Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the table of integrals on the inside back cover, perhaps combined with a substitution, to evaluate the given integrals. $$ \int \frac{\operatorname{sech} \sqrt{x}}{\sqrt{x}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( 2 \ln |\tanh(\sqrt{x}) + \operatorname{sech}(\sqrt{x})| + C \).

Step 1: Recognize the Integral Form

Look for a pattern or form that matches the integrals listed on the reference cover. In this exercise, notice that the function involves \( \operatorname{sech}(\sqrt{x}) \) and \( \frac{1}{\sqrt{x}} \), which hints towards the use of a table formula involving \( \operatorname{sech} \).

Step 2: Make an Appropriate Substitution

Let \( u = \sqrt{x} \). Then \( du = \frac{1}{2\sqrt{x}} \, dx \) or equivalently \( dx = 2u \, du \). Substituting \( u \) into the integral gives \( \int \frac{\operatorname{sech} u}{u} \cdot u \cdot 2u \, du = 2 \int \operatorname{sech} u \, du \).

Step 3: Integrate Using the Table of Integrals

This is now a standard integral which can be evaluated using a table of integrals. From the reference, the integral \( \int \operatorname{sech} u \, du = \ln |\tanh(u) + \operatorname{sech}(u)| + C \). Therefore, substituting back, we get \( 2 \ln |\tanh(\sqrt{x}) + \operatorname{sech}(\sqrt{x})| + C \).

Step 4: Finalize and Simplify the Solution

The result from the previous step is complete, but it should be written in terms of \( x \) instead of \( u \). Our final answer is \( 2 \ln |\tanh(\sqrt{x}) + \operatorname{sech}(\sqrt{x})| + C \), where C is the constant of integration.

Key Concepts

Integration Techniques

Integration is a fundamental concept in calculus. It helps us find areas under curves, among many other applications. Several techniques can be used in integration to handle different forms of functions.

In many scenarios, direct integration is not feasible, which is why advanced techniques come into play. Some common methods include:

• Substitution: Changing variables to simplify the integral.
• Integration by Parts: Using the product rule for derivatives in reverse.
• Partial Fraction Decomposition: Breaking down complex rational expressions.
• Trigonometric Identities: Applying identities to simplify trigonometric integrals.

The choice of technique often depends on recognizing specific patterns or forms in the integrand, which can sometimes be spotted using a table of integrals. When dealing with functions like \(\operatorname{sech}\), recognizing its standard integral form is crucial to apply the correct technique.

Understanding these patterns and when to use a particular technique effectively is key to mastering integration.

Substitution Method

The Substitution Method is a powerful technique in integral calculus. It simplifies integrals by changing variables to make the integrand easier to handle.

Here's how this method works:

• You choose a new variable, usually denoted by \(u\), to replace part of the integral's variable. This new variable is chosen to simplify the function.
• The derivative of this new variable \(du\) is then used to replace \(dx\) in the integrand.
• By substituting \(u\) into the integral, the expression becomes more manageable and often matches a form found in standard integral tables.

For example, in the given problem, we let \(u = \sqrt{x}\), making \(du = \frac{1}{2\sqrt{x}} \, dx\). This substitution changes the original variable and modifies the integral's complexity, simplifying the integration process. Once integrated, we substitute back the original \(x\) variable to express the result in familiar terms.

Substitution is particularly convenient when dealing with composite functions or when the integrand's derivative can be linked to another part of the function. This method is essential for anyone looking to tackle a wide range of integration problems smoothly.

Hyperbolic Functions

Hyperbolic functions, like \(\operatorname{sech}\), are analogous to trigonometric functions but for hyperbolas. They are defined using exponential functions and have unique properties that make them useful in various mathematical contexts.

Common hyperbolic functions include

• \(\sinh(x) = \frac{e^x - e^{-x}}{2}\)
• \(\cosh(x) = \frac{e^x + e^{-x}}{2}\)
• \(\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\)
• \(\operatorname{sech}(x) = \frac{1}{\cosh(x)}\)

These functions have derivatives and integrals involving their sibling functions directly, much like how derivatives and integrals for sine and cosine are related. For instance, the integral of \(\operatorname{sech}(x)\) yields \(\ln |\tanh(x) + \operatorname{sech}(x)| + C\), as utilized in the given exercise.

Understanding hyperbolic functions and their properties is essential for solving integrals that involve these expressions. They frequently show up in calculus problems, particularly those involving exponential growth, wave equations, and even some geometry problems. Recognizing these functions' standard integration forms can significantly speed up finding solutions.