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.

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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.