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Magazine Chapter 7: Problem 37
Perform the indicated integrations. \(\int \frac{e^{\tan ^{-1} 2 t}}{1+4 t^{2}} d t\)
Short Answer
Expert verified
The integral is \( \frac{1}{2} e^{\tan^{-1}(2t)} + C \).
Step by step solution
01
Identify Substitution
Notice that the integrand involves both the derivative and function of the arc tangent. Recognize that a substitution can simplify the integral. Let \( u = \tan^{-1}(2t) \). Then, the derivative \( du = \frac{2}{1+4t^2} \, dt \). Thus, \( dt = \frac{1+4t^2}{2} \, du \).
02
Simplify the Integral
Substitute \( u = \tan^{-1}(2t) \) and \( dt = \frac{1+4t^2}{2} \, du \) into the integral:\[\int \frac{e^{u}}{1+4t^2} \cdot \frac{1+4t^2}{2} \, du = \frac{1}{2} \int e^u \, du.\]The expression \( \frac{1+4t^2}{1+4t^2} \) simplifies to 1.
03
Integrate the Simplified Expression
The integral \( \frac{1}{2} \int e^u \, du \) can be computed directly as:\[\frac{1}{2} \int e^u \, du = \frac{1}{2} e^u + C,\]where \(C\) is the constant of integration.
04
Substitute Back for Original Variable
Replace \( u \) with the original function of \( t \):\[\frac{1}{2} e^u + C = \frac{1}{2} e^{\tan^{-1}(2t)} + C.\]Thus, the final answer in terms of \( t \) is \( \frac{1}{2} e^{\tan^{-1}(2t)} + C \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The Substitution Method is a fundamental integration technique that simplifies complex integrals by converting them into familiar forms. It involves substituting a part of the integrand with a new variable. This new variable is often chosen to simplify the expression, either by reducing its complexity or aligning it to match common integral forms.
- **Choosing a Substitution:** Detect parts of the integrand that appear complex or involve a composite function. In this exercise, recognizing that the expression is tied to the derivative of the arc tangent helps decide the substitution.
- **Calculating Differentials:** Once you choose a substitution, it is essential to change the differential accordingly. For example, after choosing the substitution \( u = \tan^{-1}(2t) \), calculate \( du = \frac{2}{1+4t^2} \, dt \), and rearrange as \( dt = \frac{1+4t^2}{2} \, du \).
- **Substituting Back:** After integration, revert back to the original variable to express your final result in terms of the initial variable given in the problem.
Arc Tangent Function
The Arc Tangent Function, denoted as \( \tan^{-1}(x) \) or \( \arctan(x) \), is the inverse of the tangent function within a specific range. It essentially gives the angle whose tangent is equal to a specific number. Understanding this function is crucial when it appears within integrals.
- **Properties:** The arc tangent function is continuous and differentiable, having derivatives that often appear in integration contexts. Specifically, the derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1+x^2} \).
- **Common Usage:** In integrals, the derivatives of inverse trigonometric functions are useful. Here, the presence of \( \tan^{-1}(2t) \) hints at employing its derivative form \( \frac{2}{1+4t^2} \) as part of the substitution process.
- **Range and Output:** The values of the arc tangent function range between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), thus being bounded, which fits within the needs of various integral calculations.
Exponential Function
The Exponential Function \( e^x \) is a unique and important mathematical function characterized by its base, \( e \), which is approximately 2.71828. It is renowned for the property where it is equal to its own derivative, a feature that simplifies integration.
- **Basic Properties:** The function \( e^u \) acts as a constant multiplier when differentiated, and integrating \( e^u \) results merely in \( e^u + C \).
- **Integration Simplification:** This property is very useful in calculus, especially when combined with the substitution method, because any integral involving \( e^u \) becomes straightforward. Here, \( \int e^u \, du = e^u + C \) demonstrates this clearly.
- **Growth Rate:** As a function that models growth processes, its rate of change is always proportional to its current value, which is why it's effective in natural calculations and many real-world applications.
