Problem 44 - Calculus: One and Several Variables

Question

Integrate by setting \(u=\tanh \frac{1}{2} x.\) $$\int \frac{1-e^{x}}{1+e^{x}} d x$$

Step-by-Step Solution

Answer

The short answer is: \[\int \frac{1-e^{x}}{1+e^{x}} d x = 2\tanh(\frac{1}{2}x) + C\]

Step 1: Calculate the derivative of the substitution function

We are given the substitution function \(u = \tanh(\frac{1}{2}x)\). We need to find \(\frac{d u}{d x}\) to replace \(dx\) in the integral. \[\frac{d u}{d x} = \frac{d}{d x} \tanh(\frac{1}{2}x) = \frac{1}{2} \cdot \mathrm{sech}^2(\frac{1}{2}x)\] So, \(d x = \frac{2 d u}{\mathrm{sech}^2(\frac{1}{2}x)}\).

Step 2: Rewrite the integral in terms of u

To rewrite the given integral in terms of u, we will need to replace both the numerator and the denominator with expressions involving u. Let's start by rewriting \(e^{x}\) using the definition of the hyperbolic tan function: \[e^{x} = \frac{1+\tanh(\frac{1}{2}x)}{1-\tanh(\frac{1}{2}x)} = \frac{1+u}{1-u}\] Now we can substitute this into the integral expression and rewrite it as: \[\int \frac{1-\frac{1+u}{1-u}}{1+\frac{1+u}{1-u}} \frac{2 d u}{\mathrm{sech}^2(\frac{1}{2}x)}\] Simplify the expression in the integral: \[\int \frac{1-u^2}{1-u} \frac{2 d u}{\mathrm{sech}^2(\frac{1}{2}x)}\]

Step 3: Replace sech^2(½x) with an expression involving u

Since the integral is in terms of \(u\), we now need to replace \(\mathrm{sech}^2(\frac{1}{2}x)\) with an expression in terms of \(u\). Recall that: \[\mathrm{sech}^2(x) = 1 - \tanh^2(x)\] So, \(\mathrm{sech}^2(\frac{1}{2}x) = 1 - u^2\) Substitute this into the integral: \[\int \frac{2(1-u^2) d u}{1 - u^2}\]

Step 4: Perform Integration

Now we have the integral entirely expressed in terms of \(u\), and we can perform the integration: \[\int \frac{2(1-u^2) d u}{1 - u^2} = 2 \int d u\] Now, integrate the expression with respect to \(u\): \[2 \int d u = 2u + C\]

Step 5: Replace u with the original expression

Finally, replace \(u\) with the original expression \(\tanh(\frac{1}{2}x)\) to obtain the final result in terms of \(x\): \[2u + C = 2\tanh(\frac{1}{2}x) + C\] And this is the result of the given integral.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are similar to trigonometric functions but are based on hyperbolas instead of circles. The most common hyperbolic functions are the hyperbolic sine ( anh), hyperbolic cosine ( anh), and hyperbolic tangent ( anh), among others. These functions often appear in calculus, especially when dealing with equations involving exponential growth or certain types of integrals.

• Hyperbolic functions can be expressed in terms of exponential functions. For example, the hyperbolic tangent is defined as: \( \tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \).


• They are valuable because their derivatives and integrals can be more straightforward to compute in certain situations compared to other functions.


• The derivatives of hyperbolic functions mimic those of trigonometric functions, such as \( \frac{d}{dx} \tanh(x) = \mathrm{sech}^2(x) \).

Understanding hyperbolic functions is crucial when using substitution in integrals, as they provide a way to transform complex expressions into simpler forms.

Derivative Calculation

Derivative calculation is essential in finding the slope or rate of change of a function. For integration by substitution, understanding derivatives allows us to change the variable of integration effectively.

• In the given problem, the substitution function is \(u = \tanh(\frac{1}{2}x)\). The derivative of this function is needed to express \(dx\) in terms of \(du\).


• The derivative \( \frac{d}{dx} \tanh(\frac{1}{2}x) = \frac{1}{2} \cdot \mathrm{sech}^2(\frac{1}{2}x) \) helps in replacing \(dx\) when rewriting the integral in terms of \(u\).


• Recognizing derivatives such as \( \mathrm{sech}^2(x) = 1 - \tanh^2(x) \) helps to further simplify expressions when performing substitution.

By understanding derivative calculation, transforming variables becomes a systematic process, ensuring the integral is simpler to solve.

U-Substitution

U-substitution, also known as integration by substitution, is a powerful technique for evaluating integrals. It involves replacing a variable with another expression to simplify the integral.

• First, choose a substitution that simplifies the integral. In this problem, we let \(u = \tanh(\frac{1}{2}x)\).


• The goal is to rewrite all parts of the integral, including \(dx\), in terms of \(u\). This makes solving the integral more manageable.


• Replace the expressions using known identities and derivatives. In our example, \(e^x\) is rewritten using \(u\) as \(\frac{1+u}{1-u}\).

Through u-substitution, complex expressions are converted into simpler forms, allowing for easier integration. It is especially useful when compounds involve derivatives or composition of functions.

Definite and Indefinite Integrals

Integrals can be classified as definite or indefinite, each serving a distinct purpose in calculus. Understanding the difference helps in interpreting and calculating areas under curves and general antiderivatives.

• An indefinite integral, like the one in the problem, represents a family of functions rather than a specific value. It involves integrating an expression without upper and lower limits on the integral sign.


• The result of an indefinite integral is typically expressed with an arbitrary constant \(C\), reflecting the family of potential antiderivatives, such as \(2u + C\) in our problem.


• In contrast, a definite integral computes the exact area under a curve between two points, giving a specific numerical value.

Understanding these concepts helps in knowing whether you are determining a general form of a function or calculating a specific area or total change. Properly applying these types is crucial for various calculus applications, ensuring precise results.