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Magazine Chapter 16: Problem 32
Evaluate the integral. $$\int_{\ln 2}^{\ln \left(\frac{2}{3}\right)} \frac{e^{-2 x} d x}{\sqrt{1-e^{-4 x}}}$$
Short Answer
Expert verified
The integral cannot be evaluated as \(\sin^{-1}\left(\frac{9}{4}\right)\) is undefined.
Step by step solution
01
Recognize the substitution
Notice that the integral has a square root in the denominator involving the expression \(e^{-4x}\). Recognize that using a trigonometric substitution is likely beneficial. We can use the substitution \(u = e^{-2x}\), \ hence \(du = -2e^{-2x}dx\) \ or \ \(dx = \frac{-du}{2u}\).
02
Change the bounds of integration
When \(x = \ln 2\), \(u = e^{-2 \ln 2} = \frac{1}{4}\). When \(x = \ln \left(\frac{2}{3}\right)\), \(u = e^{-2 \ln \left(\frac{2}{3}\right)} = \frac{9}{4}\). Thus, the new limits of integration are from \(u = \frac{1}{4}\) to \(u = \frac{9}{4}\).
03
Substitute and simplify the integral
The integral becomes: \[ \int_{\frac{1}{4}}^{\frac{9}{4}} \frac{u}{\sqrt{1-u^2}} \left(-\frac{du}{2u}\right) \]This simplifies to: \[ -\frac{1}{2} \int_{\frac{1}{4}}^{\frac{9}{4}} \frac{du}{\sqrt{1-u^2}} \]
04
Identify the function form
The integral \(\int \frac{du}{\sqrt{1-u^2}}\) is known to be \(\sin^{-1}(u)\), the inverse sine function. Thus, the integral becomes: \[-\frac{1}{2} \left[ \sin^{-1}(u) \right]_{\frac{1}{4}}^{\frac{9}{4}} \]
05
Evaluate the definite integral
Compute the inverse sine:\[ \left(-\frac{1}{2}\right) \times \left(\sin^{-1}\left(\frac{9}{4}\right) - \sin^{-1}\left(\frac{1}{4}\right)\right) \]Recognize that \( \sin^{-1}\left(\frac{9}{4}\right)\) is undefined because \(\frac{9}{4}\) is outside the domain of the inverse sine, which is \([-1,1]\). Hence, the integral cannot be evaluated for these bounds.
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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 mathematical technique that is often used to solve integrals containing a square root of an expression that can be related to a trigonometric identity. In this exercise, the expression under the square root is in the form of \(\sqrt{1-e^{-4x}}\), which resembles the trigonometric identity \(\sqrt{1-\sin^2(\theta)} = \cos(\theta)\). This hints at using a trigonometric substitution.
- Substitute a new variable that transforms the expression into a recognizable trigonometric form.
- In our solution, we use the substitution \(u = e^{-2x}\), simplifying the expression \(\sqrt{1-u^2}\), which is suitable for a trigonometric identity.
- Ensure to correctly change the differential \(dx\) in terms of \(du\). In this case, \(dx = \frac{-du}{2u}\).
Inverse Sine Function
The inverse sine function, also known as the arcsine function, is essential when solving integrals involving expressions of the form \(\int \frac{du}{\sqrt{1-u^2}}\). This specific integral is directly recognized as \( \sin^{-1}(u) \), the inverse sine of \(u\).
- The inverse sine function \( \sin^{-1}(x) \) is the angle whose sine is \(x\), with the principal value range of \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
- Here, \( \sin^{-1}(1) = \frac{\pi}{2} \) and \( \sin^{-1}(0) = 0 \), which should remind us of the domain \([-1, 1]\).
Integration Bounds Adjustment
Appropriately adjusting the bounds of integration is an integral part of solving definite integrals, especially after making a substitution. After substituting \(u = e^{-2x}\), the bounds of integration originally given in terms of \(x\) must also be changed accordingly to match the new variable \(u\).
- Evaluate the new limits by substituting the original bounds into the substitution expression: From \(x = \ln(2)\) and \(x = \ln(\frac{2}{3})\), calculate \(u\).
- From \(x = \ln(2)\), substitution gives \(u = \frac{1}{4}\).
- From \(x = \ln(\frac{2}{3})\), substitution gives \(u = \frac{9}{4}\).
