Question

For the following exercises, find the antiderivatives for the given functions.\(\tanh (3 x+2)\)

Short answer

The antiderivative is \( \frac{1}{3} \ln(\cosh(3x+2)) + C \).

Step-by-step solution

Identify the Integral

We need to find the antiderivative of the function \( \tanh(3x+2) \). An antiderivative of a function is another function whose derivative is the original function.

Recall the Derivative of \(\tanh(x)\)

Recall that the derivative of \( \tanh(x) \) is \( \text{sech}^2(x) \). Therefore, to find the antiderivative of \( \tanh(u) \), we need to consider \( \int \text{sech}^2(u) \, du \).

Change of Variables

Make a substitution to facilitate integration. Let \( u = 3x + 2 \), then \( du = 3 \, dx \) or \( dx = \frac{1}{3} \, du \).

Rewrite the Integral with Substitution

Substitute \( u \) and \( dx \) into the integral: \( \int \tanh(3x+2) \, dx = \int \tanh(u) \cdot \frac{1}{3} \, du \).

Solve the Integral

Now, the integral becomes \( \frac{1}{3} \int \tanh(u) \, du \). Since the integral of \( \tanh(u) \) is \( \ln(\cosh(u)) \), we have:\[ \frac{1}{3} \ln(\cosh(u)) + C \]

Substitute Back

Replace \( u = 3x + 2 \) back into the solution:\[ \frac{1}{3} \ln(\cosh(3x+2)) + C \] is the antiderivative of \( \tanh(3x+2) \).

Key concepts

Hyperbolic Functions

Hyperbolic functions are analogs to the trigonometric functions but are related to hyperbolas instead of circles. These functions include the hyperbolic sine, cosine, and tangent, denoted as \( \sinh(x) \), \( \cosh(x) \), and \( \tanh(x) \) respectively. These functions have unique properties that make them very useful in calculus and in describing certain physical phenomena.
The hyperbolic tangent function, \( \tanh(x) \), is defined as the ratio of hyperbolic sine and cosine:

• \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \)

It is important because it appears in many areas of mathematics, including the study of differential equations and complex analysis.
Its derivative, \( \text{sech}^2(x) \), is often used in integration, especially when finding antiderivatives of hyperbolic functions.

Substitution Method

The substitution method is a powerful integration technique that simplifies integrals by transforming them into more familiar forms.

• It involves changing variables so that the integral becomes easier to solve.

In the exercise, we are given the function \( \tanh(3x + 2) \). By performing a substitution, we can make the integration process more straightforward.

How Substitution Works

Here's a simplified overview of applying the substitution method:

• Identify a part of the integrand to substitute, say \( u = 3x + 2 \).
• Find \( du \), which is the derivative of \( u \) with respect to \( x \). In this case, \( du = 3 \, dx \).
• Express \( dx \) in terms of \( du \), here \( dx = \frac{1}{3} \, du \).
• Substitute \( u \) and \( dx \) back into the integral, which changes the function into an integral in terms of \( u \).

This results in a simpler integral that can be integrated using standard integration techniques.
After solving the integral, don’t forget to substitute back the original variable to obtain the antiderivative in terms of \( x \).

Integration Techniques

Integrating hyperbolic functions like \( \tanh(x) \) involves techniques that allow us to find their antiderivatives effectively. In this context, we focus on common strategies like using known antiderivatives and substitution.

• One key strategy involves recognizing the basic integral forms of hyperbolic functions. For example, the integral of \( \tanh(u) \) is \( \ln(\cosh(u)) + C \).

Additionally, understanding how to apply the chain rule in reverse is another crucial part of integration techniques.
When substituting, matching the structure of derivatives also helps simplify the integration process.
The integral \( \int \tanh(3x+2) \, dx \) is solved through following these techniques:

• Applying substitution to create a simpler integral expression.
• Recognizing known integral forms to integrate efficiently.

Finally, substituting back the original terms provides the solution in the original context of the variable, leading to the antiderivative \( \frac{1}{3} \ln(\cosh(3x+2)) + C \). Understanding these techniques helps deal with complex integrals involving hyperbolic functions.