Chapter 7: Problem 26
Use the table of integrals on the inside back cover, perhaps combined with a substitution, to evaluate the given integrals. $$ \int \frac{\operatorname{sech} \sqrt{x}}{\sqrt{x}} d x $$
Short Answer
Expert verified
The integral evaluates to \( 2 \ln |\tanh(\sqrt{x}) + \operatorname{sech}(\sqrt{x})| + C \).
Step by step solution
01
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 \( \operatorname{sech} \).
02
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 \( \int \frac{\operatorname{sech} u}{u} \cdot u \cdot 2u \, du = 2 \int \operatorname{sech} u \, du \).
03
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 \( 2 \ln |\tanh(\sqrt{x}) + \operatorname{sech}(\sqrt{x})| + C \).
04
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.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
Understanding these patterns and when to use a particular technique effectively is key to mastering integration.
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.
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:
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.
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.
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
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.
Common hyperbolic functions include
- \(\sinh(x) = \frac{e^x - e^{-x}}{2}\)
- \(\cosh(x) = \frac{e^x + e^{-x}}{2}\)
- \(\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\)
- \(\operatorname{sech}(x) = \frac{1}{\cosh(x)}\)
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.