find the substitution for the integral of sqrt(9-x^2) 
and what would the term leg or hypotneuse mean in this situation

To evaluate the integral of \(\sqrt{9 - x^2}\), we often use trigonometric substitution. Here's the step-by-step process:

### Step 1: Identify the Trigonometric Substitution
The expression under the square root is \(9 - x^2\), which resembles the Pythagorean identity \(1 - \sin^2(\theta) = \cos^2(\theta)\). Therefore, we use the substitution:
\[
x = 3\sin(\theta)
\]
This implies:
\[
dx = 3\cos(\theta) d\theta
\]

### Step 2: Rewrite the Integral
Substitute \(x = 3\sin(\theta)\) into the integral:
\[
\int \sqrt{9 - x^2} \, dx = \int \sqrt{9 - 9\sin^2(\theta)} \cdot 3\cos(\theta) \, d\theta
\]
Simplifying the square root:
\[
\sqrt{9(1 - \sin^2(\theta))} = \sqrt{9\cos^2(\theta)} = 3\cos(\theta)
\]
So, the integral becomes:
\[
\int 3\cos(\theta) \cdot 3\cos(\theta) \, d\theta = 9\int \cos^2(\theta) \, d\theta
\]

### Step 3: Simplify Using Trigonometric Identity
Use the trigonometric identity:
\[
\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}
\]
Thus, the integral becomes:
\[
9\int \frac{1 + \cos(2\theta)}{2} \, d\theta
\]
This simplifies to:
\[
\frac{9}{2} \int (1 + \cos(2\theta)) \, d\theta
\]
Integrating term by term:
\[
\frac{9}{2} \left[\theta + \frac{\sin(2\theta)}{2}\right] + C
\]

### Step 4: Back-Substitute
Recall that \(x = 3\sin(\theta)\), so \(\sin(\theta) = \frac{x}{3}\). Therefore, \(\theta = \arcsin\left(\frac{x}{3}\right)\). The final integral in terms of \(x\) is:
\[
\frac{9}{2} \left[\arcsin\left(\frac{x}{3}\right) + \frac{\sin\left(2\arcsin\left(\frac{x}{3}\right)\right)}{2}\right] + C
\]

### Explanation of "Leg" and "Hypotenuse"
In this context:
- **Leg**: When you perform the substitution \(x = 3\sin(\theta)\), you are imagining \(x\) as the opposite side (leg) of a right triangle with hypotenuse 3. The corresponding angle is \(\theta\).
- **Hypotenuse**: The hypotenuse in this situation is 3, as it represents the constant value \(9 - x^2\) under the square root when reinterpreted in the trigonometric context.

Would you like more details or have any questions?

### Related Questions
1. What is the integral of \(\sqrt{a^2 - x^2}\) using trigonometric substitution?
2. How does one derive the trigonometric identity \(\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}\)?
3. Can you explain the geometric interpretation of the substitution \(x = a\sin(\theta)\)?
4. How does one find the integral of \(\sqrt{x^2 - a^2}\) using hyperbolic substitution?
5. What is the connection between trigonometric substitution and polar coordinates?

### Tip
Always remember that trigonometric substitution is a powerful tool for integrals involving square roots of quadratic expressions. It often simplifies the integrand into a form that is easier to integrate.

Learn how to evaluate the integral of sqrt(9 - x^2) using trigonometric substitution. This detailed solution covers step-by-step trigonometric substitution, integration techniques, and the geometric interpretation of 'leg' and 'hypotenuse'. Suitable for advanced high school or college students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation