/*! 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 6 Evaluate the integrals using the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 6

Evaluate the integrals using the indicated substitutions. (a) \(\int \cos 2 x \, d x ; u=2 x\) (b) \(\int x \sec ^{2} x^{2} d x ; u=x^{2}\)

Short Answer

Expert verified
(a) \(\frac{1}{2} \sin 2x + C\); (b) \(\frac{1}{2} \tan x^2 + C\).

Step by step solution

01

Understand the Substitution

For integral (a), the substitution given is \( u = 2x \). This means we will express everything in terms of \( u \), including \( dx \). Similarly, for integral (b), the substitution given is \( u = x^2 \). We will follow the substitution method to simplify each integral.
02

Calculate \( du \) for Integral (a)

For the substitution \( u = 2x \), differentiate with respect to \( x \) to find \( du \). \[ du = 2 \, dx \] Rearrange to express \( dx \): \[ dx = \frac{1}{2} \, du \]
03

Substitute in Integral (a)

Replace \( 2x \) with \( u \) in the integral: \[ \int \cos 2x \, dx = \int \cos u \, \frac{1}{2} \, du = \frac{1}{2} \int \cos u \, du \]
04

Integrate in terms of \( u \) for (a)

Integrate \( \cos u \): \[ \int \cos u \, du = \sin u \] Thus, \[ \frac{1}{2} \int \cos u \, du = \frac{1}{2} \sin u + C \]
05

Back Substitute for (a)

Replace \( u \) back with \( 2x \): \[ \frac{1}{2} \sin u + C = \frac{1}{2} \sin 2x + C \] This completes the evaluation for integral (a).
06

Calculate \( du \) for Integral (b)

For the substitution \( u = x^2 \), differentiate with respect to \( x \) to find \( du \): \[ du = 2x \, dx \] Rearrange to express \( dx \): \[ dx = \frac{1}{2x} \, du \]
07

Substitute in Integral (b)

Replace \( x^2 \) with \( u \) and \( dx \) with \( \frac{1}{2x} \, du \) in the integral: \[ \int x \sec^2 x^2 \, dx = \int x \sec^2 u \cdot \frac{1}{2x} \, du \] Simplify: \[ \int \frac{1}{2} \sec^2 u \, du = \frac{1}{2} \int \sec^2 u \, du \]
08

Integrate in terms of \( u \) for (b)

Integrate \( \sec^2 u \): \[ \int \sec^2 u \, du = \tan u \] So, \[ \frac{1}{2} \int \sec^2 u \, du = \frac{1}{2} \tan u + C \]
09

Back Substitute for (b)

Replace \( u \) back with \( x^2 \): \[ \frac{1}{2} \tan u + C = \frac{1}{2} \tan x^2 + C \]. This completes the evaluation for integral (b).

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.

Definite Integrals
Definite integrals allow us to calculate the area under a curve within a specific interval. Unlike indefinite integrals, which include a constant of integration (\(C\)), definite integrals provide a precise numerical value. When evaluating a definite integral, the process involves applying the fundamental theorem of calculus. This theorem establishes a connection between differentiation and integration. By evaluating the antiderivative at the upper and lower bounds of the interval, we can determine the exact area. In many cases, especially involving complex functions, substitution methods are necessary to simplify the integration.

The process of substitution involves changing the variables to make the integration process easier. If you've learned integration techniques, these substitutions might remind you of changing variables in integrals, a fundamental skill in calculus. With definite integrals, remember that after integration, it's essential to plug back the original limits into the antiderivative when reversing the substitution. This is a key step to getting the correct numerical result.
Integration Techniques
Integration techniques are essential tools for solving complex integrals. One of the most powerful methods is substitution, often referred to as \( u \)-substitution. This technique involves choosing a part of the integral to substitute with a new variable, \( u \), which simplifies the integration process. Think of it as reversing the chain rule for differentiation.

To apply \( u \)-substitution effectively:
  • Select a part of the integrand that will simplify the integral when substituted.
  • Differentiate this part to express \( du \), ensuring that all pieces of the integral represent \( du \) accurately.
  • Replace the original variables and \( dx \) in the integral with \( u \) and \( du \), respectively.
  • Perform the integral in terms of \( u \), and then substitute back in the original variable after finding the antiderivative.
For example, in problem (a), we substituted \( 2x \) with \( u \) to rewrite and integrate \( \cos 2x \) as \( \cos u \). After integration, you revert \( u \) back to \( 2x \). This illustrates how substitution can transform a challenging integration problem into a manageable one.
Trigonometric Integrals
Trigonometric integrals often involve expressions like \( \sin, \cos, \tan \), or \( \sec \). These integrals take advantage of derivatives and antiderivatives associated with trigonometric functions. Integration of trigonometric functions can often be tricky, requiring advanced techniques beyond basic integration rules.

Some useful points to remember for trigonometric integrals include:
  • The integral of \( \sin u \) is \( -\cos u + C \).
  • The integral of \( \cos u \) is \( \sin u + C \).
  • The integral of \( \sec^2 u \) is \( \tan u + C \).
In the examples given, we're using the identities of these trigonometric functions. For instance, in integral (b), using the identity \( \sec^2 u \) plays a critical role. After performing the \( u \)-substitution, integrating \( \sec^2 u \) becomes straightforward because its antiderivative is a standard result, \( \tan u \).

Thus, mastering these basic trigonometric identities and their integrals will enable you to tackle more complex integrals involving trigonometric functions with confidence.

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 the Fundamental Theorem of Calculus may be applied without modification to definite integrals in which the lower limit of integration is greater than or equal to the upper limit of integration.

(a) Show that the area under the graph of \(y=x^{3}\) and over the interval \([0, b]\) is \(b^{4} / 4\) (b) Find a formula for the area under \(y=x^{3}\) over the interval \([a, b],\) where \(a \geq 0\)

(a) Prove that if \(f\) is an odd function, then $$\int_{-a}^{a} f(x) d x=0$$ and give a geometric explanation of this result. [Hint: One way to prove that a quantity \(q\) is zero is to show that \(q=-q .]\) (b) Prove that if \(f\) is an even function, then $$\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x$$ and give a geometric explanation of this result. [Hint: Split the interval of integration from \(-a\) to \(a\) into two parts at 0.]

$$\begin{aligned} \sum_{k=1}^{4}\left[(k+1)^{3}-k^{3}\right]=&\left[5^{3}-4^{3}\right]+\left[4^{3}-3^{3}\right] \\\ &+\left[3^{3}-2^{3}\right]+\left[2^{3}-1^{3}\right] \\ =& 5^{3}-1^{3}=124 \end{aligned}$$ For convenience, the terms are listed in reverse order. Note how cancellation allows the entire sum to collapse like a telescope. A sum is said to telescope when part of each term cancels part of an adjacent term, leaving only portions of the first and last terms uncanceled. Evaluate the telescoping sums in these exercises. $$\sum_{k=2}^{20}\left(\frac{1}{k^{2}}-\frac{1}{(k-1)^{2}}\right)$$

Evaluate each limit by interpreting it as a Riemann sum in which the given interval is divided into \(n\) subintervals of equal width. $$\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{n}{n^{2}+k^{2}} ;[0,1]$$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Decision 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.