Problem 25 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} d x$$

Step-by-Step Solution

Answer

Question: Evaluate the integral $$\int \frac{e^x + e^{-x}}{e^x - e^{-x}}\,dx.$$ Answer: The integral evaluates to $$-2(1-e^{-2x})-\ln|1-e^{-2x}|+C,$$ where \(C\) is the constant of integration.

Step 1: Simplify the integrand

Divide the numerator and denominator by \( e^x \). $$ \int \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} \, d x = \int \frac{\frac{e^x}{e^x}+\frac{e^{-x}}{e^x}}{\frac{e^x}{e^x}-\frac{e^{-x}}{e^x}} \, d x = \int \frac{1+e^{-2x}}{1-e^{-2x}} \, d x $$

Step 2: Make a substitution to simplify the integrand further

Let \(u=e^{-2x}\). Then, \(-2e^{-2x}\,d x=\,d u\), and the integral becomes $$ \int \frac{1+u}{1-u}(-2\,d u)$$

Step 3: Evaluate the new integral

Now, we have $$ -2 \int \frac{1+u}{1-u} \, d u $$ To solve this integral, we can split the fraction up into two parts and integrate them individually. $$ -2 \left( \int \frac{1}{1-u} \, d u + \int \frac{u}{1-u}\, d u \right) $$ Now, recall that the antiderivative of \(\int \frac{1}{1-u} \, d u\) is \(-\ln|1-u|+C_1\). And for the second term, we can make another substitution: \(v=1-u\). Then, \(-d v=d u\) and we have: $$ \int \frac{1-v}{v}(-1\,d v),$$ which is equal to \(\int \frac{1}{v} \, d v - \int d v = \ln|v| - v + C_2\).

Step 4: Combine the terms and undo the substitutions

Combine the integrals and undo the substitutions \(u\) and \(v\) to get the final answer. $$ -2 \left( -\ln |1-u| + \ln |v| -v \right) + C_3 = -2 \left( -\ln |1-e^{-2x}| + \ln |1-e^{-2x}| - (1-e^{-2x}) \right) + C_4 $$ Finally, rearrange terms to write the result as: $$ -2(1-e^{-2x})-\ln|1-e^{-2x}|+C $$

Key Concepts

Integration by Substitution

Integration by substitution is a useful technique that simplifies complex integrals by transforming them into a more familiar form. This approach involves substituting a part of the integrand with a new variable, which often turns a complicated expression into a simpler one.
It is similar to using the chain rule in differentiation, but in reverse. For example, in the given integral \[ \int \frac{1+e^{-2x}}{1-e^{-2x}} \, dx \], substitution is employed by letting \( u = e^{-2x} \).
This changes the differential \( dx \) into a more manageable form \( -2e^{-2x} \, dx = \, du \), transforming the integral considerably.

• First, identify a part of the integrand that becomes easier on substitution.
• Use derivatives to find the relationship between the original variable and the new variable.
• Replace the functions and differentials in the integral with your substitution.

By making an appropriate substitution, you can often solve integrals that were initially very challenging.

Simplifying Integrals

Simplifying integrals can make complex calculus problems much easier to tackle. It involves reducing the integrand to its simplest form, allowing for easier evaluation. In the exercise, one of the key steps is to simplify the given function \( \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} \).
This is achieved by dividing both the numerator and denominator by \( e^x \). Through this algebraic simplification, the integrand becomes \( \frac{1+e^{-2x}}{1-e^{-2x}} \).
This form is more manageable and paves the way for effective substitution.

• Locate common terms and factors in the integrand.
• Use algebraic techniques such as factoring or division to simplify complex expressions.
• Always reassess complex fractions to see if there’s a lower terms representation.

By taking these steps, integration becomes a less daunting task and opens the doors to further techniques like substitution.

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. The function \( e^x \) is a standard exponential function and appears frequently in calculus due to its unique properties. In integrals, exponential functions often present complex challenges that need simplification techniques or substitutions to solve.
The exercise involves both \( e^x \) and \( e^{-x} \), with the substitution used as \( u = e^{-2x} \). This transformation showcases how exponential functions can be reshaped during integration to suit other more solvable forms.

• Exponential functions have powerful identities, such as \( e^{a+b} = e^a e^b \), that help simplify and manipulate expressions.
• They have a unique derivative and integral: the derivative and integral of \( e^x \) is \( e^x \) itself.
• Using exponential properties, especially during substitution, greatly aids in the integration process.

Understanding how to handle exponential functions is crucial for tackling integration problems involving these distinctive expressions.