Problem 30 - Applied Calculus

Question

Find each integral by using the integral table on the inside back cover. $$ \int \frac{e^{t}}{9-e^{2 t}} d t $$

Step-by-Step Solution

Answer

The integral is \( \frac{1}{3} \text{arctanh}\left(\frac{e^{t}}{3}\right) + C \).

Step 1: Identify the Format

First, identify the integral you need to solve. The integral given is \( \int \frac{e^{t}}{9-e^{2t}} \, dt \). We try to relate this to a formula in the integral table that matches a specific integration rule.

Step 2: Use Substitution Method

Notice the denominator is \( 9 - e^{2t} \). Use the substitution \( u = e^{t} \), hence \( du = e^t \, dt \) or \( dt = \frac{1}{e^{t}} du \). Now rewrite the integral as \( \int \frac{1}{9-u^2} \, du \).

Step 3: Recognize Integral Form

Now the integral \( \int \frac{1}{9-u^2} \, du \) resembles the standard form \( \int \frac{1}{a^2 - u^2} du = \frac{1}{a} \text{arctanh}\left(\frac{u}{a}\right) + C \), with \( a = 3 \) since \( 9 = 3^2 \).

Step 4: Apply the Integral Formula

Apply the formula from Step 3: \( \int \frac{1}{9-u^2} \, du = \frac{1}{3} \text{arctanh}\left(\frac{u}{3}\right) + C \).

Step 5: Reverse the Substitution

Substitute back \( u = e^{t} \) into the result: \( \frac{1}{3} \text{arctanh}\left(\frac{e^{t}}{3}\right) + C \). This is the solution for the original integral.

Key Concepts

Substitution Method

The substitution method is a powerful technique in integral calculus that simplifies complex integrals by changing variables. It works by identifying a part of the integrand to substitute with a new variable, typically simplifying the underlying algebra. In the exercise, we faced an integral with an exponent, specifically \( \int \frac{e^{t}}{9-e^{2t}} \, dt \). Here, the substitution method helps to transform the integral into a more manageable form.

To utilize the substitution method, follow these steps:

• Identify a substitution that simplifies the integral, generally by setting \( u \) to be a function of \( x \). In this case, let \( u = e^t \).
• Express \( dt \) in terms of \( du \). Since \( du = e^t \, dt \), it follows that \( dt = \frac{1}{e^t} \, du \).
• Rewrite the integral with respect to \( u \). This transforms our integral into \( \int \frac{1}{9 - u^2} \, du \).

This method allows us to leverage integral tables or standardized forms to find a solution efficiently.

Integral Table

Integral tables are handy tools in calculus that list integrals of known functions, often displaying results of complex integration formulas. These tables provide solutions for a wide range of integral forms, enabling quicker calculations when attempting to solve integrals manually.

A typical integral table will feature standard forms such as:

• \( \int \frac{1}{a^2 - x^2} \, dx = \frac{1}{a} \text{arctanh}\left(\frac{x}{a}\right) + C \)
• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

In our exercise, after using the substitution method, the integral \( \int \frac{1}{9-u^2} \, du \) fits the specific form \( \int \frac{1}{a^2 - u^2} \, du \). Recognizing this allowed us to directly apply the formula to find the solution efficiently.

Hyperbolic Functions

Hyperbolic functions are analogues of trigonometric functions but are based on hyperbolas rather than circles. They are defined using exponential functions, providing valuable solutions in integral calculus. In our problem, we encountered the hyperbolic arctangent function (\( \text{arctanh}(x) \)), which often appears in integrals.Briefly, the hyperbolic functions include:

• Sinh: \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• Cosh: \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)
• Tanh: \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \)
• Arctanh: inverse function of \( \tanh(x) \)

In our solution, we derived \( \int \frac{1}{9-u^2} \, du = \frac{1}{3} \text{arctanh}\left(\frac{u}{3}\right) + C \) using the standard integration formula. Understanding these functions is essential, as they simplify solving integrals involving squared variables in the denominator.