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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![### Evaluating the Trigonometric Integral
**Problem Statement:**
Evaluate the trigonometric integral:
\[ \int_{0}^{\pi/2} \sin^3 x \cos^3 x \, dx \]
in two different ways:
1. **Method (a):** Splitting off one \(\cos x\) and using \( u = \sin x \).
2. **Method (b):** Splitting off one \(\sin x\) and using \( u = \cos x \).
You should get the same answer for both (a) and (b)!
---
**Detailed Steps:**
To solve the given integral, you can use two different substitutions which make solving the integral simpler.
### Method (a): Using \( u = \sin x \)
1. Split off one \(\cos x\):
\[ \int_{0}^{\pi/2} \sin^3 x \cos^3 x \, dx = \int_{0}^{\pi/2} \sin^3 x \cos^2 x \cos x \, dx \]
2. Use the substitution \( u = \sin x \) which implies \( du = \cos x \, dx \).
3. Transform the integral:
- When \( x = 0 \), \( u = \sin 0 = 0 \).
- When \( x = \frac{\pi}{2} \), \( u = \sin \frac{\pi}{2} = 1 \).
4. Substitute in the integral:
\[ \int_{0}^{1} u^3 (1 - u^2) \, du \]
5. Simplify and solve the integral:
\[ \int_{0}^{1} (u^3 - u^5) \, du \]
\[ = \left[ \frac{u^4}{4} - \frac{u^6}{6} \right]_{0}^{1} \]
\[ = \left( \frac{1}{4} - \frac{1}{6} \right) - \left( 0 - 0 \right) \]
\[ = \frac{1}{4} - \frac{1}{6} \]
\[ = \frac{3}{12} - \frac{2}{12} \]
\[ =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4d021649-84ae-494f-9309-8423bd446b81%2F56a7abf8-2f63-47f3-85e6-bf6245264f12%2Fy21fi5_processed.png&w=3840&q=75)
Transcribed Image Text:### Evaluating the Trigonometric Integral
**Problem Statement:**
Evaluate the trigonometric integral:
\[ \int_{0}^{\pi/2} \sin^3 x \cos^3 x \, dx \]
in two different ways:
1. **Method (a):** Splitting off one \(\cos x\) and using \( u = \sin x \).
2. **Method (b):** Splitting off one \(\sin x\) and using \( u = \cos x \).
You should get the same answer for both (a) and (b)!
---
**Detailed Steps:**
To solve the given integral, you can use two different substitutions which make solving the integral simpler.
### Method (a): Using \( u = \sin x \)
1. Split off one \(\cos x\):
\[ \int_{0}^{\pi/2} \sin^3 x \cos^3 x \, dx = \int_{0}^{\pi/2} \sin^3 x \cos^2 x \cos x \, dx \]
2. Use the substitution \( u = \sin x \) which implies \( du = \cos x \, dx \).
3. Transform the integral:
- When \( x = 0 \), \( u = \sin 0 = 0 \).
- When \( x = \frac{\pi}{2} \), \( u = \sin \frac{\pi}{2} = 1 \).
4. Substitute in the integral:
\[ \int_{0}^{1} u^3 (1 - u^2) \, du \]
5. Simplify and solve the integral:
\[ \int_{0}^{1} (u^3 - u^5) \, du \]
\[ = \left[ \frac{u^4}{4} - \frac{u^6}{6} \right]_{0}^{1} \]
\[ = \left( \frac{1}{4} - \frac{1}{6} \right) - \left( 0 - 0 \right) \]
\[ = \frac{1}{4} - \frac{1}{6} \]
\[ = \frac{3}{12} - \frac{2}{12} \]
\[ =
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