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...
Related questions
Question
![### Problem 16: Integration of Trigonometric Function
**Problem Statement:**
Evaluate the integral
\[ \int \tan^5 x \sec^4 x \, dx. \]
**Solution:**
Let's solve this integral by using substitution. We will use \( u = \tan x \), which implies \( du = \sec^2 x \, dx \).
Rewrite the integral in terms of \( u \):
\[ \int \tan^5 x \sec^4 x \, dx. \]
Since \( u = \tan x \), we have \( u^5 = \tan^5 x \). Also, \( \sec^4 x = \sec^2 x \cdot \sec^2 x = \left(\frac{du}{dx}\right) \cdot \sec^2 x \) and \( dx = \frac{1}{\sec^2 x} \, du \).
Therefore, the integral becomes:
\[ \int u^5 \sec^2 x \, du. \]
However, this requires modifying. Since \( du = \sec^2 x \, dx \), let's write everything in terms of \( u \):
\[ \int u^5 \sec^2 x \cdot \sec^2 x \, dx \]
\[ = \int u^5 \cdot \sec^4 x \cdot dx. \]
Note that \( \sec^2 x = 1 + \tan^2 x \), leading us to rewrite secant squared terms using \( u \):
\[ \sec^4 x = (1 + u^2)^2. \]
So we have:
\[ \int u^5 (1 + u^2)^2 \, du. \]
Now, expand \( (1 + u^2)^2 \):
\[ (1 + u^2)^2 = 1 + 2u^2 + u^4. \]
The integral thus becomes:
\[ \int u^5 (1 + 2u^2 + u^4) \, du. \]
Distribute \( u^5 \) through the polynomial:
\[ = \int (u^5 + 2u^7 + u^9) \, du. \]
Integrate each term separately:
\[ \int u^5 \, du + 2 \int u^7 \, du +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcf5ca60f-6906-44bf-a492-477ae53fc791%2F4c03b569-7b94-49e4-a13e-3c4992e26634%2Ftacyhp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 16: Integration of Trigonometric Function
**Problem Statement:**
Evaluate the integral
\[ \int \tan^5 x \sec^4 x \, dx. \]
**Solution:**
Let's solve this integral by using substitution. We will use \( u = \tan x \), which implies \( du = \sec^2 x \, dx \).
Rewrite the integral in terms of \( u \):
\[ \int \tan^5 x \sec^4 x \, dx. \]
Since \( u = \tan x \), we have \( u^5 = \tan^5 x \). Also, \( \sec^4 x = \sec^2 x \cdot \sec^2 x = \left(\frac{du}{dx}\right) \cdot \sec^2 x \) and \( dx = \frac{1}{\sec^2 x} \, du \).
Therefore, the integral becomes:
\[ \int u^5 \sec^2 x \, du. \]
However, this requires modifying. Since \( du = \sec^2 x \, dx \), let's write everything in terms of \( u \):
\[ \int u^5 \sec^2 x \cdot \sec^2 x \, dx \]
\[ = \int u^5 \cdot \sec^4 x \cdot dx. \]
Note that \( \sec^2 x = 1 + \tan^2 x \), leading us to rewrite secant squared terms using \( u \):
\[ \sec^4 x = (1 + u^2)^2. \]
So we have:
\[ \int u^5 (1 + u^2)^2 \, du. \]
Now, expand \( (1 + u^2)^2 \):
\[ (1 + u^2)^2 = 1 + 2u^2 + u^4. \]
The integral thus becomes:
\[ \int u^5 (1 + 2u^2 + u^4) \, du. \]
Distribute \( u^5 \) through the polynomial:
\[ = \int (u^5 + 2u^7 + u^9) \, du. \]
Integrate each term separately:
\[ \int u^5 \, du + 2 \int u^7 \, du +
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 1 images
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
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
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
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
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning