/*! 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 4 Evaluate the integrals by making... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 4

Evaluate the integrals by making appropriate \(u\) -substitutions and applying the formulas reviewed in this section. $$\int 4 x \tan \left(x^{2}\right) d x$$

Short Answer

Expert verified
The integral is \(2 \log|\sec(x^2)| + C\).

Step by step solution

01

Choose the Appropriate Substitution

To simplify the integral, we start by identifying a substitution that will make the integration easier. Notice that the argument of the tangent function is \(x^2\). Therefore, set \(u = x^2\).
02

Differentiate to Find du

Differentiate \(u = x^2\) with respect to \(x\) to find \(du\). We get \(\frac{du}{dx} = 2x\). This rearranges to \(du = 2x \, dx\).
03

Solve for dx

From the expression \(du = 2x \, dx\), solve for \(dx\). We get \(dx = \frac{du}{2x}\).
04

Substitute into the Integral

Substitute \(u = x^2\) and \(dx = \frac{du}{2x}\) into the integral: \[\int 4x \tan(x^2) \, dx = \int 4x \tan(u) \frac{du}{2x}\].
05

Simplify the Integral

Simplify the integral: \[\int 4x \tan(u) \frac{du}{2x} = \int 2 \tan(u) \, du\] by canceling \(x\) and simplifying the constant factor.
06

Integrate with respect to u

The integral \(\int 2 \tan(u) \, du\) can be simplified using the antiderivative formula for \(\tan(u)\), which is \(\log|\sec(u)|\). Therefore, \(\int 2 \tan(u) \, du = 2 \log|\sec(u)| + C\), where \(C\) is the integration constant.
07

Substitute Back for x

Replace \(u\) with \(x^2\) to go back to the variable \(x\). The final answer is \(2 \log|\sec(x^2)| + C\).

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.

u-substitution
U-substitution is a technique used in integral calculus to simplify the process of finding antiderivatives. The main idea is to substitute a part of the integrand with a new variable, usually denoted as \( u \). This replacement makes the integral easier to evaluate. Here's how you can effectively apply this method:

  • Identify a section of the integrand that, when substituted, will simplify the integral process. Typically, this involves parts that are difficult to integrate directly.
  • Assign \( u \) to this section. For example, if you have \( x^2 \) as part of your integrand, consider setting \( u = x^2 \).
  • Differentiate \( u \) with respect to \( x \) to find \( du \). This helps in replacing \( dx \) in the integral. Following the example, \( du = 2x \, dx \).
  • Rearrange the expression for \( du \) to solve for \( dx \). This might involve dividing or multiplying both sides by constants or expressions in \( x \).
By introducing \( u \), the integral can now possibly turn into a more straightforward form, often appearing in standard integrals' tables, making it easier to solve.
integration techniques
Integration techniques are strategies used to find integrals, particularly when basic integration formulas aren't directly applicable. Various methods exist, with u-substitution being one of the most common:

  • U-Substitution: As discussed, replaces a complex part of the function with a variable \( u \) to simplify the integral.
  • Integration by Parts: Useful for products of functions, it is based on the product rule for differentiation. It transforms an integral of products into simpler parts.
  • Partial Fraction Decomposition: Used for rational functions, it expresses the integrand as a sum of simpler fractions that are easier to integrate.
  • Trigonometric Integrals and Substitution: Applies to integrands involving trigonometric functions, often simplifying the integral using trigonometric identities or substitution methods.
All these techniques aim to transform a difficult integral into one that is easier to evaluate using known formulas or simpler algebra. The choice of method depends on the form of the integrand.
antiderivative
The antiderivative, also known as the indefinite integral, is a fundamental concept in calculus. It involves finding a function whose derivative matches the given function. Recognizing antiderivatives is crucial for solving integrals:

  • An antiderivative of a function \( f(x) \) is a function \( F(x) \) such that \( F'(x) = f(x) \).
  • Notation: The antiderivative is represented as \( \int f(x) \, dx = F(x) + C \), where \( C \) is the constant of integration, reflecting the multiple possible vertical shifts of \( F(x) \).
  • While basic rules exist, such as the power rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), more complex functions require different integration techniques.
In the case of the exercise, the antiderivative formula for \( \tan(u) \) was used, resulting in \( 2 \, \log|\sec(u)| + C \). Identifying and applying the correct antiderivative formula are key steps in integrating functions.
definite integrals
Definite integrals are used to calculate the net area under a curve between two specific points. They provide not just the antiderivative, but also an evaluated result:

  • The definite integral of a function \( f(x) \) from \( a \) to \( b \) is denoted as \( \int_{a}^{b} f(x) \, dx \).
  • The process typically involves finding the antiderivative \( F(x) \) of \( f(x) \) and evaluating it at the endpoints: \( F(b) - F(a) \).
  • It's crucial to ensure the continuity of \( f(x) \) over the interval \([a, b]\) for the integral to be defined.
This concept is fundamental in calculus for solving problems related to areas, accumulated quantities, and even physics for motion-related applications. While the exercise provided involves an indefinite integral, transitioning to definite integrals involves extra evaluation steps but retains the foundational concept of calculating integrals.

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

Evaluate the integral. $$\int \frac{2 x^{5}-x^{3}-1}{x^{3}-4 x} d x$$

Find a substitution that can be used to integrate rational functions of \(\sinh x\) and \(\cosh x\) and use your substitution to evaluate $$\int \frac{d x}{2 \cosh x+\sinh x}$$ without expressing the integrand in terms of \(e^{x}\) and \(e^{-x}.\)

(a) Make an appropriate \(u\) -substitution of the form \(u=x^{1 / n}\) \(u=(x+a)^{1 / n},\) or \(u=x^{n},\) and then use the Endpaper Integral Table to evaluate the integral. (b) If you have a CAS, use it to evaluate the integral, and then confirm that the result is equivalent to the one that you found in part (a). $$\int \frac{1}{x \sqrt{x^{3}-1}} d x$$

(a) Show that \(\int \csc x \, d x=-\ln |\csc x+\cot x|+c\) (b) Show that the result in part (a) can also be written as $$ \int \csc x \, d x=\ln |\csc x-\cot x|+C $$ and $$ \int \csc x \, d x=\ln \left|\tan \frac{1}{2} x\right|+C $$

Evaluate the integrals that converge. $$\int_{0}^{\pi / 4} \frac{\sec ^{2} x}{1-\tan x} d x$$
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Statistics

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.