sec 3 x tan x (sec ³x sec³x secrx S see of secxu S u ли secx u du dx let a = tan x du = sec² x dx dx= 1 Sec² x du -How to get rid of this x?

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
icon
Related questions
Question
How can I get rid of the highlighted x in this u-substitution problem?
## Integration by Substitution Example

This example demonstrates how to use substitution for the integration of \( \int \sec^3{x} \tan{x} \, dx \).

### Step-by-Step Solution:

1. Begin with the integral:
    \[
    \int \sec^3{x} \tan{x} \, dx
    \]

2. Introduce a substitution:

    Let \( u = \tan{x} \) which implies that \( du = \sec^2{x} \, dx \).

    Rearranging for \( dx \), we get:
    \[
    dx = \frac{1}{\sec^2{x}} \, du
    \]

3. Substitute \( u \) and \( dx \) into the integral:

    Since \( \sec^3{x} = \sec{x} \cdot \sec^2{x} \), the integral becomes:
    \[
    \int \frac{\sec^3{x} \cdot u}{\sec^2{x}} \, du
    \]

4. Simplify the expression:

    \[
    \int \sec{x} \cdot u \, du
    \]

    At this point, there is a remaining \( \sec{x} \) term, despite substituting \( dx \).

5. Identifying the problem:

    There is a highlighted text in the image indicating: 

    *How to get rid of this \( x \)?*
    
    To handle the extra \(\sec{x}\) term in \( \sec{x} \cdot u \, du \), note that \(\sec{x} = \sqrt{u^2 + 1}\) because \(\sec^2{x} = 1 + \tan^2{x}\).

### Summary

To complete the integral, substitute \(\sec{x}\) using the relation to \( u \):

\[
\sec{x} = \sqrt{1 + u^2}
\]

Transforming the integral fully in terms of \( u \):

\[
\int \sqrt{1 + u^2} \, u \, du
\]

The final result requires further simplification or a different technique to integrate the expression completely.
Transcribed Image Text:## Integration by Substitution Example This example demonstrates how to use substitution for the integration of \( \int \sec^3{x} \tan{x} \, dx \). ### Step-by-Step Solution: 1. Begin with the integral: \[ \int \sec^3{x} \tan{x} \, dx \] 2. Introduce a substitution: Let \( u = \tan{x} \) which implies that \( du = \sec^2{x} \, dx \). Rearranging for \( dx \), we get: \[ dx = \frac{1}{\sec^2{x}} \, du \] 3. Substitute \( u \) and \( dx \) into the integral: Since \( \sec^3{x} = \sec{x} \cdot \sec^2{x} \), the integral becomes: \[ \int \frac{\sec^3{x} \cdot u}{\sec^2{x}} \, du \] 4. Simplify the expression: \[ \int \sec{x} \cdot u \, du \] At this point, there is a remaining \( \sec{x} \) term, despite substituting \( dx \). 5. Identifying the problem: There is a highlighted text in the image indicating: *How to get rid of this \( x \)?* To handle the extra \(\sec{x}\) term in \( \sec{x} \cdot u \, du \), note that \(\sec{x} = \sqrt{u^2 + 1}\) because \(\sec^2{x} = 1 + \tan^2{x}\). ### Summary To complete the integral, substitute \(\sec{x}\) using the relation to \( u \): \[ \sec{x} = \sqrt{1 + u^2} \] Transforming the integral fully in terms of \( u \): \[ \int \sqrt{1 + u^2} \, u \, du \] The final result requires further simplification or a different technique to integrate the expression completely.
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning