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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Question
![**Integration by Parts**
For the following integral, which part would be assigned the \( u \) and the \( dv \)?
\[
\int x^2 \sin(2x) \, dx
\]
In the integration by parts method, we need to determine which component of the integrand should be \( u \) and which should be \( dv \). Generally, we choose \( u \) and \( dv \) based on the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions), where the logarithmic functions take precedence, followed by inverse trigonometric functions, algebraic functions, trigonometric functions, and finally exponential functions.
In this integral:
- \( x^2 \) is an algebraic function.
- \( \sin(2x) \) is a trigonometric function.
According to the LIATE rule:
- Let \( u = x^2 \) (the algebraic function)
- Let \( dv = \sin(2x) \, dx \) (the trigonometric function)
Proceed with these assignments to apply the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
First, compute \( du \) and \( v \):
- \( u = x^2 \) implies \( du = 2x \, dx \)
- \( dv = \sin(2x) \, dx \)
To find \( v \), integrate \( dv \):
\[ v = \int \sin(2x) \, dx \]
Using the substitution method:
- Let \( w = 2x \), hence \( dw = 2 \, dx \), which implies \( dx = \frac{dw}{2} \)
- Substitute back:
\[ \int \sin(2x) \, dx = \int \sin(w) \cdot \frac{dw}{2} = \frac{1}{2} \int \sin(w) \, dw = -\frac{1}{2} \cos(w) = -\frac{1}{2} \cos(2x) \]
Thus, \( v = -\frac{1}{2} \cos(2x) \).
Now, apply the integration by parts formula using the derived values:
\[ \int x^2 \sin(2x)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0ea55b65-6a30-46ce-96b8-2b732a599bb3%2F5e42f3c3-e5f6-4ee6-bdb1-af7dbb46ac31%2Fefiy22h_processed.png&w=3840&q=75)
Transcribed Image Text:**Integration by Parts**
For the following integral, which part would be assigned the \( u \) and the \( dv \)?
\[
\int x^2 \sin(2x) \, dx
\]
In the integration by parts method, we need to determine which component of the integrand should be \( u \) and which should be \( dv \). Generally, we choose \( u \) and \( dv \) based on the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions), where the logarithmic functions take precedence, followed by inverse trigonometric functions, algebraic functions, trigonometric functions, and finally exponential functions.
In this integral:
- \( x^2 \) is an algebraic function.
- \( \sin(2x) \) is a trigonometric function.
According to the LIATE rule:
- Let \( u = x^2 \) (the algebraic function)
- Let \( dv = \sin(2x) \, dx \) (the trigonometric function)
Proceed with these assignments to apply the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
First, compute \( du \) and \( v \):
- \( u = x^2 \) implies \( du = 2x \, dx \)
- \( dv = \sin(2x) \, dx \)
To find \( v \), integrate \( dv \):
\[ v = \int \sin(2x) \, dx \]
Using the substitution method:
- Let \( w = 2x \), hence \( dw = 2 \, dx \), which implies \( dx = \frac{dw}{2} \)
- Substitute back:
\[ \int \sin(2x) \, dx = \int \sin(w) \cdot \frac{dw}{2} = \frac{1}{2} \int \sin(w) \, dw = -\frac{1}{2} \cos(w) = -\frac{1}{2} \cos(2x) \]
Thus, \( v = -\frac{1}{2} \cos(2x) \).
Now, apply the integration by parts formula using the derived values:
\[ \int x^2 \sin(2x)
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