Problem 49 - Thomas Calculus

Question

Use a substitution to change the integral into one you can find in the table. Then evaluate the integral. \(\int \frac{3 d r}{\sqrt{9 r^{2}-1}}\)

Step-by-Step Solution

Answer

The integral evaluates to \(\sec^{-1}(3r) + C\).

Step 1: Identify the substitution

For integrals involving the form \(\sqrt{a^2r^2 - b^2}\), a common substitution is \(r = \frac{b}{a}\sec(\theta)\). In our case, rewrite the integral as \( \int \frac{3 \, dr}{\sqrt{(3r)^2 - 1}} \). This suggests the substitution \(r = \frac{1}{3}\sec(\theta)\).

Step 2: Differentiate the substitution

Differentiate \(r = \frac{1}{3}\sec(\theta)\) with respect to \(\theta\) to find \(dr\). This gives \(dr = \frac{1}{3} \sec(\theta)\tan(\theta) \, d\theta\).

Step 3: Substitute and simplify the integral

Replace \(r\) and \(dr\) in the integral with the expressions obtained in the previous steps. The integral becomes \( \int \frac{3 \cdot \frac{1}{3} \sec(\theta)\tan(\theta) \, d\theta}{\sqrt{(3 \cdot \frac{1}{3}\sec(\theta))^2 - 1}} \). Simplify it to get \( \int \sec(\theta)\tan(\theta) d\theta\).

Step 4: Simplify the expression inside the square root

Evaluate the square root expression \(\sqrt{\sec^2(\theta) - 1}\) to obtain \(\tan(\theta)\). Thus, the integral simplifies further to \( \int d\theta \).

Step 5: Integrate

The integral \( \int d\theta = \theta + C\), where \(C\) is the constant of integration. Now, substitute back for \(\theta\) using \(r = \frac{1}{3}\sec(\theta)\), which gives \(\theta = \sec^{-1}(3r)\).

Step 6: Substitute back to the original variable

Finally substitute \(\theta = \sec^{-1}(3r)\) back, giving \(\theta + C = \sec^{-1}(3r) + C\). Thus, the evaluated integral is \(\sec^{-1}(3r) + C\).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used to evaluate integrals, especially when they involve expressions like \( \sqrt{a^2r^2 - b^2} \). This method uses trigonometric identities to simplify these expressions, making your task much more manageable.
A common substitution used is replacing \( r \) with \( \frac{b}{a}\sec(\theta) \). This choice leverages the identity \( \sec^2(\theta) - 1 = \tan^2(\theta) \), transforming the integral into a simpler form.
You follow these steps:

• Identify a suitable trigonometric substitution. For expressions like \( \sqrt{9r^2 - 1} \), use \( r = \frac{1}{3}\sec(\theta) \).
• Differentiate \( r \) with respect to \( \theta \) to find \( dr \).
• Substitute \( r \) and \( dr \) back into the integral, transforming it accordingly.

This substitution effectively changes the variable of integration from \( r \) to \( \theta \), simplifying the problem using trigonometric identities and making the integral easier to solve.

Definite and Indefinite Integrals

Integrals come in two main types: definite and indefinite. Understanding the difference is crucial for proper application and satisfying the mathematical aim.
**Indefinite Integrals:**

• Represent a family of functions and include a constant \( C \), which stands for any real number.
• For example, integrating \( \int x \, dx \) gives \( \frac{x^2}{2} + C \), where \( C \) represents the constant of integration.

**Definite Integrals:**

• Calculate the net area under a curve between two specific points, producing a number rather than a function.
• Represented as \( \int_{a}^{b} f(x) \, dx \), they yield the difference \( F(b) - F(a) \), where \( F \) is an antiderivative of \( f \).

In our exercise, you deal with an indefinite integral, which signifies finding a general solution rather than a specific value. The presence of \( + C \) signifies the range of possibilities depending on initial conditions or further constraints.

Inverse Trigonometric Functions

Inverse trigonometric functions, like \( \sec^{-1}(x) \), are crucial in evaluating integrals arising from trigonometric substitutions.
These functions help revert back from trigonometric expressions to your original variable, enabling meaningful solutions to evaluated integrals.
**Functions Commonly Used:**

• \( \sin^{-1}(x) \) - Arcsine
• \( \cos^{-1}(x) \) - Arccosine
• \( \tan^{-1}(x) \) - Arctangent
• \( \sec^{-1}(x) \) - Arcsecant, as used in this problem

In our specific exercise, after simplifying the integral, we arrive at \( \theta \), which when inverted using the substitution \( r = \frac{1}{3}\sec(\theta) \), transitions back to \( \theta = \sec^{-1}(3r) \).
These inverse trigonometric functions are pivotal not just for switching back but also in connecting the arc measurements to real-world applications, ensuring your integrals make sense in practical scenarios.