Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Perform the indicated integrations. $$ \int \frac{6 e^{x}}{\sqrt{1-e^{2 x}}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(6e^x + C\).

Step 1: Identify the substitution

Here, a substitution can simplify the integral. Notice the expression under the square root: \(1 - e^{2x}\). This suggests using a trigonometric substitution. Specifically, let \( e^{x} = \sin(u) \), then \( dx = \frac{1}{e^{x}} \cos(u) \, du \).

Step 2: Substitute and simplify

Substitute \( e^x = \sin(u) \) and \( dx = \frac{1}{e^{x}} \cos(u) \, du \) into the integral:\[\int \frac{6 e^{x}}{\sqrt{1-e^{2x}}} \, dx = \int \frac{6 \sin(u)}{\sqrt{1-\sin^2(u)}} \cdot \frac{\cos(u)}{\sin(u)} \, du\]Simplifying, we get:\[6 \int \cos(u) \, du\]

Step 3: Integrate the simpler expression

The integral \(6 \int \cos(u) \, du\) is straightforward and becomes:\[6 \sin(u) + C\]

Step 4: Reverse the substitution

Recall the substitution \( e^{x} = \sin(u) \). We originally had \( u = \arcsin(e^{x}) \). Therefore, \( \sin(u) = e^{x} \). Our integrated result becomes:\[6 e^{x} + C\]

Step 5: Present the final answer

As a result, the integral is:\[\int \frac{6 e^{x}}{\sqrt{1-e^{2x}}} \, dx = 6 e^{x} + C\]

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used to simplify integrals by leveraging known trigonometric identities. It is especially useful when dealing with integrals involving square roots, as in our problem. The basic idea is to substitute a trigonometric function for a variable to simplify the expression. In this exercise, we faced the expression \(1 - e^{2x} \) under a square root. The substitution \( e^{x} = \sin(u) \) was chosen because it makes the expression under the square root become \( \cos(u) \), thanks to the identity \( \cos^2(u) = 1 - \sin^2(u) \). When you select a trigonometric substitution like \( e^{x} = \sin(u) \), you also adjust the differential \( dx \) to match the new variable \( du \). Here, differentiating gives \( dx = \frac{1}{e^{x}} \cos(u) \, du \), which is critical for the transformation of the original integral into one that's easier to solve. This step-by-step substitution helps to bridge the gap from a complex-looking integral to a form that is straightforward to integrate.

Indefinite Integral

An indefinite integral represents a family of functions whose derivative is the integrand. This means when you find an indefinite integral, you are essentially determining the original function given its rate of change (derivative). In mathematical notation, this is expressed as \( \int f(x) \, dx = F(x) + C \), where \( F(x) \) is the antiderivative of \( f(x) \), and \( C \) is the constant of integration. In this exercise, the given integral \( \int \frac{6 e^{x}}{\sqrt{1-e^{2x}}} \, dx \) was reduced through substitutions to a simpler form \( 6 \int \cos(u) \, du \). Finding its indefinite integral involves computing the antiderivative of \( \cos(u) \), which is \( \sin(u) \). Thus, the solution to the integral is \( 6 \sin(u) + C \). The constant \( C \) is crucial because the process of differentiation eliminates constants, leaving their reverse identification during integration necessary for a complete solution.

Calculus Problem Solving

Calculus problem solving involves systematically transforming a problem until it becomes manageable, following clear steps and logical reasoning. Each integration problem often requires understanding which techniques to apply, such as substitution methods, partial fractions, or trigonometric identities. A clear path to solving calculus problems often involves:

• Identifying the problem's structure and deciding on a suitable technique or substitution to simplify.
• Transforming the variables and equations while keeping the problem's ultimate requirements in focus.
• Integrating simple expressions and linking back to the original variables.
• Including constants of integration to ensure comprehensive solutions.

In our exercise, the problem-solving approach began by recognizing the benefit of trigonometric substitution to simplify the complex square root in the denominator. The equations were methodically transformed into simpler terms, integrated, and then reversed back to the original variable context. This thorough process ensures that even seemingly complicated calculus problems are broken into small, solvable parts, demonstrating effective problem-solving in calculus.