Problem 32 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the indefinite integral. $$\int \frac{e^{x}-e^{-x}}{\left(e^{x}+e^{-x}\right)^{3 / 2}} d x$$

Step-by-Step Solution

Answer

The short answer to the indefinite integral \(\int \frac{e^{x}-e^{-x}}{\left(e^{x}+e^{-x}\right)^{3 / 2}} d x\) is \(-2\left( e^x + e^{-x}\right)^{-\frac{1}{2}} + C\).

Step 1: Identify the substitution

Let's make the substitution \(u = e^x + e^{-x}\). We will then express the integrand in terms of \(u\) to simplify it.

Step 2: Find the derivative of the substitution

To express \(dx\) in terms of \(du\), we need to find the derivative of the substitution with respect to x: $$\frac{d u}{d x} = \frac{d}{d x}\left(e^x + e^{-x}\right)$$ Using the chain rule and properties of exponentials, we obtain: $$\frac{d u}{d x} = e^x - e^{-x}$$ Therefore, we have $$d x=\frac{d u}{e^x-e^{-x}}$$

Step 3: Rewrite the integral in terms of u

Now, we substitute our expressions for u and dx into the original integral: $$\int \frac{e^x-e^{-x}}{\left(e^{x}+e^{-x}\right)^{3 / 2}} d x=\int \frac{1}{u^{3/2}}\: du$$

Step 4: Evaluate the integral in terms of u

Let's evaluate the integral in terms of \(u\): $$\int \frac{1}{u^{3/2}}\: du$$ Applying the power rule for integration, we have: $$-\int u^{-3/2}\: du = -2u^{-1/2} + C$$ where \(C\) is the constant of integration.

Step 5: Substitute back to x

Finally, we substitute back our original expression for \(u\) in terms of \(x\): $$-2u^{-1/2} + C = -2\left( e^x + e^{-x}\right)^{-\frac{1}{2}} + C$$ Thus, the indefinite integral is: $$\int \frac{e^x-e^{-x}}{\left(e^{x}+e^{-x}\right)^{3 / 2}} d x = -2\left( e^x + e^{-x}\right)^{-\frac{1}{2}} + C$$

Key Concepts

Integration Techniques

Integration is a fundamental aspect of calculus, primarily used to find areas under curves. When faced with complex integrals, there are various strategies known as integration techniques to simplify and solve them. Among the methods are:

• Substitution: Transform the integral by substituting a complex part with a simpler variable, making the integration straightforward.
• Integration by Parts: Useful when integrals are products of functions, it applies the reverse of the product rule of differentiation.
• Partial Fractions: Applies primarily to rational functions to break them into simpler fractions before integrating.
• Trigonometric Integrals: Focuses on integrals involving trigonometric functions using trigonometric identities as transformations.
• Numerical Integration: Used for functions unable to be integrated analytically, using methods like Simpsons' Rule or the Trapezoidal Rule.

In the provided exercise, we adopted the substitution method to simplify the given integral. Breaking down the expression with substitution makes it more manageable, displaying the power of integration techniques to solve complex problems easily.

Substitution Method

The substitution method is a powerful strategy in integration, especially when dealing with complex algebraic expressions or functions. It involves replacing a part of an integrand, the expression being integrated, with a new variable, simplifying the integral.

Steps to Perform Substitution

• Choose the Substitution: Identify a part of the integrand to replace with a single variable, usually a function of the variable of integration.
• Derive Relation: Differentiate the chosen substitution to express the differential of the original integrand variable in terms of the new variable.
• Substitute in the Integral: Replace the original variable and differential in the integral with the new ones obtained from the substitution.
• Solve the New Integral: Integrate the simpler function with respect to the new variable.
• Back Substitute: Convert the result back to the original variable to finalize the solution.

In our example, identifying the substitution, finding its derivative, and substituting effectively transformed the complex original integral into a simpler one, showing why the substitution method is an integral part of solving comprehensive integrals.

Exponential Functions

Exponential functions, represented as forms of \(e^x\) where \(e\) is the base of the natural logarithm, are fundamental in continuous growth and decay processes across various scientific fields.

Characteristics of Exponential Functions

Exponential functions have unique properties making them significant:

• Rapid Growth/Decay: The rate of increase or decrease is proportional to the function's current value, leading to exponential growth or decay.
• Derivatives and Integrals: The derivative of \(e^x\) is itself, and similarly, the integral of \(e^x\) maintains this form, making them straightforward to differentiate and integrate.
• Complex Expressions: Functions like \(e^x + e^{-x}\) or \(e^x - e^{-x}\) often appear, requiring careful manipulation, as seen in the exercise.

Utilizing properties of exponential functions, we constructed transformations simplifying the expression. In integration exercises, recognizing these patterns efficiently helps evaluate the integral, demonstrating why exponential functions are pivotal in differential and integral calculus.