Chapter 29: Problem 36
$$ \text { Determine the following: } $$ $$ \int_{0}^{\pi / 3} \frac{\sin x}{(1+\cos x)^{2}} d x $$
Short Answer
Expert verified
The value of the integral is \( \frac{1}{6} \).
Step by step solution
01
Identify the substitution
For the integral \( \int_{0}^{\pi / 3} \frac{\sin x}{(1+\cos x)^{2}} \, dx \), we will use the substitution \( u = 1 + \cos x \). This choice simplifies the integration process. Notice that the derivative \( du = -\sin x \, dx \) will also help in rewriting the integral.
02
Change limits of integration
Since \( u = 1 + \cos x \), evaluate the bounds of integration. When \( x = 0 \), \( u = 1 + \cos(0) = 2 \). When \( x = \pi/3 \), \( u = 1 + \cos(\pi/3) = 1.5 \). Thus, the limits of integration change from \( x = 0 \) to \( x = \pi/3 \) into \( u = 2 \) to \( u = 1.5 \).
03
Transform the integral
Substitute \( u = 1 + \cos x \) and \( du = -\sin x \, dx \) into the integral. Therefore, \( \sin x \, dx = -du \). The integral becomes:\[\int_{2}^{1.5} \frac{-1}{u^{2}} \, du = \int_{1.5}^{2} \frac{1}{u^{2}} \, du\]
04
Integrate using simple power rule
The integral \( \int \frac{1}{u^{2}} \, du \) can be expressed using the power rule as \( \int u^{-2} \, du \). Integrating, we get:\[\int u^{-2} \, du = -u^{-1} = -\frac{1}{u}\]
05
Evaluate the definite integral
Substitute the limits of integration into the antiderivative:\[\left[ -\frac{1}{u} \right]_{1.5}^{2} = \left( -\frac{1}{2} \right) - \left( -\frac{1}{1.5} \right)\]Simplify:\[-\frac{1}{2} + \frac{2}{3} = \frac{-3 + 4}{6} = \frac{1}{6}\]
06
Confirm the conclusion
The evaluated value of the original integral is \( \frac{1}{6} \). Ensuring that all steps, substitutions, and calculations are correct confirms this answer.
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.
Definite Integral
The definite integral is a fundamental concept in calculus. It represents the signed area under a function's curve within a specific interval. For the integral \( \int_{a}^{b} f(x) \, dx \), the values \( a \) and \( b \) are the bounds of integration. These bounds determine the specific range for which the area is calculated. The process essentially sums up infinitesimal slices across the interval.
- Lower and upper bounds: These are the integral's limits, symbolizing the region over which the function is integrated.
- Net area: Unlike indefinite integrals, definite integrals account for the direction, yielding positive and negative values based on the function's position relative to the x-axis.
Trigonometric Substitution
Trigonometric substitution is a technique used to simplify the integration of expressions involving square roots, fractions, or both. It's especially useful when dealing with functions containing terms like \( \sin x \) or \( \cos x \). By substituting trigonometric identities or parts of the integral, the expression becomes easier to integrate.
For instance, to solve \( \int_{0}^{\pi / 3} \frac{\sin x}{(1+\cos x)^{2}} \, dx \), we used the substitution \( u = 1 + \cos x \). This choice transforms the integral into a more manageable form by simplifying the function's denominator.
For instance, to solve \( \int_{0}^{\pi / 3} \frac{\sin x}{(1+\cos x)^{2}} \, dx \), we used the substitution \( u = 1 + \cos x \). This choice transforms the integral into a more manageable form by simplifying the function's denominator.
- Identify substitutions: Choose a substitution that simplifies the integral. Here, \( u = 1 + \cos x \) aligns terms with \( \sin x \), making the expression simpler.
- Change variables: Derivatives \( du = -\sin x \, dx \) help convert the integral into a new variable space, easing the integration process.
Power Rule Integration
Power rule integration is a foundational technique in calculus that aids in evaluating integrals involving powers of a variable. It states that for any real number \( n eq -1 \), the integral \( \int x^n \, dx \) is \( \frac{x^{n+1}}{n+1} + C \), with \( C \) being the constant of integration.
In the context of our exercise, once the integral \( \int \frac{1}{u^{2}} \, du \) is transformed into the form \( \int u^{-2} \, du \), the power rule can be directly applied.
In the context of our exercise, once the integral \( \int \frac{1}{u^{2}} \, du \) is transformed into the form \( \int u^{-2} \, du \), the power rule can be directly applied.
- General form: Convert expressions, if necessary, to a straightforward power form \( x^n \).
- Application: The expression \( \int u^{-2} \, du \) becomes \( -u^{-1} = -\frac{1}{u} \) using the power rule.
Change of Variables
Change of variables, often referred to as substitution in integration, is crucial for transforming integrals into simpler forms. This technique involves introducing a new variable to replace an existing one, making the integral easier to solve.
In the example of \( \int_{0}^{\pi / 3} \frac{\sin x}{(1+\cos x)^{2}} \, dx \), changing variables by setting \( u = 1 + \cos x \), we effectively redirect the problem into a more solvable path.
In the example of \( \int_{0}^{\pi / 3} \frac{\sin x}{(1+\cos x)^{2}} \, dx \), changing variables by setting \( u = 1 + \cos x \), we effectively redirect the problem into a more solvable path.
- Establish new limits: When changing variables, it's crucial to adjust the integral's limits to match the new variable. Here, when \( x = 0 \), \( u = 2 \), and when \( x = \frac{\pi}{3} \), \( u = 1.5 \).
- Rewrite the integral: Substitute \( dx \) and adjust the differential appropriately. With \( du = -\sin x \, dx \), the transformation facilitates straightforward integration.