Problem 398 - Calculus Volume 1

Question

For the following exercises, find the antiderivatives for the given functions. $$x^{3} \tanh \left(x^{4}\right)$$

Step-by-Step Solution

Answer

The antiderivative is \( \frac{1}{4} \ln|\cosh(x^4)| + C \).

Step 1: Understand the Problem

We need to find the antiderivative (also known as the indefinite integral) of the function \( f(x) = x^3 \tanh(x^4) \). This requires integrating \( x^3 \tanh(x^4) \) with respect to \( x \).

Step 2: Choose a Substitution

We see that the argument of the \( \tanh \) function, \( x^4 \), suggests using substitution. Let \( u = x^4 \), so \( \frac{du}{dx} = 4x^3 \), or \( du = 4x^3 \, dx \). Therefore, \( x^3 \, dx = \frac{1}{4} \, du \).

Step 3: Substitute Variables

Substitute the expressions for \( u \) and \( du \) into the integral:\[ \int x^3 \tanh(x^4) \, dx = \int \tanh(u) \cdot \frac{1}{4} \, du = \frac{1}{4} \int \tanh(u) \, du \]

Step 4: Integrate the New Function

The antiderivative of \( \tanh(u) \) is \( \ln|\cosh(u)| + C \), where \( C \) is the constant of integration. So we have:\[ \frac{1}{4} \int \tanh(u) \, du = \frac{1}{4} [ \ln|\cosh(u)| + C ] = \frac{1}{4} \ln|\cosh(u)| + \, C' \]

Step 5: Substitute Back the Original Variable

Recall that \( u = x^4 \). Substitute \( u \) back into the expression:\[ \frac{1}{4} \ln|\cosh(x^4)| + C' \]

Step 6: Conclude the Solution

The antiderivative of the function \( x^3 \tanh(x^4) \) with respect to \( x \) is \( \frac{1}{4} \ln|\cosh(x^4)| + C \), where \( C \) is the integration constant.

Key Concepts

Substitution Method

The substitution method is a powerful tool used in calculus to simplify the process of integration, particularly when dealing with composite functions. The underlying idea is to transform a difficult integral into a simpler one by changing the variables.

In the original exercise, we faced the challenge of integrating the function \(x^3 \tanh(x^4)\). To facilitate this, we used substitution to replace parts of the function with a new variable:

• Start by identifying a part of the function to substitute. Here, \(u = x^4\) was chosen because it simplifies the integration of the hyperbolic function \(\tanh(x^4)\).
• Compute the derivative of your new variable with respect to the original one: \(\frac{du}{dx} = 4x^3\).
• Rearrange for \(dx\) and express in terms of \(du\): \(du = 4x^3 \, dx\), or \(x^3 \, dx = \frac{1}{4} \, du\).
• Substitute back into the integral to get \(\frac{1}{4} \int \tanh(u) \, du\).

The substitution method transforms the original problem into one that is much easier to solve, paving the way for integration.

Indefinite Integral

An indefinite integral, often referred to as an antiderivative, involves finding a function whose derivative matches a given function. Unlike definite integrals, which calculate the exact area under a curve, indefinite integrals represent families of functions without specific bounds.

In solving the exercise, we were tasked with finding the indefinite integral of \(x^3 \tanh(x^4)\). The process aimed to identify a function\( F(x) \) such that \(F'(x) = x^3 \tanh(x^4)\).

• The use of substitution simplifies the original expression, making the integration process more straightforward.
• Post-substitution, it is common to encounter simple integrals. Here, \(\int \tanh(u) \, du\) was tackled with ease compared to its original form.
• Don't forget the constant \(C\) of integration, signifying that indefinite integrals represent a family of functions.

In this exercise, the constant \(C\) is necessary to account for any vertical shifts of the resulting function.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but are expressed in terms of exponential functions. They often appear in calculus due to their unique properties and relationship with differential equations and integrals.

The function in our exercise, \( \tanh(x^4) \), is the hyperbolic tangent function, which can be expressed as \(\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\), where \(\sinh\) and \(\cosh\) are the hyperbolic sine and cosine functions, respectively.

• Hyperbolic functions behave similarly to their trigonometric counterparts but with differences in their derivative forms.
• The integral of \(\tanh(u)\), encountered here, results in \(\ln|\cosh(u)| + C\), highlighting a peculiar property of hyperbolic functions.
• These functions are especially relevant in hyperbolic geometry and in solving hyperbolic differential equations.

Understanding hyperbolic functions can expand one's mathematical toolkit, enabling the solution of complex integrals like the one tackled in this exercise.