/*! 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 27 Evaluate the integral. \(\int_... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 27

Evaluate the integral. \(\int_{0}^{1 / \sqrt{3}} \frac{t^{2}-1}{t^{4}-1} d t\)

Short Answer

Expert verified
The integral evaluates to \(\pi/6\).

Step by step solution

01

Factor the Denominator

The denominator of the integrand is \(t^4 - 1\). This expression can be factored using the difference of squares method: \(t^4 - 1 = (t^2 - 1)(t^2 + 1)\). So, \(\frac{t^2 - 1}{t^4 - 1}\) can be rewritten as \(\frac{t^2 - 1}{(t^2 - 1)(t^2 + 1)}\).
02

Simplify the Fraction

In the expression \(\frac{t^2 - 1}{(t^2 - 1)(t^2 + 1)}\), cancel the \(t^2 - 1\) terms to simplify it to \(\frac{1}{t^2 + 1}\).
03

Integrate the Simplified Expression

Now that we have \(\int \frac{1}{t^2 + 1} \, dt\), recognize that \(\frac{1}{t^2 + 1}\) is the derivative of the arctangent function. Therefore, the antiderivative is \(\tan^{-1}(t) + C\).
04

Evaluate the Definite Integral

Evaluate \(\left[\tan^{-1}(t)\right]_{0}^{1/\sqrt{3}}\). This means calculating \(\tan^{-1}(1/\sqrt{3}) - \tan^{-1}(0)\).
05

Solve the Expression

\(\tan^{-1}(0) = 0\) because the tangent of 0 is 0. For \(\tan^{-1}(1/\sqrt{3})\), recognize that \(\tan(\pi/6) = 1/\sqrt{3}\), so \(\tan^{-1}(1/\sqrt{3}) = \pi/6\). Therefore, the integral evaluates to \(\pi/6\).

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.

Factorization
Factorization is a crucial step when working with integrals, especially with polynomial expressions. It involves breaking down a complex expression into simpler factors, making it easier to handle and solve. For example, in the expression \( t^4 - 1 \), we can apply the difference of squares. This technique works because any term of the form \( a^2 - b^2 \) can be factored into \( (a-b)(a+b) \).
This is helpful because it simplifies the original expression to \( (t^2 - 1)(t^2 + 1) \), providing a pathway to reduce the complexity of the integral.
  • Step 1: Notice that \( t^4 - 1 \) mimics the difference of squares \( (t^2)^2 - 1^2 \).
  • Step 2: Factor it into \( (t^2 - 1)(t^2 + 1) \), which further simplifies calculations.
By factorizing, you reduce expressions to their basic components, which is invaluable for both integration and algebraic manipulation.
Definite Integral
A definite integral calculates the area under a curve between certain bounds. It's written in the notation \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration. In the case of our problem, these bounds are 0 and \( 1/\sqrt{3} \).
Definite integrals help us find exact numerical values rather than indefinite integrals that provide a general form with a constant of integration \( C \).
  • Understanding the bounds: \( t = 0 \) to \( t = 1/\sqrt{3} \)
  • Evaluate the simplified integral \( \int \frac{1}{t^2+1} \, dt \) over these bounds.
In summary, converting a function into a definite integral allows you to determine specific numerical results, which is useful in a range of mathematical contexts, including physics and engineering.
Arctangent Function
The arctangent function, denoted as \( \tan^{-1}(x) \) or \( \arctan(x) \), is the inverse of the tangent function. It provides the angle whose tangent is a given number. This function is vital in the integration process.
In the problem, after simplification, we arrive at integrating \( \frac{1}{t^2 + 1} \), recognized as the derivative of the arctangent. Thus, its antiderivative is \( \tan^{-1}(t) + C \).
  • Recognize \( \int \frac{1}{t^2 + 1} \, dt = \tan^{-1}(t) + C \)
  • Use the arctangent to evaluate the integral between bounds.
  • For example, \( \tan^{-1}(1/\sqrt{3}) \) results in \( \pi/6 \) because \( \tan(\pi/6) = 1/\sqrt{3} \).
Understanding the role of arctangent in integration enriches your mathematical toolkit, allowing you to effectively solve integrals involving inverse trigonometric functions.

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

Determine whether each integral is convergent or divergent. Evaluate those that are convergent. \(\int_{0}^{\infty} \frac{e^{x}}{e^{2 x}+3} d x\)

\(35-36\) Verify by differentiation that the formula is correct. \(\int \cos ^{3} x d x=\sin x-\frac{1}{3} \sin ^{3} x+C\)

Sketch the region and find its area (if the area is finite). \(S=\\{(x, y) | x \leqslant 1,0 \leqslant y \leqslant e^{x}\\}\)

Evaluate the integral, given that \(\int_{0}^{\infty} e^{-x^{2}} d x=\frac{1}{2} \sqrt{\pi}\) \(\int_{0}^{\infty} e^{-x^{2} / 2} d x\)

Use a computer algebra system to evaluate the integral. Compare the answer with the result of using tables. If the answers are not the same, show that they are equivalent. \(\int \tan ^{5} x d x\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Geometry

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.