Question

Evaluate each integral. $$\int \operatorname{sech}^{2} w \tanh w d w$$

Short answer

Question: Evaluate the integral \(\int \operatorname{sech}^2 w \tanh w dw\). Answer: The integral evaluates to \(\frac{1}{2}(\tanh w)^2 + C\).

Step-by-step solution

Identify the substitution

Let's define a function that will make the integral easier to evaluate. Notice that the derivative of the hyperbolic tangent is the square of the hyperbolic secant: $$\frac{d}{dw}\tanh w = \operatorname{sech}^2 w$$ So, we'll set: $$v = \tanh w$$ Now, find the derivative of \(v\) with respect to \(w\): $$\frac{dv}{dw} = \operatorname{sech}^2 w$$

Substitute into the integral

Rewrite the integral using \(v\) and \(dv\) instead of \(\tanh w\) and \(dw\), using the substitution we just made: $$\int \operatorname{sech}^2 w \tanh w dw = \int v dv$$

Evaluate the new integral

Now, we have a simple integral to evaluate: $$\int v dv = \frac{1}{2}v^2 + C$$

Substitute back the original variable

We made the substitution \(v = \tanh w\) at the beginning, so we'll substitute back our original variable: $$\frac{1}{2}v^2 + C = \frac{1}{2}(\tanh w)^2 + C$$

Write the final result

Our evaluated integral is: $$\int \operatorname{sech}^2 w \tanh w dw = \frac{1}{2}(\tanh w)^2 + C$$

Key concepts

Hyperbolic Functions

Hyperbolic functions are analogs of the trigonometric functions, but they are based on hyperbolas instead of circles. Two of the primary hyperbolic functions are the hyperbolic sine, denoted as \( \sinh(x) \), and the hyperbolic cosine, denoted as \( \cosh(x) \). These functions have many properties similar to their trigonometric counterparts, such as being used to define other hyperbolic functions.

Some key hyperbolic functions include:

• \( \tanh(x) \): the hyperbolic tangent, defined as \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \)
• \( \operatorname{sech}(x) \): the hyperbolic secant, defined as \( \operatorname{sech}(x) = \frac{1}{\cosh(x)} \)
• \( \operatorname{csch}(x) \): the hyperbolic cosecant, defined as \( \operatorname{csch}(x) = \frac{1}{\sinh(x)} \)

These functions are critical in various physics and engineering applications where exponential growth and decay describe real-world phenomena.

One of the unique aspects of hyperbolic functions is their derivatives, which exhibit simple patterns. For example, the derivative of the hyperbolic tangent \( \tanh(x) \) is \( \operatorname{sech}^2(x) \). This property proves useful in many calculus problems, including the integration problem we are dealing with in this exercise.

Substitution Method

The substitution method is a crucial technique in calculus, especially when evaluating integrals. It involves substituting a part of the integral with a new variable, which simplifies the original integral. This method is especially useful when dealing with composite functions or when recognizing patterns within an integral.

Here’s a simplified approach to using substitution when integrating:

• Identify a part of the integral that can be substituted with a new variable, say \( u \).
• Calculate the derivative of this new variable with respect to the original variable, \( \frac{du}{dx} \).
• Replace the identified part and its differential in the integral with \( u \) and \( du \).
• Evaluate the simplified integral.
• Substitute back the original variable into the result.

This method was employed in our exercise to simplify the integral \( \int \operatorname{sech}^{2} w \tanh w \, dw \) by setting \( v = \tanh w \). Noticing that \( \frac{dv}{dw} = \operatorname{sech}^2 w \), meant we could substitute \( dv \) for \( \operatorname{sech}^2 w \, dw \), leading to an easier integral \( \int v \, dv \).

Integration Techniques

Integration techniques help solve different types of integrals, making complex problems more approachable. Among the various techniques are substitution and integration by parts, with substitution being particularly beneficial when an integrand has a derivative present, making it easy to reformulate the integral.

The substitution technique helps transform complicated integrals into simpler forms that are more straightforward to solve. Here’s an important fact to consider when choosing an integration technique:

• If the derivative of a function within your integrand also appears within the integral, substitution might simplify the problem.
• Utilizing patterns within the integrand can guide choosing the right substitution.

In our specific case, recognizing that \( \frac{d}{dw} \tanh w = \operatorname{sech}^2 w \) guided our substitution choice. The presence of \( \operatorname{sech}^2 w \) as the derivative simplified the integral to \( \int v \, dv \), a straightforward polynomial integral.

Thus, mastering various integration techniques expands the possibilities for solving a broader range of integral calculus problems, and repeated practice helps in recognizing when and how to use each technique effectively.