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 Statement:**
Evaluate the integral
\[
\int_{-\pi}^{\pi} \cos^2(9x) \, dx.
\]
**Explanation:**
The problem involves integrating the square of the cosine function with a multiple angle, \(9x\), over the symmetric interval \([-π, π]\).
A common technique to solve integrals of squared trigonometric functions is to use the power-reduction identity:
\[
\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}.
\]
In this case, substitute \(\theta = 9x\), so:
\[
\cos^2(9x) = \frac{1 + \cos(18x)}{2}.
\]
Substitute this identity back into the integral:
\[
\int_{-\pi}^{\pi} \cos^2(9x) \, dx = \int_{-\pi}^{\pi} \frac{1 + \cos(18x)}{2} \, dx.
\]
This becomes:
\[
\frac{1}{2} \int_{-\pi}^{\pi} (1 + \cos(18x)) \, dx = \frac{1}{2} \left( \int_{-\pi}^{\pi} 1 \, dx + \int_{-\pi}^{\pi} \cos(18x) \, dx \right).
\]
Calculate each integral separately.
1. The integral of a constant over \([-π, π]\):
\[
\int_{-\pi}^{\pi} 1 \, dx = x \Big|_{-\pi}^{\pi} = \pi - (-\pi) = 2\pi.
\]
2. The integral of \(\cos(18x)\) over a symmetric interval:
The integral of \(\cos(kx)\) from \([-a, a]\) is zero if \(k \neq 0\), since cosine is an even function and its values will cancel out.
By observation:
\[
\int_{-\pi}^{\pi} \cos(18x) \, dx = 0.
\]
Finally, plug these results back into the original equation:
\[
\frac{1}{2} (2\pi + 0) = \frac{1}{2} \cdot 2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F678fef96-64be-4096-a74a-c4eb0a405ee7%2F8574f19a-39b9-4839-a657-02f41a609a80%2F3l6avk6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Evaluate the integral
\[
\int_{-\pi}^{\pi} \cos^2(9x) \, dx.
\]
**Explanation:**
The problem involves integrating the square of the cosine function with a multiple angle, \(9x\), over the symmetric interval \([-π, π]\).
A common technique to solve integrals of squared trigonometric functions is to use the power-reduction identity:
\[
\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}.
\]
In this case, substitute \(\theta = 9x\), so:
\[
\cos^2(9x) = \frac{1 + \cos(18x)}{2}.
\]
Substitute this identity back into the integral:
\[
\int_{-\pi}^{\pi} \cos^2(9x) \, dx = \int_{-\pi}^{\pi} \frac{1 + \cos(18x)}{2} \, dx.
\]
This becomes:
\[
\frac{1}{2} \int_{-\pi}^{\pi} (1 + \cos(18x)) \, dx = \frac{1}{2} \left( \int_{-\pi}^{\pi} 1 \, dx + \int_{-\pi}^{\pi} \cos(18x) \, dx \right).
\]
Calculate each integral separately.
1. The integral of a constant over \([-π, π]\):
\[
\int_{-\pi}^{\pi} 1 \, dx = x \Big|_{-\pi}^{\pi} = \pi - (-\pi) = 2\pi.
\]
2. The integral of \(\cos(18x)\) over a symmetric interval:
The integral of \(\cos(kx)\) from \([-a, a]\) is zero if \(k \neq 0\), since cosine is an even function and its values will cancel out.
By observation:
\[
\int_{-\pi}^{\pi} \cos(18x) \, dx = 0.
\]
Finally, plug these results back into the original equation:
\[
\frac{1}{2} (2\pi + 0) = \frac{1}{2} \cdot 2
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