arcsin function in integrals involving trigonometric substitutions?
4. What is the significance of the constant of integration, \( C \)?
5. How does trigonometric substitution simplify expressions involving \( \sqrt{a^2 - x^2} \)?

#### Tip
For integrals involving \( \sqrt{a^2 - x^2} \), trigonometric substitution is often the most efficient approach.

how do you calculate the integral of the function f(x) = sqrt(5 - x^2)

Step-by-step solution for calculating the integral of f(x) = sqrt(5 - x^2) using trigonometric substitution. This example explores key calculus concepts and techniques, including trigonometric substitution, integration, and the use of trigonometric identities. Ideal for undergraduate calculus students.

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