Problem 47 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(41-60\) $$\int \operatorname{sech}^{2}\left(x-\frac{1}{2}\right) d x$$

Step-by-Step Solution

Answer

The integral is \( \tanh(x - \frac{1}{2}) + C \).

Step 1: Identify the Integral Form

The given integral is \( \int \operatorname{sech}^{2} \left( x - \frac{1}{2} \right) \, dx \). Recognize that \( \operatorname{sech}^{2}(x) \) is the derivative of \( \tanh(x) \).

Step 2: Use Substitution Method

Let \( u = x - \frac{1}{2} \). Then \( du = dx \), which means that the substitution updates the integral to \( \int \operatorname{sech}^{2}(u) \, du \).

Step 3: Integrate the Resulting Expression

Since \( \frac{d}{du}(\tanh(u)) = \operatorname{sech}^{2}(u) \), the integral becomes \( \int \operatorname{sech}^{2}(u) \, du = \tanh(u) + C \), where \( C \) is the constant of integration.

Step 4: Substitute Back

Substitute \( u = x - \frac{1}{2} \) back into the resulting expression. Thus, we have \( \tanh(x - \frac{1}{2}) + C \) as the integral of the original function.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions, but they are based on hyperbolas instead of circles. A couple of these functions include \(\sinh(x)\) and \(\cosh(x)\), which are related to the exponential function. They provide a natural way to describe certain types of geometric shapes and physical phenomena. For example:

• \(\sinh(x) = \frac{e^x - e^{-x}}{2}\)
• \(\cosh(x) = \frac{e^x + e^{-x}}{2}\)

Unlike their circular counterparts, these functions extend the properties of exponential functions. This means they are continuously growing or decaying rather than oscillating.
Hyperbolic functions often pop up in calculus, especially with certain types of integrals, because they have easily manageable derivatives and integrals like trigonometric functions.
In the context of this exercise, the specific hyperbolic function we focus on is \(\operatorname{sech}(x)\). It plays a crucial role in understanding the integral involved.

Substitution Method

The substitution method in calculus is a technique that simplifies the process of integration. It is analogous to the reverse of the chain rule used in differentiation. This method involves replacing a part of the integral with a new variable to make integration easier.
To use substitution effectively:

• Identify a part of the integrand that can be substituted with a single variable, typically simplifying the integral.
• Express the differential \(dx\) in terms of the new variable, \(du\).
• Perform the integration with respect to the new variable.
• Substitute back to the original variable after integration.

In our original exercise, by letting \(u = x - \frac{1}{2}\), the integral was reduced to a simpler form \(\int \operatorname{sech}^{2}(u) \, du\).
This substitution is simple because it directly aligns with the derivative of the hyperbolic tangent function, \(\tanh(u)\), making the integration process straightforward.

Sech Function

The \( \operatorname{sech}(x) \) function is the hyperbolic secant, which is comparable to the secant function in trigonometry but applicable to hyperbolic angles. It is defined as the reciprocal of the hyperbolic cosine function:

• \( \operatorname{sech}(x) = \frac{1}{\cosh(x)} \)

This function is significant in calculus due to its direct association with the derivative of the hyperbolic tangent function, \( \tanh(x) \).
In the provided exercise, recognizing that \( \operatorname{sech}^{2}(x) \) is the derivative of \( \tanh(x) \) is key to solving the integral effortlessly. The relationship is:

• \( \frac{d}{dx} (\tanh(x)) = \operatorname{sech}^{2}(x) \)

Hence, integrating \( \operatorname{sech}^{2}(u) \) with respect to \( u \) results in \( \tanh(u) + C \), where \( C \) is the constant of integration. Understanding this reciprocal relationship is crucial in tackling calculus problems involving hyperbolic functions.