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Magazine Chapter 16: Problem 27
Evaluate the integral. $$\int_{-\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{\sin x}{\sqrt{3+\cos x}} d x$$
Short Answer
Expert verified
The integral evaluates to \( 2\sqrt{\frac{7}{2}} - 2\sqrt{\frac{5}{2}} \).
Step by step solution
01
Choose a substitution
We start with the substitution \( u = 3 + \cos x \). This substitution simplifies the expression inside the square root. To proceed, we also need to find the differential. Differentiating both sides with respect to \( x \), we get \( du = -\sin x \, dx \) or equivalently \( dx = \frac{-du}{\sin x} \).
02
Change the limits of integration
We change the limits of integration based on our substitution. For \( x = -\frac{\pi}{3} \), \( u = 3 + \cos\left(-\frac{\pi}{3}\right) = 3 + \frac{1}{2} = \frac{7}{2} \). For \( x = \frac{2\pi}{3} \), \( u = 3 + \cos\left(\frac{2\pi}{3}\right) = 3 - \frac{1}{2} = \frac{5}{2} \). Thus, the limits change from \( x = -\frac{\pi}{3} \) to \( u = \frac{7}{2} \) and \( x = \frac{2\pi}{3} \) to \( u = \frac{5}{2} \).
03
Simplify and integrate
Substituting back into the integral, we have:\[\int_{\frac{7}{2}}^{\frac{5}{2}} \left(\frac{\sin x}{\sqrt{u}}\right) \left(\frac{-du}{\sin x}\right) = \int_{\frac{7}{2}}^{\frac{5}{2}} \frac{-du}{\sqrt{u}}.\]The \( \sin x \) terms cancel out, and we are left with an integral involving \( u \). This simplifies to:\[-\int_{\frac{7}{2}}^{\frac{5}{2}} u^{-1/2} \, du.\]
04
Evaluate the definite integral
The integral \( \int u^{-1/2} \, du \) is equal to \( 2u^{1/2} \), giving us the evaluated expression:\[-\left[ 2u^{1/2} \right]_{\frac{7}{2}}^{\frac{5}{2}}.\]Calculate the result by substituting the limits:\[= -\left(2 \sqrt{\frac{5}{2}} - 2 \sqrt{\frac{7}{2}} \right) = 2 \left( \sqrt{\frac{7}{2}} - \sqrt{\frac{5}{2}} \right).\]This evaluates to \( 2\sqrt{\frac{7}{2}} - 2\sqrt{\frac{5}{2}} \).
05
Simplify the result
The expression \( 2\sqrt{\frac{7}{2}} - 2\sqrt{\frac{5}{2}} \) doesn't simplify further into a cleaner form. Approximate numeric values may be used for practical purposes if needed, but the analytical result stands as shown.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Substitution
Trigonometric substitution is a technique used to simplify integrals by replacing variables with trigonometric functions. In our integral, the substitution chosen was \( u = 3 + \cos x \).
This choice helps in simplifying the integral by converting the complicated expression under the square root into a simpler form.
- The main goal of this technique is to make the integral easier to solve by taking advantage of trigonometric identities and properties. - To apply it, identify a part of the integral that resembles a trigonometric identity or is more complex when involving sine or cosine.
In this case, \( u = 3 + \cos x \) simplifies the square root, and the substitution method transforms the integral into a form that can easily cancel out terms, like \( \sin x \), leading to a smoother path towards finding the solution.
This choice helps in simplifying the integral by converting the complicated expression under the square root into a simpler form.
- The main goal of this technique is to make the integral easier to solve by taking advantage of trigonometric identities and properties. - To apply it, identify a part of the integral that resembles a trigonometric identity or is more complex when involving sine or cosine.
In this case, \( u = 3 + \cos x \) simplifies the square root, and the substitution method transforms the integral into a form that can easily cancel out terms, like \( \sin x \), leading to a smoother path towards finding the solution.
Limits of Integration
Adjusting the limits of integration is crucial when performing substitutions in definite integrals. Initially, the limits were \( x = -\frac{\pi}{3} \) to \( x = \frac{2\pi}{3} \). With the substitution \( u = 3 + \cos x \), these limits need to be recalculated for the new variable of integration, \( u \).
- Convert each original limit to the new variable: - For \( x = -\frac{\pi}{3} \), calculate \( u = 3 + \cos \left( -\frac{\pi}{3} \right) = \frac{7}{2} \). - For \( x = \frac{2\pi}{3} \), compute \( u = 3 + \cos \left( \frac{2\pi}{3} \right) = \frac{5}{2} \).
These adjusted limits, now \( u = \frac{7}{2} \) to \( u = \frac{5}{2} \), reflect the transformation of the integral necessary for consistent evaluation in terms of the new variable.
- Convert each original limit to the new variable: - For \( x = -\frac{\pi}{3} \), calculate \( u = 3 + \cos \left( -\frac{\pi}{3} \right) = \frac{7}{2} \). - For \( x = \frac{2\pi}{3} \), compute \( u = 3 + \cos \left( \frac{2\pi}{3} \right) = \frac{5}{2} \).
These adjusted limits, now \( u = \frac{7}{2} \) to \( u = \frac{5}{2} \), reflect the transformation of the integral necessary for consistent evaluation in terms of the new variable.
Differential Calculus
Within the substitution method, differential calculus plays a key role in adjusting the differential when we change the variable of integration. By differentiating \( u = 3 + \cos x \), we get \( du = -\sin x \, dx \).
- This differential \( du \) helps replace \( dx \) in the integral, resulting in \( dx = \frac{-du}{\sin x} \). - This substitution is vital in ensuring the integral is expressed exclusively in terms of \( u \).
By differentiating the substitution equation, both \( du \) and \( dx \) are corrected to maintain the integrity of the integral. It results in the cancellation of \( \sin x \) and further dramatizes the importance of proper differentiations, ensuring our transformations are accurately implemented.
- This differential \( du \) helps replace \( dx \) in the integral, resulting in \( dx = \frac{-du}{\sin x} \). - This substitution is vital in ensuring the integral is expressed exclusively in terms of \( u \).
By differentiating the substitution equation, both \( du \) and \( dx \) are corrected to maintain the integrity of the integral. It results in the cancellation of \( \sin x \) and further dramatizes the importance of proper differentiations, ensuring our transformations are accurately implemented.
Square Root Simplification
The heart of this exercise is simplifying the integral that contains a square root, specifically \( \sqrt{3+\cos x} \).
- Through substitution, \( u = 3 + \cos x \), the square root becomes \( \sqrt{u} \), which is a much simpler expression.
- This simplification helps convert the integral to a less complex form, which can typically be solved without additional substitutions or transformations.
Once simplified, the integral reduces to a format like \( \int u^{-1/2} \, du \), allowing for straightforward integration using basic power rule knowledge. Such simplifications are crucial in solving integrals more efficiently, especially when dealing with trigonometric expressions that initially appear more complex.
- Through substitution, \( u = 3 + \cos x \), the square root becomes \( \sqrt{u} \), which is a much simpler expression.
- This simplification helps convert the integral to a less complex form, which can typically be solved without additional substitutions or transformations.
Once simplified, the integral reduces to a format like \( \int u^{-1/2} \, du \), allowing for straightforward integration using basic power rule knowledge. Such simplifications are crucial in solving integrals more efficiently, especially when dealing with trigonometric expressions that initially appear more complex.
