Chapter 7: Problem 65
Evaluate the integrals in Exercises \(47-70\) $$ \int_{-\pi / 2}^{\pi / 2} \frac{2 \cos \theta d \theta}{1+(\sin \theta)^{2}} $$
Short Answer
Expert verified
The integral evaluates to \( \pi \).
Step by step solution
01
Identify the Integral and Substitution
The given integral is \( \int_{-\pi / 2}^{\pi / 2} \frac{2 \cos \theta \, d\theta}{1+(\sin \theta)^{2}} \). We can simplify the problem by using a trigonometric identity or substitution. Let's use the substitution \( u = \sin \theta \), which implies \( du = \cos \theta \, d\theta \).
02
Change the Limits of Integration
When \( \theta = -\pi/2 \), \( u = \sin(-\pi/2) = -1 \). When \( \theta = \pi/2 \), \( u = \sin(\pi/2) = 1 \). Therefore, the limits of integration with respect to \( u \) are from \(-1\) to \(1\).
03
Substitute and Transform the Integral
Substituting into the integral, we have \( \int_{-1}^{1} \frac{2 \, du}{1 + u^{2}} \). This transformation relies on the identity \( \cos \theta \, d\theta = du \).
04
Recognize the Integral Form
The integral \( \int \frac{du}{1 + u^{2}} \) is a standard form that is equal to \( \arctan(u) \). Thus, our integral becomes \( 2 \int_{-1}^{1} \frac{du}{1 + u^{2}} = 2[\arctan(u)]_{-1}^{1} \).
05
Evaluate the Integral at the Limits
Substitute the limits of integration: \( 2 [\arctan(1) - \arctan(-1)] \). Since \( \arctan(1) = \frac{\pi}{4} \) and \( \arctan(-1) = -\frac{\pi}{4} \), it follows that \( \arctan(1) - \arctan(-1) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} \).
06
Final Calculation
Therefore, the integral evaluates to \( 2 \times \frac{\pi}{2} = \pi \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Substitution
Trigonometric substitution is a powerful method to simplify integrals involving expressions such as \( a^2 - x^2 \), \( a^2 + x^2 \), or \( x^2 - a^2 \). In this context, it helps convert complex trigonometric functions into simpler, more manageable forms.
For the integral \( \int_{-\pi / 2}^{\pi / 2} \frac{2 \cos \theta d \theta}{1+(\sin \theta)^{2}} \), using a substitution can transform the original variables to make integration easier.
An effective substitution here is \( u = \sin \theta \). This means \( du = \cos \theta \, d\theta \), allowing conversion of the integral of a trigonometric function to a function of \( u \) only.
For the integral \( \int_{-\pi / 2}^{\pi / 2} \frac{2 \cos \theta d \theta}{1+(\sin \theta)^{2}} \), using a substitution can transform the original variables to make integration easier.
An effective substitution here is \( u = \sin \theta \). This means \( du = \cos \theta \, d\theta \), allowing conversion of the integral of a trigonometric function to a function of \( u \) only.
- This changes the integral limits, translating \( \theta = -\pi/2 \) and \( \pi/2 \) to \( u = -1 \) and \( 1 \), respectively.
- It simplifies the integral to \( \int_{-1}^{1} \frac{2 \, du}{1 + u^{2}} \).
Integral Calculus
Integral calculus is centered around the concept of integrals, which are used to find areas, volumes, and central points. It plays a crucial role in understanding and solving problems involving continuous functions.
This problem involves a definite integral, \( \int_{-\pi / 2}^{\pi / 2} \frac{2 \cos \theta d \theta}{1+ (\sin \theta)^{2}} \), which requires evaluation from one point to another.
Integrals can be classified as definite or indefinite, where a definite integral provides a numerical value representing an area under a curve between specified limits.
The substitution reduces the integral to a simpler form, \( \int_{-1}^{1} \frac{2 \ du}{1 + u^{2}} \), involving only basic calculus skills to evaluate the new limits.
This problem involves a definite integral, \( \int_{-\pi / 2}^{\pi / 2} \frac{2 \cos \theta d \theta}{1+ (\sin \theta)^{2}} \), which requires evaluation from one point to another.
Integrals can be classified as definite or indefinite, where a definite integral provides a numerical value representing an area under a curve between specified limits.
The substitution reduces the integral to a simpler form, \( \int_{-1}^{1} \frac{2 \ du}{1 + u^{2}} \), involving only basic calculus skills to evaluate the new limits.
- Recognizing and transforming the integral into a standard form is key in solving it efficiently.
- Definite integrals, like this one, have direct applications in computing physical quantities and are foundational in the analysis of mathematical models.
Standard Integral Forms
Standard integral forms serve as a toolkit in calculus, simplifying the process of evaluating integrals that match certain common patterns.
Once an integral is reduced and transformed, like \( \int_{-1}^{1} \frac{2 \, du}{1 + u^{2}} \), recognizing it as a standard form is crucial.
In this instance, \( \int \frac{du}{1 + u^{2}} \) matches the well-known form that evaluates to \( \arctan(u) \). This allows for a straightforward computation.
Recognizing and using standard integral forms ensures calculations are quick and accurate, allowing you to focus on more complex aspects of a problem.
Once an integral is reduced and transformed, like \( \int_{-1}^{1} \frac{2 \, du}{1 + u^{2}} \), recognizing it as a standard form is crucial.
In this instance, \( \int \frac{du}{1 + u^{2}} \) matches the well-known form that evaluates to \( \arctan(u) \). This allows for a straightforward computation.
- By applying the result \( \arctan(u) \), the definite integral becomes \( 2[\arctan(u)]_{-1}^{1} \).
- Each integral form has associated rules and results, familiarizing with them accelerates problem-solving.
Recognizing and using standard integral forms ensures calculations are quick and accurate, allowing you to focus on more complex aspects of a problem.