Textbook Question
In Exercises 1–4, show that each function y=f(x) is a solution of the accompanying differential equation.
3. y = 1/x ∫(from 1 to x) e^t/t dt, x²y' + xy = e^x
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.3.51
Evaluate the integrals in Exercises 33–54.
51. ∫ from ln(π/6) to ln(π/2) 2e^v cos(e^v) dv
Verified step by step guidance1
Identify the integral to evaluate: \(\int_{\ln(\frac{\pi}{6})}^{\ln(\frac{\pi}{2})} 2e^{v} \cos(e^{v}) \, dv\).
Use the substitution method by letting \(u = e^{v}\). Then, compute the differential: \(du = e^{v} dv\), which implies \(dv = \frac{du}{u}\).
Rewrite the integral in terms of \(u\). Note that \(e^{v} = u\), so \(2e^{v} \cos(e^{v}) dv = 2u \cos(u) \cdot \frac{du}{u} = 2 \cos(u) du\).
Change the limits of integration according to the substitution: when \(v = \ln(\frac{\pi}{6})\), \(u = e^{\ln(\frac{\pi}{6})} = \frac{\pi}{6}\); when \(v = \ln(\frac{\pi}{2})\), \(u = e^{\ln(\frac{\pi}{2})} = \frac{\pi}{2}\).
Now, the integral becomes \(\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} 2 \cos(u) \, du\). Proceed to integrate \(2 \cos(u)\) with respect to \(u\).
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
2mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definite Integrals
A definite integral calculates the net area under a curve between two specific limits. It involves evaluating the integral function at the upper and lower bounds and subtracting these values. Understanding the limits of integration and their role in the evaluation process is essential.
Recommended video:
05:43
Definition of the Definite Integral
Substitution Method
The substitution method simplifies integrals by changing variables to transform the integral into a more manageable form. It often involves identifying a function and its derivative within the integral, allowing for easier integration. This technique is crucial when dealing with composite functions like e^v inside trigonometric functions.
Recommended video:
07:33Euler's Method
Integration of Composite Functions
Integrating composite functions requires recognizing inner and outer functions and applying appropriate techniques such as substitution. For example, integrating expressions like cos(e^v) involves treating e^v as the inner function and using substitution to simplify the integral.
Recommended video:
3:48Evaluate Composite Functions - Special Cases
Related Practice
Textbook Question
Evaluate the integrals in Exercises 41–60.
47. ∫sech²(x - 1/2)dx
Textbook Question
In Exercises 57–70, use logarithmic differentiation to find the derivative of y with respect to the given independent variable.
59. y = √(t/(t+1))
Textbook Question
In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
35. y=arccsc(e^t)
Textbook Question
In Exercises 13–24, find the derivative of y with respect to the appropriate variable.
21. y = ln(cosh v) - 1/2 tanh²v
Textbook Question
Rewrite the expressions in Exercises 5–10 in terms of exponentials and simplify the results as much as you can.
6. sinh(2ln x)