Problem 396 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

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

Step 1: Review Familiar Antiderivative Forms

Recall that the hyperbolic tangent function, \( \tanh(x) \), is the derivative of the hyperbolic logarithm \( \ln(\cosh(x)) \). This piece of knowledge is useful for finding the antiderivative of similar forms.

Step 2: Identify the Inner Function

Identify the inner function of \( f(x) = \tanh(3x+2) \), which is \( g(x) = 3x+2 \). This is part of the chain rule in reverse (integration by substitution).

Step 3: Set Up the Integration by Substitution

Let \( u = 3x + 2 \). Calculate the derivative \( \frac{du}{dx} = 3 \). Then, solve for \( dx \): \( dx = \frac{du}{3} \).

Step 4: Substitute and Transform the Integral

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

Step 5: Integrate the Simplified Expression

Knowing that the integral of \( \tanh(u) \) is \( \ln(\cosh(u)) \), integrate to get \( \frac{1}{3} \ln(\cosh(u)) + C \), where \( C \) is the constant of integration.

Step 6: Substitute Back the Original Variable

Replace \( u \) with \( 3x + 2 \) to get the final antiderivative: \( \frac{1}{3} \ln(\cosh(3x+2)) + C \). This reverts the integration dilation back to the original terms.

Key Concepts

Integration by Substitution

Integration by substitution is a powerful technique in calculus, used to simplify the process of finding antiderivatives. It is often applied when dealing with composite functions, where one function is inside another. The idea is to transform a complex integral into a simpler form by identifying a part of the integrand as a new variable, often denoted as \( u \).

Here's how it works:

• Identify the inner function of the integrand. In our example, for \( \tanh(3x + 2) \), the inner function is \( g(x) = 3x + 2 \).
• Set \( u = g(x) \), making the derivative easy to find. Here, \( u = 3x + 2 \), so \( \frac{du}{dx} = 3 \).
• Express \( dx \) in terms of \( du \): \( dx = \frac{du}{3} \).
• Substitute \( u \) and \( dx \) into the integral to get a simpler one, `\int \tanh(u) \frac{du}{3}`.

Integration by substitution helps by rewiring the problem into something easier to integrate.

Hyperbolic Functions

Hyperbolic functions, including \( \tanh(x) \), are analogs of trigonometric functions but for hyperbolas instead of circles. They have similar properties and identities.

The hyperbolic tangent function, \( \tanh(x) \), is defined as \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \) where \( \sinh(x) \) and \( \cosh(x) \) are hyperbolic sine and hyperbolic cosine functions, respectively. These functions are vital in various areas, including calculus and differential equations.

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

Understanding these functions is crucial since they appear frequently in integration, as with \( \tanh(x) \), whose antiderivative is \( \ln(\cosh(x)) \).

Chain Rule in Calculus

The chain rule is a fundamental theorem in calculus used for differentiating composite functions. When reversing this process during integration (e.g., `integration by substitution`), we essentially apply the chain rule backwards.

The chain rule states that if you have a composite function \( f(g(x)) \), then its derivative is \( f'(g(x)) \cdot g'(x) \). For integration purposes, you find "what function, when derived, gives us the integrand."

For example, when integrating \( \tanh(3x + 2) \), recognizing the inner function \( g(x) = 3x + 2 \) allows substitution. This substitution, followed by reversing integration, simplifies handling the integrations of more complex expressions by disentangling their layers.

Constant of Integration

In calculus, when finding an antiderivative, a constant of integration \( C \) must always be added to the solution. This constant accounts for any vertical shifts in the original function whose derivative results in the integrand.

The reason is due to the nature of differentiation. When a function \( F(x) \) is derived, any constant value added to it will disappear (as the derivative of a constant is zero). Therefore, the general antiderivative includes this unknown constant, \( F(x) + C \).

• Ensures all possible antiderivatives are covered.
• Critical when considering initial conditions in differential equations.

By incorporating \( C \), we acknowledge all potential original functions, providing a complete solution set.