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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Am I doing the u-substitution correctly?
![### Problem 16: Integration of \( \frac{\sin(\sqrt{x})}{\sqrt{x}} \)
To solve the integral \( \int \frac{\sin(\sqrt{x})}{\sqrt{x}} \, dx \), we can use substitution. Let's follow through the steps:
1. **Substitution**:
Let \( u = \sqrt{x} \).
Then, \( u^2 = x \).
Differentiating both sides with respect to \( x \),
\[
du = \frac{1}{2} x^{-\frac{1}{2}} \, dx = \frac{1}{2} u^{-1} \, dx
\]
This implies:
\[
dx = 2u \, du
\]
2. **Rewriting the Integral**:
Substitute \( u = \sqrt{x} \) and \( dx = 2u \, du \),
\[
\int \frac{\sin(\sqrt{x})}{\sqrt{x}} \, dx = \int \frac{\sin(u)}{u} \cdot 2u \, du
\]
The \( u \)'s cancel out, leaving:
\[
\int 2 \sin(u) \, du
\]
3. **Integration**:
Integrate \( 2 \sin(u) \),
\[
\int 2 \sin(u) \, du = -2 \cos(u) + C
\]
where \( C \) is the constant of integration.
4. **Substitute Back \( u = \sqrt{x} \)**:
\[
-2 \cos(u) + C = -2 \cos(\sqrt{x}) + C
\]
Thus, the integral \( \int \frac{\sin(\sqrt{x})}{\sqrt{x}} \, dx \) evaluates to:
\[
-2 \cos(\sqrt{x}) + C
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5362cbea-9427-4b2b-99cb-57a208edd853%2F31adab66-717a-4252-b38c-a377e44e5a4e%2Fohsjhlf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 16: Integration of \( \frac{\sin(\sqrt{x})}{\sqrt{x}} \)
To solve the integral \( \int \frac{\sin(\sqrt{x})}{\sqrt{x}} \, dx \), we can use substitution. Let's follow through the steps:
1. **Substitution**:
Let \( u = \sqrt{x} \).
Then, \( u^2 = x \).
Differentiating both sides with respect to \( x \),
\[
du = \frac{1}{2} x^{-\frac{1}{2}} \, dx = \frac{1}{2} u^{-1} \, dx
\]
This implies:
\[
dx = 2u \, du
\]
2. **Rewriting the Integral**:
Substitute \( u = \sqrt{x} \) and \( dx = 2u \, du \),
\[
\int \frac{\sin(\sqrt{x})}{\sqrt{x}} \, dx = \int \frac{\sin(u)}{u} \cdot 2u \, du
\]
The \( u \)'s cancel out, leaving:
\[
\int 2 \sin(u) \, du
\]
3. **Integration**:
Integrate \( 2 \sin(u) \),
\[
\int 2 \sin(u) \, du = -2 \cos(u) + C
\]
where \( C \) is the constant of integration.
4. **Substitute Back \( u = \sqrt{x} \)**:
\[
-2 \cos(u) + C = -2 \cos(\sqrt{x}) + C
\]
Thus, the integral \( \int \frac{\sin(\sqrt{x})}{\sqrt{x}} \, dx \) evaluates to:
\[
-2 \cos(\sqrt{x}) + C
\]
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