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
\(\frac{1}{2} e^{\tan^{-1}(2t)} + C\)
Step by step solution
01
Identify the Substitution
Recognize that the integral involves the inverse tangent function, often requiring a substitution. Here, set: \[u = \tan^{-1}(2t),\quad \text{then}\quad du = \frac{2}{1+4t^2} dt.\]
02
Solve for dt
Rearrange the expression for the differential: \[ du = \frac{2}{1+4t^2} dt \quad \Rightarrow \quad dt = \frac{1+4t^2}{2}du.\]
03
Prepare the Substitution
Substitute the expressions found into the integral:\[ \int \frac{e^{u}}{1+4t^2} \cdot \frac{1+4t^2}{2} \ du.\]
04
Simplify the Integral
Observe that the \(1+4t^2\) terms cancel:\[ \int \frac{e^u}{1+4t^2} \cdot \frac{1+4t^2}{2} \ du = \frac{1}{2} \int e^u \, du.\]
05
Integrate with Respect to u
Integrate the simplified expression:\[ \frac{1}{2} \int e^u \, du = \frac{1}{2} e^u + C,\] where \(C\) is the constant of integration.
06
Back-Substitute to t
Substitute back the original variable to obtain the final answer:\[ \frac{1}{2} e^{\tan^{-1}(2t)} + C.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inverse Trigonometric Functions
Inverse trigonometric functions are mathematical functions that reverse the action of the corresponding trigonometric functions. For example, the inverse tangent, denoted as \( \tan^{-1}(x) \), gives an angle whose tangent is \( x \). This means that if you know the tangent of an angle, the inverse tangent function can help you find that angle.
In the context of integration, inverse trigonometric functions can often simplify the integration process. This is because these functions are frequently involved in calculus problems where they help express relationships in terms of angles, making the integral easier to evaluate.
In the given exercise, the function \( \tan^{-1}(2t) \) appears. Recognizing when and how to use such functions is key to performing successful substitutions. The inverse tangent function plays a crucial role in reformatting the given function into a simpler form that can be integrated with ease.
In the context of integration, inverse trigonometric functions can often simplify the integration process. This is because these functions are frequently involved in calculus problems where they help express relationships in terms of angles, making the integral easier to evaluate.
In the given exercise, the function \( \tan^{-1}(2t) \) appears. Recognizing when and how to use such functions is key to performing successful substitutions. The inverse tangent function plays a crucial role in reformatting the given function into a simpler form that can be integrated with ease.
Substitution Method
The substitution method is a powerful technique for evaluating integrals, especially when dealing with more complex functions. Think of it as changing the variable of integration to make an integral simpler to solve.
To perform substitution successfully, you need to decide what the new variable \( u \) will be. Often, it's a component of the expression within the integral. In this case, setting \( u = \tan^{-1}(2t) \) simplifies the integration process. You then compute \( du \), the differential of \( u \), with respect to the original variable, which helps to transform the entire integral into a variable form easier to integrate.
After substitution, the integral's differential parts are expressed only in terms of \( u \). Solving integrals using substitution not only simplifies steps but also helps you gain deeper insight into how different parts of the integral relate to each other.
To perform substitution successfully, you need to decide what the new variable \( u \) will be. Often, it's a component of the expression within the integral. In this case, setting \( u = \tan^{-1}(2t) \) simplifies the integration process. You then compute \( du \), the differential of \( u \), with respect to the original variable, which helps to transform the entire integral into a variable form easier to integrate.
After substitution, the integral's differential parts are expressed only in terms of \( u \). Solving integrals using substitution not only simplifies steps but also helps you gain deeper insight into how different parts of the integral relate to each other.
Exponential Functions
Exponential functions often appear in calculus and integration exercises. These functions have the form \( e^x \), where \( e \) is the base of the natural logarithm, approximately equal to 2.71828. It is called an 'exponential' function because it involves raising \( e \) to the power of \( x \).
Exponential functions are unique because their derivatives and integrals include the same exponential expression, which simplifies calculations. For instance, the integral of \( e^x \) is simply \( e^x + C \), where \( C \) is a constant.
In the original exercise, transforming the inverse tangent function as part of the exponent of an exponential function simplifies into an integral of \( e^u \), which is straightforward to evaluate. Understanding the behavior of exponential functions allows for efficient integration and solution of calculus problems.
Exponential functions are unique because their derivatives and integrals include the same exponential expression, which simplifies calculations. For instance, the integral of \( e^x \) is simply \( e^x + C \), where \( C \) is a constant.
In the original exercise, transforming the inverse tangent function as part of the exponent of an exponential function simplifies into an integral of \( e^u \), which is straightforward to evaluate. Understanding the behavior of exponential functions allows for efficient integration and solution of calculus problems.
Definite and Indefinite Integrals
In calculus, integrals are used to calculate areas, volumes, and other sums. There are two primary types: definite and indefinite integrals.
Definite integrals compute the precise area under a curve between two specified limits and result in a real number. The notation typically involves upper and lower bounds. Indefinite integrals, like the one in the problem, determine a family of functions whose derivative results in the original function. It includes a constant of integration \( C \) because there are infinitely many antiderivatives possible for a given function.
The indefinite integral process seen in the exercise involves finding a general form of the original expression without specific bounds. This provides a general solution to understand how the function behaves over an interval. Each technique used, like substitution, assists in achieving a simplified integrable form to find these antiderivatives efficiently.
Definite integrals compute the precise area under a curve between two specified limits and result in a real number. The notation typically involves upper and lower bounds. Indefinite integrals, like the one in the problem, determine a family of functions whose derivative results in the original function. It includes a constant of integration \( C \) because there are infinitely many antiderivatives possible for a given function.
The indefinite integral process seen in the exercise involves finding a general form of the original expression without specific bounds. This provides a general solution to understand how the function behaves over an interval. Each technique used, like substitution, assists in achieving a simplified integrable form to find these antiderivatives efficiently.
