/*! 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 54 In Exercises \(17-56,\) find the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 54

In Exercises \(17-56,\) find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int\left(1-\cot ^{2} x\right) d x$$

Short Answer

Expert verified
The most general antiderivative is \(2x - \cot x + C\).

Step by step solution

01

Recognize Trigonometric Identity

We start by recognizing a trigonometric identity. Recall that \( \cot^2 x = \csc^2 x - 1 \), which means \( 1 - \cot^2 x = 1 - (\csc^2 x - 1) = 2 - \csc^2 x \). Therefore, the integral can be rewritten as \( \int (2 - \csc^2 x) \, dx \).
02

Split the Integral

Split the integral \( \int (2 - \csc^2 x) \, dx \) into two separate integrals: \( \int 2 \, dx - \int \csc^2 x \, dx \).
03

Integrate Each Term

Now integrate each term separately. The integral \( \int 2 \, dx = 2x + C_1 \), where \( C_1 \) is a constant, and \( \int \csc^2 x \, dx = -\cot x + C_2 \), where \( C_2 \) is another constant.
04

Combine Results

Combine the results of the integrals: \( 2x + C_1 - \cot x + C_2 \). This simplifies to \( 2x - \cot x + C \), where \( C = C_1 + C_2 \), as both are arbitrary constants and can be combined into one.
05

Check by Differentiation

Differentiate the result \( 2x - \cot x + C \) with respect to \( x \). The derivative is \( 2 + \csc^2 x \). Recalling \( \csc^2 x = 1 + \cot^2 x \), we substitute to get \( 2 - (1 + \cot^2 x) = 1 - \cot^2 x \), confirming our original integrand.

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.

Trigonometric Identities
Trigonometric identities are essential tools in calculus, as they allow for the simplification of complex expressions. They are equations that relate various trigonometric functions to one another. In the exercise, we use the identity \( \cot^2 x = \csc^2 x - 1 \). This identity helps transform the original integral expression \( 1 - \cot^2 x \) into a more manageable form. By rewriting it as \( 2 - \csc^2 x \), we simplify the integration process. Recognizing and applying such identities can make an integral easier to solve and are pivotal in solving trigonometric integrals.
Indefinite Integrals
Indefinite integrals, also known as antiderivatives, are a fundamental concept in calculus. They represent the family of functions whose derivative is the integrand. When solving indefinite integrals, the result includes an integration constant, often denoted as \( C \), because integration is the reverse process of differentiation, which eliminates constants.
In the given exercise, integrating the function \( 2 - \csc^2 x \) involves finding the antiderivatives of \( 2 \) and \( \csc^2 x \). Each term is integrated separately and the final solution is expressed as \( 2x - \cot x + C \), where \( C \) is the sum of all constant terms from each integration step.
Differentiation
Differentiation is the process of finding the rate at which a function changes. It's essentially the reverse of integration. To verify the correctness of an antiderivative, students can differentiate their result to see if it returns the original integrand. This step is crucial to ensure the accuracy of the integral calculation.
For instance, in this exercise, after calculating the antiderivative \( 2x - \cot x + C \), differentiating it results in \( 2 + \csc^2 x \), which simplifies back to the original function \( 1 - \cot^2 x \). This checks the correctness of the integral solution.
Integration Techniques
Integration techniques are strategies used to simplify and solve integrals. Common techniques include integration by parts, substitution, and recognizing standard integral forms, among others. Selecting the appropriate technique largely depends on the form of the function you're integrating.
In this exercise, the integration strategy involved recognizing a trigonometric identity to simplify the initial integrand. Then, splitting the integral into simpler parts allowed straightforward integration of each component. Understanding various techniques enables more efficient problem-solving and is key to mastering calculus.

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

How we cough \begin{equation} \begin{array}{l}{\text { a. When we cough, the trachea (windpipe) contracts to increase }} \\ {\text { the velocity of the air going out. This raises the questions of }} \\ {\text { how much it should contract to maximize the velocity and }} \\ {\text { whether it really contracts that much when we cough. }}\\\\{\text { Under reasonable assumptions about the elasticity of the }} \\ {\text { tracheal wall and about how the air near the wall is slowed by }} \\ {\text { friction, the average flow velocity } v \text { can be modeled by the }} \\ {\text { equation }}\end{array} \end{equation} $$v=c\left(r_{0}-r\right) r^{2} \mathrm{cm} / \mathrm{sec}, \quad \frac{r_{0}}{2} \leq r \leq r_{0}$$ \begin{equation} \begin{array}{l}{\text { where } r_{0} \text { is the rest radius of the trachea in centimeters and }} \\ {c \text { is a positive constant whose value depends in part on the }} \\ {\text { length of the trachea. }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { Show that } v \text { is greatest when } r=(2 / 3) r_{0} \text { ; that is, when }} \\ {\text { the trachea is about } 33 \% \text { contracted. The remarkable fact is }} \\ {\text { that } X \text { -ray photographs confirm that the trachea contracts }} \\ {\text { about this much during a cough. }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { b. Take } r_{0} \text { to be } 0.5 \text { and } c \text { to be } 1 \text { and graph } v \text { over the interval }} \\ {0 \leq r \leq 0.5 . \text { Compare what you see with the claim that } v \text { is }} \\ {\text { at a maximum when } r=(2 / 3) r_{0} .}\end{array} \end{equation}

Verify the formulas in Exercises 57-62 by differentiation. $$\int \sec ^{2}(5 x-1) d x=\frac{1}{5} \tan (5 x-1)+C$$

Show that if \(r(x)=6 x\) and \(c(x)=x^{3}-6 x^{2}+15 x\) are your revenue and cost functions, then the best you can do is break even (have revenue equal cost).

Solve the initial value problems in Exercises \(71-90\) $$\frac{d^{2} s}{d t^{2}}=\frac{3 t}{8} ; \quad\left.\frac{d s}{d t}\right|_{t=4}=3, \quad s(4)=4$$

Liftoff from Earth A rocket lifts off the surface of Earth with a constant acceleration of 20 \(\mathrm{m} / \mathrm{sec}^{2}\) . How fast will the rocket be going 1 min later?
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

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.