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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![### Problem 9: Trigonometric Substitution
Use a trigonometric substitution to show that:
\[ \int \frac{dx}{(1+x^2)^{3/2}} = \frac{x}{\sqrt{1+x^2}} + C. \]
Be sure to clearly show all steps.
---
**Steps for the Solution:**
1. **Substitution:**
Let \( x = \tan(\theta) \). Therefore, \( dx = \sec^2(\theta) d\theta \).
2. **Replacement in the Integral:**
Substitute \( x = \tan(\theta) \) into \( (1 + x^2) \):
\[
1 + x^2 = 1 + \tan^2(\theta) = \sec^2(\theta).
\]
Thus, the integral becomes:
\[
\int \frac{\sec^2(\theta) d\theta}{(\sec^2(\theta))^{3/2}}.
\]
3. **Simplifying the Integral:**
Simplify \( (\sec^2(\theta))^{3/2} \):
\[
(\sec^2(\theta))^{3/2} = (\sec(\theta))^3.
\]
Therefore, the integral becomes:
\[
\int \frac{\sec^2(\theta) d\theta}{\sec^3(\theta)} = \int \frac{d\theta}{\sec(\theta)} = \int \cos(\theta) d\theta.
\]
4. **Integrate:**
\[
\int \cos(\theta) d\theta = \sin(\theta) + C.
\]
5. **Back-Substitution:**
Recall \( x = \tan(\theta) \). We need to express \(\sin(\theta)\) in terms of \(x\):
\[
\sin(\theta) = \frac{\tan(\theta)}{\sqrt{1+\tan^2(\theta)}} = \frac{x}{\sqrt{1+x^2}}.
\]
6. **Final Answer:**
\[
\sin(\theta) + C = \frac{x}{\sqrt{1+x^2}} + C.
\]
Thus, we have shown that:
\[
\int \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff015db4-0da8-42fc-9d60-01a672218c35%2Fbedcf447-50b7-4f05-84db-bbf83266a479%2Fcwfgxvr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 9: Trigonometric Substitution
Use a trigonometric substitution to show that:
\[ \int \frac{dx}{(1+x^2)^{3/2}} = \frac{x}{\sqrt{1+x^2}} + C. \]
Be sure to clearly show all steps.
---
**Steps for the Solution:**
1. **Substitution:**
Let \( x = \tan(\theta) \). Therefore, \( dx = \sec^2(\theta) d\theta \).
2. **Replacement in the Integral:**
Substitute \( x = \tan(\theta) \) into \( (1 + x^2) \):
\[
1 + x^2 = 1 + \tan^2(\theta) = \sec^2(\theta).
\]
Thus, the integral becomes:
\[
\int \frac{\sec^2(\theta) d\theta}{(\sec^2(\theta))^{3/2}}.
\]
3. **Simplifying the Integral:**
Simplify \( (\sec^2(\theta))^{3/2} \):
\[
(\sec^2(\theta))^{3/2} = (\sec(\theta))^3.
\]
Therefore, the integral becomes:
\[
\int \frac{\sec^2(\theta) d\theta}{\sec^3(\theta)} = \int \frac{d\theta}{\sec(\theta)} = \int \cos(\theta) d\theta.
\]
4. **Integrate:**
\[
\int \cos(\theta) d\theta = \sin(\theta) + C.
\]
5. **Back-Substitution:**
Recall \( x = \tan(\theta) \). We need to express \(\sin(\theta)\) in terms of \(x\):
\[
\sin(\theta) = \frac{\tan(\theta)}{\sqrt{1+\tan^2(\theta)}} = \frac{x}{\sqrt{1+x^2}}.
\]
6. **Final Answer:**
\[
\sin(\theta) + C = \frac{x}{\sqrt{1+x^2}} + C.
\]
Thus, we have shown that:
\[
\int \
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