Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-54, perform the indicated integrations. \(\int \frac{e^{\tan ^{-1} 2 t}}{1+4 t^{2}} d t\)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{2} e^{\tan^{-1}(2t)} + C \).

Step 1: Identify the Substitution

We notice that the integral is of the form \( \int \frac{e^{g(t)}}{g'(t)} dt \),which suggests a substitution. Here, let’s set \( u = \tan^{-1}(2t) \). The derivative is \( du = \frac{1}{1+4t^2} \cdot 2 dt \), so \( dt = \frac{1}{2} (1 + 4t^2) du \).

Step 2: Substitute into the Integral

Substitute \( u = \tan^{-1}(2t) \) and its derivative into the integral:\[\int \frac{e^u}{1 + 4t^2} \times 1 \cdot dt = \int \frac{e^u}{1 + 4t^2} \cdot \frac{1}{2(1+4t^2)} du \]Simplifying gives:\[\int \frac{1}{2} e^u du \]

Step 3: Integrate the New Expression

The new integral is a straightforward exponential integral. Integrate\[\int \frac{1}{2} e^u du = \frac{1}{2} e^u + C \]where \( C \) is the constant of integration.

Step 4: Substitute Back to Original Variable

Replace \( u \) back with \( \tan^{-1}(2t) \):\[\frac{1}{2} e^{\tan^{-1}(2t)} + C \]This is the solution in terms of the original variable \( t \).

Key Concepts

Substitution Method

The substitution method in calculus is a powerful tool for simplifying the process of integration. It involves replacing a complicated part of the integral with a simpler expression, making the integral easier to solve. The key to using substitution is to identify a function and its derivative within the integral.
In this context, we recognize that the integral has a form that suggests substitution can be useful. The function inside the exponent, \( e^{\tan^{-1}(2t)} \), connects neatly with its derivative when expressed in terms of \( u = \tan^{-1}(2t) \).
Here's the process:

• Choose a substitution: Set \( u = \tan^{-1}(2t) \).
• Find the derivative: \( du = \frac{1}{1+4t^2} \cdot 2 dt \).
• Solve for \( dt \): \( dt = \frac{1}{2} (1 + 4t^2) du \).

By substituting \( u \) and \( du \) back into the integral, we transform it into a much simpler expression. This method is particularly effective when an integral matches the pattern of \( \frac{g'(x)}{g(x)} \), as this fits well with the division structure, simplifying subsequent steps.

Exponential Integration

Integrating exponential functions, such as \( e^u \), is one of the more straightforward processes in calculus. Exponential functions have consistent rules when it comes to integration. The base of the natural logarithm, \( e \), is unique in that its derivative is itself. This property extends to the integration process.
When we faced the integral expression \( \int \frac{1}{2} e^u du \), after substitution, it is a simple task to integrate.

• Integrate the simplified function: \( \int \frac{1}{2} e^u du \).
• The antiderivative of \( e^u \) is \( e^u \).
• So, \( \int \frac{1}{2} e^u du = \frac{1}{2} e^u + C \).

Here, \( C \) is the constant of integration, reminding us to include it in indefinite integrals to account for all possibilities of the original function. This consistent pattern makes exponential integration less daunting and highly predictive.

Inverse Trigonometric Functions

Inverse trigonometric functions, like \( \tan^{-1}(x) \), are crucial for solving more complex integration problems. These functions help us relate angles to their respective tangent values and are often pivotal in transformation steps involving substitution.
In the given problem, \( \tan^{-1}(2t) \) acted as a bridge that allowed us to connect the initial complex expression to a simpler one. The role of \( \tan^{-1}(x) \) in calculus extends to finding derivatives and solving equations involving right angles.
Understanding these functions involves:

• Recognizing their derivative form, such as \( \frac{d}{dx} [\tan^{-1}(x)] = \frac{1}{1+x^2} \).
• Applying them systematically to integrate trigonometric expressions.

By establishing \( u = \tan^{-1}(2t) \), the problem enabled us to use its derivative neatly, simplifying both the differentiation and integration steps.

Integration Techniques

Various integration techniques exist to tackle different types of integrals efficiently. Each method has its uses depending on the form of the integral. For the given problem, a combination of substitution and exponential integration came into play.
Core integration techniques include:

• Substitution method: Simplifies integrals using a change of variable.
• Integration by parts: Useful for products of functions.
• Partial fraction decomposition: Helps with rational functions.

By combining these approaches, like substitution with exponential integration, complex integrals can be accurately and efficiently solved. The ability to choose the right method is largely dependent on recognizing patterns and forms of integrals, an essential skill in calculus. Each technique complements the others, offering a toolkit that caters to diverse mathematical expressions, enhancing problem-solving abilities.