Chapter 8: Problem 49
Use a substitution to change the integral into one you can find in the table. Then evaluate the integral. \(\int \frac{3 d r}{\sqrt{9 r^{2}-1}}\)
Short Answer
Expert verified
The integral evaluates to \(\sec^{-1}(3r) + C\).
Step by step solution
01
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 \(r = \frac{1}{3}\sec(\theta)\).
02
Differentiate the substitution
Differentiate \(r = \frac{1}{3}\sec(\theta)\) with respect to \(\theta\) to find \(dr\). This gives \(dr = \frac{1}{3} \sec(\theta)\tan(\theta) \, d\theta\).
03
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 \( \int \sec(\theta)\tan(\theta) d\theta\).
04
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 \( \int d\theta \).
05
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 \(\theta = \sec^{-1}(3r)\).
06
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 \(\sec^{-1}(3r) + C\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
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:**
**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.
- 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 \( f \).
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:**
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.
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
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.