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...
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![### Evaluate the Integral
Given the integral to evaluate:
\[ \int_{0}^{\frac{\pi}{10}} 11 \sqrt{1 + \cos{10x}} \, dx \]
Let's solve this step by step.
### Step-by-Step Solution:
1. **Substitution**:
Let \( u = 10x \). Then \( du = 10 \, dx \) or \( dx = \frac{du}{10} \).
2. **Change Limits**:
When \( x = 0 \), \( u = 0 \).
When \( x = \frac{\pi}{10} \), \( u = \pi \).
3. **Transform the Integral**:
Substitute the variables and limits:
\[ \int_{0}^{\pi} 11 \sqrt{1 + \cos{u}} \cdot \frac{du}{10} \]
\[ = \frac{11}{10} \int_{0}^{\pi} \sqrt{1 + \cos{u}} \, du \]
4. **Simplify**:
Use trigonometric identities to simplify \( \sqrt{1 + \cos{u}} \):
\[ 1 + \cos{u} = 2 \cos^2{\left(\frac{u}{2}\right)} \]
Therefore,
\[ \sqrt{1 + \cos{u}} = \sqrt{2} \left| \cos{\left(\frac{u}{2}\right)} \right| \]
Since \( u \) ranges from 0 to \( \pi \), \( \cos{\left(\frac{u}{2}\right)} \) is non-negative. Thus,
\[ \sqrt{1 + \cos{u}} = \sqrt{2} \cos{\left(\frac{u}{2}\right)} \]
5. **Integrate**:
Substitute back in the integral:
\[ = \frac{11}{10} \int_{0}^{\pi} \sqrt{2} \cos{\left(\frac{u}{2}\right)} \, du \]
\[ = \frac{11 \sqrt{2}}{10} \int_{0}^{\pi} \cos{\left(\frac{u}{2}\right)}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1c5e517d-57db-4b65-baa5-8a292d97b7f5%2Fcfe8e2bd-f29b-4148-b321-c35ea76e7cdb%2F6mqmh86_processed.png&w=3840&q=75)
Transcribed Image Text:### Evaluate the Integral
Given the integral to evaluate:
\[ \int_{0}^{\frac{\pi}{10}} 11 \sqrt{1 + \cos{10x}} \, dx \]
Let's solve this step by step.
### Step-by-Step Solution:
1. **Substitution**:
Let \( u = 10x \). Then \( du = 10 \, dx \) or \( dx = \frac{du}{10} \).
2. **Change Limits**:
When \( x = 0 \), \( u = 0 \).
When \( x = \frac{\pi}{10} \), \( u = \pi \).
3. **Transform the Integral**:
Substitute the variables and limits:
\[ \int_{0}^{\pi} 11 \sqrt{1 + \cos{u}} \cdot \frac{du}{10} \]
\[ = \frac{11}{10} \int_{0}^{\pi} \sqrt{1 + \cos{u}} \, du \]
4. **Simplify**:
Use trigonometric identities to simplify \( \sqrt{1 + \cos{u}} \):
\[ 1 + \cos{u} = 2 \cos^2{\left(\frac{u}{2}\right)} \]
Therefore,
\[ \sqrt{1 + \cos{u}} = \sqrt{2} \left| \cos{\left(\frac{u}{2}\right)} \right| \]
Since \( u \) ranges from 0 to \( \pi \), \( \cos{\left(\frac{u}{2}\right)} \) is non-negative. Thus,
\[ \sqrt{1 + \cos{u}} = \sqrt{2} \cos{\left(\frac{u}{2}\right)} \]
5. **Integrate**:
Substitute back in the integral:
\[ = \frac{11}{10} \int_{0}^{\pi} \sqrt{2} \cos{\left(\frac{u}{2}\right)} \, du \]
\[ = \frac{11 \sqrt{2}}{10} \int_{0}^{\pi} \cos{\left(\frac{u}{2}\right)}
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