Question

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals. \(\int \frac{d t}{\left(e^{t}-e^{-t}\right)^{2}}\)

Short answer

\( \int \frac{d t}{\left(e^{t}-e^{-t}\right)^{2}} = -\frac{1}{e^t - e^{-t}} + C \)

Step-by-step solution

Identify the Substitution

To simplify the integral, we start by using the substitution \( u = e^t - e^{-t} \). This transforms our variable of integration from \( t \) to \( u \). Now, we differentiate to find \( dt \) in terms of \( du \). Since \( du = (e^t + e^{-t}) dt \), we rearrange to get \( dt = \frac{du}{e^t + e^{-t}} \). Also, note that \( e^t - e^{-t} = u \) suggests simple expressions for \( e^t \) and \( e^{-t} \).

Express \( e^t \) and \( e^{-t} \) in terms of \( u \)

Notice that \( (e^t)^2 - 1 + (e^{-t})^2 = 2 \). So, if \( r = e^t \), then \( u = r - \frac{1}{r} \) implies \( ru = r^2 - 1 \), giving us the quadratic equation \( r^2 - ur - 1 = 0 \). To solve for \( r \), which is \( e^t \), we use the quadratic formula: \( r = \frac{u \pm \sqrt{u^2 + 4}}{2} \).

Conversion of Integral

Substitute \( u = e^t - e^{-t} \) into the integral. We have \( dt = \frac{du}{e^t + e^{-t}} = \frac{du}{u} \) for the integral. Now, the integral becomes \( \int \frac{du}{u^2} \) which is much easier to solve.

Solve the Integral

The integral \( \int \frac{du}{u^2} \) is a simple power integral. It equals \(-\frac{1}{u} + C\), where \( C \) is the integration constant.

Back Substitute

Now, we substitute back \( u = e^t - e^{-t} \) to return to the original variable. The integral becomes \(-\frac{1}{e^t - e^{-t}} + C\).

Key concepts

Substitution Method

In integral calculus, the substitution method is a crucial technique. It aids in simplifying complex integrals by changing variables. When confronted with difficult integrals, we often look for substitutions that transform them into a more manageable form. For instance, in the given exercise, a substitution is made to convert the integral \( \int \frac{d t}{(e^{t}-e^{-t})^{2}} \) into a more straightforward expression. Here, we let \( u = e^t - e^{-t} \). This represents a strategic choice, making differentiation and integration processes simpler. Upon differentiating, we find the relation \( du = (e^t + e^{-t}) \, dt \). Hence, the original integral transforms into an integral with respect to \( u \), leading us to define \( dt \) in terms of \( du \): \[ dt = \frac{du}{e^t + e^{-t}}. \]This substitution reveals a simpler integral to evaluate. Understanding when and how to apply the substitution method is key to solving complex problems in integral calculus. It leads to transformed equations that are much easier to integrate.

Partial Fractions

The partial fractions decomposition is a valuable method when dealing with rational functions in integrals. Rational functions are quotients of polynomials, and decomposing them can simplify the integration process. In the context of our exercise, once we transform the original integral using substitution, it becomes crucial to analyze its new form. Partial fractions are usually applied to simplify such expressions further, especially when the denominator can be factored into linear or quadratic factors. While the demonstrated solution doesn't explicitly apply partial fractions after substitution, understanding this technique is valuable. If the resulting integral from substitution were more complex, partial fractions could be the next step. It involves expressing the integrand as a sum of simpler fractions, each of which is easier to integrate. Grasping partial fractions benefits students by offering enhanced tools to tackle various rational integrals, breaking them into more manageable parts.

Rational Functions

Rational functions are expressions that represent the ratio of two polynomials. They are central in many calculus problems, particularly in integral calculus. In our exercise, the transformation step converts a complex trigonometric-like integral into an integral of a simpler rational function. Understanding rational functions is key because once expressed in terms of rational functions, many advanced techniques such as substitution and partial fractions become applicable.For instance, after using the substitution \( u = e^t - e^{-t} \), the integral takes the rational form \( \int \frac{du}{u^2} \). This transformation dramatically eases the problem, allowing straightforward application of power integral rules.Being adept with rational functions helps students recognize when and how to apply specific techniques for integration, improving problem-solving skills and efficiency.

Power Integrals

Power integrals are a fundamental concept in calculus, representing the integration of functions of the form \( x^n \). Learning to deal with power integrals is essential for calculus students. In the exercise, the integral becomes \( \int \frac{du}{u^2} \) after substitution. This is a classic power integral, which falls under the category of \( \int x^n \, dx \). Here, \( n = -2 \), so we apply the power rule for integration: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \text{ for } n eq -1. \]For \( n = -2 \), this simplifies to \[ \int \frac{du}{u^2} = -\frac{1}{u} + C. \]Understanding power integrals helps students quickly solve such problems by applying a systematic formula. This knowledge is foundational, enabling students to handle more complex integrals by breaking them down into simpler components.