Problem 45 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int \tanh \frac{x}{7} d x$$

Step-by-Step Solution

Answer

The integral is \( 7 \ln(\cosh \frac{x}{7}) + C \).

Step 1: Understand the Integral

We need to evaluate the integral \( \int \tanh \frac{x}{7} \, dx \). The function \( \tanh \frac{x}{7} \) is the hyperbolic tangent of \( \frac{x}{7} \).

Step 2: Use the Identity for Hyperbolic Tangent

Recall the identity \( \tanh u = \frac{e^u - e^{-u}}{e^u + e^{-u}} \), where \( u = \frac{x}{7} \). This means \( \tanh \frac{x}{7} = \frac{e^{x/7} - e^{-x/7}}{e^{x/7} + e^{-x/7}} \). However, a more direct approach to integration is using a standard result for the derivative of \( \ln(\cosh u) \), where \( \frac{d}{du} \ln(\cosh u) = \tanh u \).

Step 3: Apply Substitution

Let \( u = \frac{x}{7} \). Then, \( du = \frac{1}{7} dx \) or equivalently, \( dx = 7\, du \). Substituting these into the integral gives \( \int \tanh(u) \times 7 \, du \).

Step 4: Integrate the Function

The integral \( \int \tanh(u) \, du = \ln(\cosh u) + C \), where \( C \) is the constant of integration. Therefore, the integral becomes \( 7 \times \ln(\cosh u) + C \).

Step 5: Substitute Back in U Terms

Now substitute \( u = \frac{x}{7} \) back into the integral. This gives us \( 7 \times \ln(\cosh \frac{x}{7}) + C \).

Step 6: Conclusion

Hence, the evaluated integral of \( \int \tanh \frac{x}{7} \, dx \) is \( 7 \ln(\cosh \frac{x}{7}) + C \), where \( C \) is the constant of integration.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions, but they are based on hyperbolas rather than circles. These functions include hyperbolic sine, hyperbolic cosine, and hyperbolic tangent: \(\sinh\), \(\cosh\), and \(\tanh\), respectively. Each has a specific relation to exponential functions.
- The hyperbolic tangent, \(\tanh(u)\), is defined as \((e^u - e^{-u})/(e^u + e^{-u})\).- It is similar to the trigonometric function tangent, \(\tan(u)\), but serves different purposes in calculus and geometry.In calculus, hyperbolic functions are useful because they simplify derivatives and integrals involving exponential terms. For instance, the derivative of \(\ln(\cosh(u))\) is \(\tanh(u)\). This identity often helps in evaluating integrals that involve \(\tanh\). Knowing these relationships can greatly ease the process of solving complex calculus problems.

Substitution Method

The substitution method, also known as u-substitution, is a technique used in integration to simplify a given integral. This technique resembles the chain rule for differentiation. It involves changing the variable to make the function easier to handle.
To successfully apply substitution:

• Identify a function within the integral, usually a composite function, and set this function equal to \(u\).
• Follow by differentiating \(u\) to find \(du\) and then solve for \(dx\).

In this exercise, we use the substitution \(u = \frac{x}{7}\). The differential works out to \(dx = 7 du\). This substitution simplifies the original integral \(\int \tanh \frac{x}{7} dx\) to \(\int 7 \tanh u \, du\), which is then readily integrable. This method is fundamental for dealing with complex derivatives and integrals.

Indefinite Integrals

Indefinite integrals are integrals that do not have specified limits of integration. They represent a family of functions and include a constant of integration \(C\), which accounts for the fact that there are infinitely many antiderivatives.
- To compute an indefinite integral, identify the function to integrate and perform the integration process.- The result will show the general solution, representing the antiderivative.In this exercise, after applying the substitution method, we find that \(\int \tanh u \, du = \ln(\cosh u) + C\). Substituting back \(u = \frac{x}{7}\), the indefinite integral becomes \(7 \ln(\cosh \frac{x}{7}) + C\). Indefinite integrals are crucial in calculus as they help find antiderivatives, which is a step towards solving differential equations and assessing accumulation functions.