Chapter 7: Problem 53
Perform the indicated integrations. $$ \int \frac{d t}{t \sqrt{2 t^{2}-9}} $$
Short Answer
Expert verified
The integral evaluates to \( \sec^{-1}\left(\frac{\sqrt{2}t}{3}\right) + C \).
Step by step solution
01
Identify the Type of Integral
The integral \( \int \frac{d t}{t \sqrt{2 t^{2}-9}} \) is a rational function of \( t \) multiplied by a square root of a quadratic expression. This suggests that a trigonometric substitution might be helpful.
02
Make a Suitable Substitution
To simplify the expression under the square root, use the substitution \( t = \frac{3}{\sqrt{2}} \sec(\theta) \). This implies \( dt = \frac{3}{\sqrt{2}} \sec(\theta) \tan(\theta) \, d\theta \). The expression \( 2t^2 - 9 \) becomes \( \frac{9}{2}\tan^2(\theta) \).
03
Rewrite the Integral in Terms of Theta
Substitute \( t \) and \( dt \) in the integral: \( \int \frac{\frac{3}{\sqrt{2}} \sec(\theta) \tan(\theta) \, d\theta}{\frac{3}{\sqrt{2}} \sec(\theta) \sqrt{\frac{9}{2} \tan^2(\theta)}} \). This simplifies to \( \int \frac{\sec(\theta) \tan(\theta) \, d\theta}{\sec(\theta) \sqrt{\tan^2(\theta)}} \).
04
Simplify the Integral
Since \( \sqrt{\tan^2(\theta)} = \tan(\theta) \) for values of \( \theta \) where \( \tan(\theta) \) is positive, the integral simplifies to \( \int \frac{\sec(\theta) \tan(\theta)}{\sec(\theta) \tan(\theta)} \, d\theta = \int 1 \, d\theta = \theta + C \).
05
Convert Back to Variable t
Recall that \( t = \frac{3}{\sqrt{2}} \sec(\theta) \), so \( \sec(\theta) = \frac{\sqrt{2}t}{3} \). Thus, \( \theta = \sec^{-1}\left(\frac{\sqrt{2}t}{3}\right) \). Therefore, the integral evaluates to \( \sec^{-1}\left(\frac{\sqrt{2}t}{3}\right) + C \).
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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 strategy used in integration when dealing with integrals that involve square roots of quadratic expressions. The objective is to transform the integrand into a form that is easier to integrate. In the case of the original exercise, the integral \( \int \frac{d t}{t \sqrt{2 t^{2}-9}} \), a trigonometric substitution helps simplify the complicated square root expression.
To execute this substitution, you first need to identify a substitution that matches the form \( a^2 \pm x^2 \). For our exercise, using \( t = \frac{3}{\sqrt{2}} \sec(\theta) \) was ideal because the expression under the square root \( 2t^2 - 9 \) is a quadratic form. This substitution method leverages trigonometric identities to reduce complex expressions to simpler, often linear forms.
Here's a guide on when to apply which substitution based on the quadratic expression under the square root:
To execute this substitution, you first need to identify a substitution that matches the form \( a^2 \pm x^2 \). For our exercise, using \( t = \frac{3}{\sqrt{2}} \sec(\theta) \) was ideal because the expression under the square root \( 2t^2 - 9 \) is a quadratic form. This substitution method leverages trigonometric identities to reduce complex expressions to simpler, often linear forms.
Here's a guide on when to apply which substitution based on the quadratic expression under the square root:
- For expressions of the type \( a^2 - x^2 \), use \( x = a \sin(\theta) \)
- For \( a^2 + x^2 \), use \( x = a \tan(\theta) \)
- For \( x^2 - a^2 \), use \( x = a \sec(\theta) \)
Rational Functions
Rational functions are functions that can be expressed as the quotient of two polynomials. The integral in the exercise is an example of a rational function being integrated with another function type, a square root in this case. Understanding how rational functions behave helps in deciding the most suitable integration technique to apply.
Before directly integrating a rational function, always consider simplifying the expression. Check if partial fraction decomposition might be useful. However, in our exercise, the presence of a square root of a quadratic expression suggested using a trigonometric substitution instead.
Key points to remember about rational functions in integration:
Before directly integrating a rational function, always consider simplifying the expression. Check if partial fraction decomposition might be useful. However, in our exercise, the presence of a square root of a quadratic expression suggested using a trigonometric substitution instead.
Key points to remember about rational functions in integration:
- They can often be simplified by factoring or breaking down into partial fractions.
- Integration might require clever substitutions if additional terms like square roots are present.
- Always verify if the function fits a form that allows direct integration methods to be applied, such as trigonometric substitutions in this case.
Definite and Indefinite Integrals
Integration is a fundamental concept in calculus, used to find areas under curves or solve differential equations. It comes in two forms: definite and indefinite integrals. In this exercise, the task involves an indefinite integral, meaning we are seeking a general form of the antiderivative of the function.
Indefinite integrals do not have limits of integration, which means they include an arbitrary constant \( C \) when the integration is completed. This constant represents the family of all antiderivatives that differ by a constant. For the integral \( \int \frac{d t}{t \sqrt{2 t^{2}-9}} \), after applying trigonometric substitution, we find that the antiderivative is \( \sec^{-1}\left(\frac{\sqrt{2}t}{3}\right) + C \).
Indefinite integrals do not have limits of integration, which means they include an arbitrary constant \( C \) when the integration is completed. This constant represents the family of all antiderivatives that differ by a constant. For the integral \( \int \frac{d t}{t \sqrt{2 t^{2}-9}} \), after applying trigonometric substitution, we find that the antiderivative is \( \sec^{-1}\left(\frac{\sqrt{2}t}{3}\right) + C \).
Definite vs. Indefinite Integrals
- Definite Integrals: Compute the exact area under a curve between two points, providing a numerical value.
- Indefinite Integrals: Provide the general antiderivative, representing an entire family of functions.