Problem 18
Question
Use Wallis's Formulas to evaluate the integral. $$ \int_{0}^{\pi / 2} \sin ^{7} x d x $$
Step-by-Step Solution
Verified Answer
The integral \( \int_{0}^{\pi / 2} \sin ^{7} x d x = \frac{16\pi}{35} \)
1Step 1: Recall Wallis's Formula Definition
Wallis's formulas define that the integration of \( \sin^{m}(x) \) or \( \cos^{m}(x) \) from 0 to \( \pi/2 \) is given by: \( \frac{(m-1)!!}{m!!} \cdot \frac{\pi}{2} \) if m is odd and \( \frac{(m-1)!!}{m!!} \) if m is even, where !! represents the double factorial operation.
2Step 2: Apply Wallis's Formula
Since \( m=7 \) is odd, we will use the formula specified for odd integrals. \( \int_{0}^{\pi / 2} \sin ^{7} x d x = \frac{6!!}{7!!} \cdot \frac{\pi}{2} \)
3Step 3: Compute the Double Factorial
Compute the double factorial. Recall that the double factorial is the product of all the integers from 1 to n that have the same parity (oddness or evenness) as n. Thus, \( 6!! = 6 \cdot 4 \cdot 2 \cdot 1 = 48 \) and \( 7!! = 7 \cdot 5 \cdot 3 \cdot 1 = 105 \)
4Step 4: Substitute and Simplify the Equation
Substitute \( 6!! \) and \( 7!! \) into the equation: \( \frac{6!!}{7!!} \cdot \frac{\pi}{2} = \frac{48}{105} \cdot \frac{\pi}{2} = \frac{16\pi}{35} \)
Key Concepts
Double FactorialIntegration of Trigonometric FunctionsOdd and Even Functions
Double Factorial
Double factorial is a mathematical expression that involves the product of integers of the same parity (odd or even) leading up to a specified number, n. It is denoted with two exclamation marks, !!.
For example:
For example:
- For an even number, such as 6, the double factorial is calculated as: \[6!! = 6 \times 4 \times 2 = 48 \]
- For an odd number, like 7, it is: \[7!! = 7 \times 5 \times 3 = 105 \]
Integration of Trigonometric Functions
Integrating trigonometric functions, such as sine or cosine raised to a power, can be made simpler using Wallis's Formula. This method is designed for integrals of the form \[\int_{0}^{\pi / 2} \sin^{m}(x) \, dx\] or \[\int_{0}^{\pi / 2} \cos^{m}(x) \, dx\].
Using Wallis's Formula, you can estimate these integrals effectively without requiring standard integration techniques.
Using Wallis's Formula, you can estimate these integrals effectively without requiring standard integration techniques.
- For odd powers, the integration is expressed as: \[\frac{(m-1)!!}{m!!} \cdot \frac{\pi}{2}\]
- For even powers, it simplifies to: \[\frac{(m-1)!!}{m!!}\]
Odd and Even Functions
Functions can be categorized into odd and even based on their symmetry. This classification is pivotal in mathematics, particularly in defining integration limits and simplifying calculations.
- **Odd Functions** are symmetrical about the origin. They satisfy: \[f(-x) = -f(x)\]. Examples include \( \sin(x) \) and \( x^3 \).- **Even Functions** show symmetry about the y-axis. They meet the condition: \[f(-x) = f(x)\]. Common examples are \( \cos(x) \) and \( x^2 \).
This symmetry makes dealing with integrals straightforward in some cases:- For even functions over symmetric limits around zero, the integral simplifies, possibly doubling the area from 0 to the limit.- For odd functions over symmetric limits around zero, the integral equals zero since the areas cancel each other.
Understanding these properties can help predict the behavior of integrals and aid in utilizing formulas like Wallis's.
- **Odd Functions** are symmetrical about the origin. They satisfy: \[f(-x) = -f(x)\]. Examples include \( \sin(x) \) and \( x^3 \).- **Even Functions** show symmetry about the y-axis. They meet the condition: \[f(-x) = f(x)\]. Common examples are \( \cos(x) \) and \( x^2 \).
This symmetry makes dealing with integrals straightforward in some cases:- For even functions over symmetric limits around zero, the integral simplifies, possibly doubling the area from 0 to the limit.- For odd functions over symmetric limits around zero, the integral equals zero since the areas cancel each other.
Understanding these properties can help predict the behavior of integrals and aid in utilizing formulas like Wallis's.
Other exercises in this chapter
Problem 17
Find the integral. (Note: Solve by the simplest method-not all require integration by parts.) $$ \int\left(x^{2}-1\right) e^{x} d x $$
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Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{0}^{\infty}(x-1) e^{-x} d x $$
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Use integration tables to find the integral. $$ \int \frac{e^{x}}{\left(1-e^{2 x}\right)^{3 / 2}} d x $$
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In Exercises \(7-26,\) evaluate the limit, using \(L\) 'Hôpital's Rule if necessary. (In Exercise \(12, n\) is a positive integer.) \(\lim _{x \rightarrow \inft
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