Chapter 28: Problem 21
Integrate each of the given functions. $$\int \frac{e^{\tan ^{-1} 2 x}}{8 x^{2}+2} d x$$
Short Answer
Expert verified
\( \frac{1}{2} e^{\tan^{-1}(2x)} + C \)
Step by step solution
01
Simplify the Integral
Observe the form of the function in the integral: \( \int \frac{e^{\tan^{-1} 2x}}{8x^{2}+2} \, dx \). Notice that the denominator can be rewritten as \( 2(4x^2 + 1) \). Recognize that the derivative of \( \tan^{-1}(2x) \) is \( \frac{2}{4x^2+1} \), which suggests substitution might be useful.
02
Substitution
Use substitution to simplify the integral. Let \( u = \tan^{-1}(2x) \). Then, the derivative \( \frac{du}{dx} = \frac{2}{4x^2 + 1} \), or \( du = \frac{2}{4x^2 + 1} \, dx \). Rearrange for \( dx \): \( dx = \frac{du}{\frac{2}{4x^2+1}} = \frac{(4x^2 + 1)\, du}{2} \). Substitute into the integral: \( \int \frac{e^u}{2} \, du \).
03
Integrate using Substitution
Now the integral becomes \( \int \frac{e^u}{2} \,du \), which simplifies to \( \frac{1}{2} \int e^u \, du \). The integral of \( e^u \) is itself, \( e^u \). So, \( \frac{1}{2} \int e^u \, du = \frac{1}{2} e^u + C \), where \( C \) is the integration constant.
04
Substitute Back for Original Variable
Replace \( u \) with the original function \( \tan^{-1}(2x) \). Therefore, the integrated function is \( \frac{1}{2} e^{\tan^{-1}(2x)} + C \). This is the integral of the given function in terms of \( x \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution in Integration
Substitution is a potent method in integration that simplifies the process by transforming a complex integral into a more manageable form. It's similar to a clever "change of variables" technique. This method is particularly useful when dealing with composite functions or when the derivative of a part of the function is also present within the integral.
In our exercise, we used substitution by letting the variable \( u \) equal \( \tan^{-1}(2x) \). This choice was strategic because the derivative of \( \tan^{-1}(2x) \), which is \( \frac{2}{4x^2+1} \), appeared in the denominator of the integrand.
In our exercise, we used substitution by letting the variable \( u \) equal \( \tan^{-1}(2x) \). This choice was strategic because the derivative of \( \tan^{-1}(2x) \), which is \( \frac{2}{4x^2+1} \), appeared in the denominator of the integrand.
- First, we express \( dx \) in terms of \( du \) to substitute values correctly in the integral.
- Then, we perform the integration on the new variable, greatly simplifying the calculus.
- Once integrated, we substitute back the original variable to get the solution in terms of \( x \).
Inverse Trigonometric Functions in Integration
Inverse trigonometric functions, such as \( \tan^{-1}(x) \), play a crucial role in calculus, especially in integration. These functions reverse the trigonometric functions to find angles, based on known ratios.
In our integration problem, \( \tan^{-1}(2x) \) directly influences the manipulation of the integral.
In our integration problem, \( \tan^{-1}(2x) \) directly influences the manipulation of the integral.
- It substitutes into a simpler form that aligns with the capabilities of basic integration rules, converting the integral into a form that can be smoothly tackled.
- This function also has derivatives that are easier to manage, aiding in the substitution process as it connects the integral to a form that is integrable. The derivative \( \frac{2}{4x^2+1} \), was a key observation that led us to use substitution effortlessly.
Exponential Functions in Integration
Exponential functions, written in the form \( e^{x} \), are a foundational concept of calculus, especially when dealing with integrals and derivatives. They possess unique properties that make them very amenable to integration.
In the step-by-step solution provided for our problem, the exponential function \( e^{\tan^{-1}(2x)} \) was seamlessly integrated thanks to its inherent simplicity.
In the step-by-step solution provided for our problem, the exponential function \( e^{\tan^{-1}(2x)} \) was seamlessly integrated thanks to its inherent simplicity.
- Exponential functions are unique because they are their own derivatives and integrals. Therefore, integrating functions of the form \( e^u \) directly simplifies to \( e^u \), plus a constant of integration \( C \).
- In practical terms, this means when our integral transformed into \( \int e^{u} \, du \), it was straightforward to integrate.
- Once integrated, the solution \( \frac{1}{2} e^{u} + C \) ultimately allowed us to revert back to the original variable through substitution, offering the final solution in terms of \( x \).