Problem 53 - Calculus Early Transcendentals: Pearson New International Edition

Question

Perform the indicated integrations. $$ \int \frac{d t}{t \sqrt{2 t^{2}-9}} $$

Step-by-Step Solution

Answer

The integral evaluates to \( \sec^{-1}\left(\frac{\sqrt{2}t}{3}\right) + C \).

Step 1: 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.

Step 2: 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 \( \frac{9}{2}\tan^2(\theta) \).

Step 3: 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 \( \int \frac{\sec(\theta) \tan(\theta) \, d\theta}{\sec(\theta) \sqrt{\tan^2(\theta)}} \).

Step 4: Simplify the Integral

Since \( \sqrt{\tan^2(\theta)} = \tan(\theta) \) for values of \( \theta \) where \( \tan(\theta) \) is positive, the integral simplifies to \( \int \frac{\sec(\theta) \tan(\theta)}{\sec(\theta) \tan(\theta)} \, d\theta = \int 1 \, d\theta = \theta + C \).

Step 5: 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 \( \sec^{-1}\left(\frac{\sqrt{2}t}{3}\right) + C \).

Key Concepts

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:

• 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 \( x = a \sec(\theta) \)

This particular method can greatly simplify integration, transforming them into more manageable trigonometric integrals.

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:

• 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.

By recognizing the nature of the rational terms, a path to simplify and integrate becomes clearer, turning an initially complex problem into a more straightforward process.

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 \( \sec^{-1}\left(\frac{\sqrt{2}t}{3}\right) + C \).

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.

Understanding when and how to apply these two types of integration allows for solving a variety of problems in mathematics and applied fields.