Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-16, perform the indicated integrations. \(\int \frac{x^{2} d x}{\sqrt{16-x^{2}}}\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Integral

The integral \( \int \frac{x^2 dx}{\sqrt{16-x^2}} \) is not a standard form but looks like a trigonometric substitution candidate due to the \( \sqrt{16-x^2} \).

Step 2: Choose the Appropriate Trigonometric Substitution

For integrals involving \( \sqrt{a^2 - x^2} \), use the substitution \( x = a \sin{\theta} \). Here, let \( x = 4 \sin{\theta} \), so \( dx = 4 \cos{\theta} d\theta \).

Step 3: Substitute and Simplify the Integral

Substituting \( x = 4\sin\theta \) and \( dx = 4\cos\theta d\theta \) into the integral gives us: \[\int \frac{(4\sin\theta)^2 \cdot 4\cos\theta d\theta}{\sqrt{16 - (4\sin\theta)^2}}\]The denominator simplifies to \( \sqrt{16 - 16\sin^2\theta} = \sqrt{16\cos^2\theta} = 4\cos\theta \).

Step 4: Simplify Further

Substituting the simplifications from Step 3, the integral reduces to: \[\int \frac{16\sin^2\theta \cdot 4\cos\theta}{4\cos\theta} d\theta = \int 16\sin^2\theta d\theta\]The \(\cos\theta\) terms cancel out.

Step 5: Convert to a Solvable Trigonometric Integral

Replace \( \sin^2{\theta} \) using the identity \( \sin^2{\theta} = \frac{1 - \cos{2\theta}}{2} \): \[\int 16\left( \frac{1 - \cos{2\theta}}{2} \right) d\theta = \int 8(1 - \cos{2\theta}) d\theta\]Simplifying yields: \[= \int 8 d\theta - \int 8\cos{2\theta} d\theta\]

Step 6: Evaluate the Simplified Integral

Evaluate each component separately:- The integral \( \int 8 d\theta = 8\theta \).- The integral \( \int 8\cos{2\theta} d\theta \) becomes \( 4\sin{2\theta} \) using the identity \( \int \cos{ax} dx = \frac{\sin{ax}}{a} \).Thus, the entire integral becomes: \( 8\theta - 4\sin{2\theta} + C \).

Step 7: Reverse the Trigonometric Substitution

Recall \( x = 4\sin\theta \) which gives \( \sin\theta = \frac{x}{4} \), and \( \theta = \arcsin{\frac{x}{4}} \). Also, use \( \cos{2\theta} = 1 - 2\sin^2\theta \), so \( \sin{2\theta} = 2\sin\theta\cos\theta = 2\sin\theta\sqrt{1 - \sin^2\theta} = \frac{2x\sqrt{16-x^2}}{16} \).

Step 8: Write the Final Expression

Substituting \( \theta \) and \( \sin{2\theta} \) back, we get:\[ 8\arcsin{\left(\frac{x}{4}\right)} - \frac{x\sqrt{16-x^2}}{2} + C \].

Key Concepts

Integration Techniques

Integration is a fundamental concept in calculus, serving as a tool to calculate areas, volumes, and more. A relevant technique in this context is **trigonometric substitution**, which is useful for integrals involving square roots, particularly with expressions like \( \sqrt{a^2 - x^2} \). Here are some common integration techniques:

• **Substitution**: Useful when you can transform the variable into a simpler form.
• **Integration by Parts**: Based on the product rule for differentiation, it splits the integral into two simpler parts.
• **Partial Fraction Decomposition**: Useful when dealing with rational functions, breaking them into simpler fractions.
• **Trigonometric Substitution**: Involves using trigonometric identities to simplify integrals. For example, the substitution \( x = a \sin{\theta} \) transforms \( \sqrt{a^2 - x^2} \) into a manageable form.

In this exercise, recognizing the expression \( \sqrt{16-x^2} \) suggests using trigonometric substitution. We substitute \( x = 4 \sin{\theta} \), making calculations smoother. This approach converts the integral into a simpler trigonometric form, enabling straightforward evaluation.

Definite Integrals

Definite integrals are used to calculate the exact area under a curve between two points. Unlike indefinite integrals, they produce a numerical value rather than a function, enabled by evaluating the integral across given limits.
When dealing with a definite integral, you often use the fundamental theorem of calculus, which connects differentiation with integration. It allows computation of the integral by finding an antiderivative and evaluating it at the upper and lower limits.
In the context of this problem, while the example given is an indefinite integral, understanding definite integrals is crucial. If you were to calculate the definite version, integrating from one point to another, you would follow these steps:

• **Setup**: Identify bounds of integration on the transformed variable \( \theta \), which depends on the bounds of \( x \).
• **Compute**: Find the antiderivative using trigonometric identities and substitution.
• **Evaluate**: Substitute the limits back through the inverse trigonometric functions, replacing \( \theta \) with \( x \).

Trigonometric Identities

Trigonometric identities simplify expressions when integrating, especially with trigonometric substitutions. They transform complex trigonometric forms into familiar or easily integrable expressions.
A few key identities are frequently used:

• **Pythagorean identity**: \( \sin^2{\theta} + \cos^2{\theta} = 1 \)
• **Double angle identities**: \( \sin{2\theta} = 2\sin{\theta}\cos{\theta} \), and \( \cos{2\theta} = \cos^2{\theta} - \sin^2{\theta} \)
• **Power reduction identity**: \( \sin^2{\theta} = \frac{1 - \cos{2\theta}}{2} \), useful for simplifying higher powers of sine or cosine before integration.

In our scenario, the identity \( \sin^2{\theta} = \frac{1 - \cos{2\theta}}{2} \) helps to reduce the squared sine term to a linear form, making integration achievable. Meanwhile, \( \sin{2\theta} \) and \( \cos{2\theta} \) open pathways to substitute back in terms of \( x \) after integration.