Problem 3 - Calculus - AP Edition

Question

Evaluate the integral. $$\int \frac{x^{2}}{\sqrt{16-x^{2}}} d x$$

Step-by-Step Solution

Answer

\( 8\arcsin\left(\frac{x}{4}\right) - \frac{x\sqrt{16-x^2}}{2} + C \)

Step 1: Recognize the Integral Form

The integral \( \int \frac{x^{2}}{\sqrt{16-x^{2}}} \, dx \) contains a square root in the denominator and can be approached using trigonometric substitution. The presence of \( \sqrt{16-x^2} \) suggests using the substitution \( x = 4\sin\theta \) because \( 16 - x^2 \) will then transform into a trigonometric identity, \( 16\cos^2\theta \).

Step 2: Apply Trigonometric Substitution

Substitute \( x = 4\sin\theta \), so that \( dx = 4\cos\theta \, d\theta \). Also, observe \( \sqrt{16-x^2} = \sqrt{16-16\sin^2\theta} = 4\cos\theta \). Replace these into the integral to get:\[ \int \frac{(4\sin\theta)^2}{4\cos\theta} \cdot 4\cos\theta \, d\theta = \int 16\sin^2\theta \, d\theta \]

Step 3: Simplify the Integral

The integral becomes \( \int 16\sin^2\theta \, d\theta \). Use the identity \( \sin^2\theta = \frac{1 - \cos(2\theta)}{2} \) to rewrite the integral:\[ \int 16\left(\frac{1 - \cos(2\theta)}{2}\right) \, d\theta = \int 8(1 - \cos(2\theta)) \, d\theta \]This simplifies to \( 8\int (1 - \cos(2\theta)) \, d\theta \).

Step 4: Evaluate the Integral Using Trigonometric Identity

Now, evaluate the integral:\[ 8\int (1 - \cos(2\theta)) \, d\theta = 8\left(\theta - \frac{1}{2}\sin(2\theta)\right) + C \]where \( C \) is the constant of integration.

Step 5: Back Substitution

Given \( x = 4\sin\theta \), we have \( \sin\theta = \frac{x}{4} \). Use the original substitution to express \( \theta \) in terms of \( x \):\[ \theta = \arcsin\left(\frac{x}{4}\right) \]So, \( \sin(2\theta) = 2\sin\theta\cos\theta = 2\left(\frac{x}{4}\right)\cos\theta \). And since \( \cos\theta = \frac{\sqrt{16-x^2}}{4} \), it follows that:\[ \sin(2\theta) = \frac{x\sqrt{16-x^2}}{8} \]

Step 6: Write the Final Answer

Substitute back all the terms into the integral solution:\[ 8\left(\arcsin\left(\frac{x}{4}\right) - \frac{x\sqrt{16-x^2}}{16}\right) + C \]Thus, the evaluated integral is:\( 8\arcsin\left(\frac{x}{4}\right) - \frac{x\sqrt{16-x^2}}{2} + C \)

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique used in calculus to simplify the integration of expressions involving square roots. The idea is to convert algebraic expressions into trigonometric ones using identities, which are often easier to integrate. For example, in the integral \( \int \frac{x^2}{\sqrt{16-x^2}} \, dx \), the denominator contains \( \sqrt{16 - x^2} \), resembling the Pythagorean identity \( 1 = \sin^2\theta + \cos^2\theta \). By substituting \( x = 4\sin\theta \), we transform the expression under the square root into \( 16\cos^2\theta \).
This substitution transforms the variables and changes the integral into a simpler trigonometric form. Specifically, \( dx = 4\cos\theta \, d\theta \) and \( \sqrt{16-x^2} = 4\cos\theta \). As a result, our integral becomes \( \int 16\sin^2\theta \, d\theta \), which is significantly easier to handle than the original integration form. This process provides a pathway to apply trigonometric identities to further simplify and evaluate the integral.

Definite Integrals

Definite integrals calculate the area under a curve between two specified points. While the original exercise concerns an indefinite integral, knowing the concept of definite integrals is crucial for understanding integration's application in real-world contexts. A definite integral is denoted with limits; for example, \( \int_{a}^{b} f(x) \, dx \) represents the area from \( x = a \) to \( x = b \).
This contrasts with indefinite integrals, which represent a family of functions that include a constant of integration \( C \). Recognizing when to use definite vs. indefinite integrals is key to problem-solving in calculus. Although this exercise primarily demonstrates indefinite integration, similar techniques apply when evaluating definite integrals, often following a similar process of substitution and simplification.

Indefinite Integrals

Indefinite integrals focus on finding a general function whose derivative is the given integrand. They do not involve specific limits; hence, they include the constant of integration \( C \). In our exercise, the goal is to find an antiderivative for \( \int \frac{x^2}{\sqrt{16-x^2}} \, dx \).
Starting with trigonometric substitution, the integral is simplified to \( \int 16\sin^2\theta \, d\theta \). Using the identity \( \sin^2\theta = \frac{1 - \cos(2\theta)}{2} \), the integral becomes more straightforward: \( 8\int (1 - \cos(2\theta)) \, d\theta \). The exploration of identities simplifies the integration process, resulting in an expression \( 8\left(\theta - \frac{1}{2}\sin(2\theta)\right) + C \).
Finally, back substitution allows converting the solution back in terms of \( x \), providing the final indefinite integral solution \( 8\arcsin\left(\frac{x}{4}\right) - \frac{x\sqrt{16-x^2}}{2} + C \). Indefinite integrals thus provide powerful tools for solving complex equations and finding functions based on rates of change.