/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Evaluate the integral. \(\int ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 5

Evaluate the integral. \(\int \frac{t}{t^{4}+2} d t\)

Short Answer

Expert verified
The integral evaluates to \(\frac{1}{2\sqrt{2}} \tan^{-1}(\frac{t^2}{\sqrt{2}}) + C.\)

Step by step solution

01

Identify substitution variables

For the integral \(\int \frac{t}{t^{4}+2} \, dt\), we notice that the derivative of \(t^4+2\) is related to the numerator. This suggests a substitution. Let's set \(u = t^4 + 2\), then \(\frac{du}{dt} = 4t^3\), or \(du = 4t^3 dt\). However, we adjust for the term in the integral. Since it's just \(t\), we try a simpler substitution: set \(u = t^2\). Then \(du = 2t \, dt\), which means \(t \, dt = \frac{1}{2} \, du\).
02

Substitute and simplify the integral

With the substitution \(u = t^2\), we rewrite the original integral: \(\int \frac{t}{t^4 + 2} \, dt = \int \frac{1}{2} \cdot \frac{1}{u^2 + 2} \, du\). Thus, the integral becomes \(\frac{1}{2} \int \frac{1}{u^2 + 2} \, du\).
03

Recognize the integral form

The integral \(\int \frac{1}{u^2 + 2} \, du\) resembles the standard form \(\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C\). Here, \(a^2 = 2\), so \(a = \sqrt{2}\). Therefore, the integral is \(\int \frac{1}{u^2 + (\sqrt{2})^2} \, du = \frac{1}{\sqrt{2}} \tan^{-1}(\frac{u}{\sqrt{2}}) + C\).
04

Integrate and back substitute

Inserting the recognized form into our equation, we have \(\frac{1}{2} \cdot \frac{1}{\sqrt{2}} \tan^{-1}(\frac{u}{\sqrt{2}}) + C\). Thus, \(\frac{1}{2\sqrt{2}} \tan^{-1}(\frac{u}{\sqrt{2}}) + C\). Substitute back \(u = t^2\), giving us:\[ \frac{1}{2\sqrt{2}} \tan^{-1}(\frac{t^2}{\sqrt{2}}) + C. \]

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 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration Techniques
Integration techniques are a variety of methods used to solve integrals, which are fundamental in calculus. The purpose of these techniques is to transform a complex integral into a form that is easier to evaluate.
  • Substitution Method: This is one of the most common techniques. It involves changing the variables to simplify the integral, often using a substitution that makes the integral resemble a known form.
  • Integration by Parts: Useful when dealing with the product of functions, it is based on the product rule for differentiation.
  • Partial Fraction Decomposition: Used specifically for rational functions, where the function is expressed as a sum of simpler fractions.
In the example provided, the substitution method was used to simplify the integral by letting \( u = t^2 \). This change reduced the complexity and allowed the use of a standard form integral.
Trigonometric Substitution
Trigonometric substitution is a technique that uses trigonometric identities to transform integrals, especially those involving square roots, into a form that involves trigonometric functions. This can simplify the process of integration significantly.
  • It is particularly useful for integrals containing expressions like \( \sqrt{a^2 - x^2} \), \( \sqrt{a^2 + x^2} \), and \( \sqrt{x^2 - a^2} \).
  • Common substitutions include \( x = a \sin\theta \), \( x = a \tan\theta \), and \( x = a \sec\theta \).
In the exercise, the integral was transformed into a standard form that resembles a trigonometric integral, albeit indirectly. While a direct trigonometric substitution was not used, the form matched that of an integral involving the arctan function, showcasing the versatility of these methods.
Inverse Trigonometric Functions
Inverse trigonometric functions serve as the reverse of the regular trigonometric functions and are essential for evaluating certain integrals. These functions include \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \), among others.
  • Tangent Inverse Function (\(\tan^{-1}(x)\)): This function is often used in integrals that involve \(x^2 + a^2\), leading to the integral of the form \( \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C \).
  • These integrals offer a neat solution compared to dealing with complex algebraic expressions.
In the provided solution, the integral resolved to \( \frac{1}{\sqrt{2}} \tan^{-1}(\frac{u}{\sqrt{2}}) \) because it directly matched the form required for the application of inverse trigonometric functions.
Definite and Indefinite Integrals
Integrals are categorized into two types: definite and indefinite. Understanding the difference and application of each is crucial in calculus.
  • Indefinite Integrals: These represent a family of functions and include a constant \(C\), which accounts for the vertical shift. Solutions are expressed in the form \( F(x) + C \).
  • Definite Integrals: These calculate the net area under a curve between two specific limits. They are written as \( \int_{a}^{b} f(x) \, dx \) and result in a numerical value.
  • The choice between definite and indefinite usually depends upon what the problem demands, whether it is finding an area or solving a general integral equation.
The exercise deals with an indefinite integral, as seen in the solution, since the result included a constant \(C\), indicating an indefinite form.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Explain why each of the following integrals is improper. (a) \( \displaystyle \int_1^2 \frac{x}{x - 1}\ dx \) (b) \( \displaystyle \int_0^\infty \frac{1}{1 + x^3}\ dx \) (c) \( \displaystyle \int_{-\infty}^\infty x^2 e^{-x^2}\ dx \) (d) \( \displaystyle \int_0^{\frac{\pi}{4}} \cot x\ dx \)

Suppose that \( 0 \le f(t) \le Me^{at} \) and \( 0 \le f^\prime (t) \le Ke^{at} \) for \( t \ge 0 \), where \( f^\prime \) is continuous. If the Laplace transform of \( f(t) \) is \( F(s) \) and the Laplace transform of \( f^\prime (t) \) is \( G(s) \) , show that \( G(s) = s F(s) - f(0) \) \( s > a \)

If \( f(t) \) is continuous for \( t \ge 0 \), the Laplace transform of \( f \) is the function \( F \) defined by $$ F(s) = \int_0^\infty f(t) e^{-st}\ dt $$ and the domain of \( F \) is the set consisting of all numbers s for which the integral converges. Find the Laplace transforms of the following functions. (a) \( f(t) = 1 \) (b) \( f(t) = e^t \) (c) \( f(t) = t \)

Use the Comparison Theorem to determine whether the integral is convergent or divergent. \( \displaystyle \int_1^\infty \frac{x + 1}{\sqrt{x^4 - x}}\ dx \)

Determine whether each integral is convergent or divergent. Evaluate those that are convergent. \( \displaystyle \int_0^\infty \sin \theta e^{\cos \theta}\ d \theta \)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.