Problem 90 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int \frac{e^{x} \sin ^{-1} e^{x}}{\sqrt{1-e^{2 x}}} d x$$

Step-by-Step Solution

Answer

Evaluate the integral as \( e^x \sin^{-1} e^x + \sqrt{1 - e^{2x}} + C \).

Step 1: Identify the Substitution

Notice that the integrand has an expression involving \( e^x \) and \( \sin^{-1} e^x \). Both suggest using a substitution to simplify the integral. Let's set \( u = e^x \). Then, \( du = e^x \cdot dx = u \cdot dx \), which implies \( dx = \frac{du}{u} \).

Step 2: Apply the Substitution

Substituting \( u = e^x \) into the integral gives:\[ \int \frac{u \sin^{-1} u}{\sqrt{1-u^2}} \cdot \frac{du}{u} \]This simplifies to:\[ \int \frac{\sin^{-1} u}{\sqrt{1-u^2}} du \]

Step 3: Recognize the Integral Form

The integral \( \int \frac{\sin^{-1} u}{\sqrt{1-u^2}} du \) is a standard form that can be easily evaluated. Recognize that the antiderivative of \( \frac{\sin^{-1} u}{\sqrt{1-u^2}} \) with respect to \( u \) is \( u \cdot \sin^{-1} u + \sqrt{1-u^2} + C \).

Step 4: Back-Substitute

Substitute back the original \( u = e^x \) into the antiderivative:\[ e^x \sin^{-1} e^x + \sqrt{1 - (e^{x})^2} + C \]

Step 5: Finalize the Solution

The integral evaluates to:\[ e^x \sin^{-1} e^x + \sqrt{1 - e^{2x}} + C \] where \( C \) is the constant of integration.

Key Concepts

Substitution Method

Integration using the substitution method is a powerful technique that can greatly simplify complex integrals. In our exercise, the presence of functions like \( e^x \) and \( \sin^{-1} e^x \) hints at substitution. We choose \( u = e^x \), which conveniently simplifies the expression under the integral sign. This method transforms both the integrand and the differential, making the problem more manageable.

• Start by identifying a part of the integrand that can be replaced with a simpler variable.
• Determine the derivative of the chosen substitution variable (\( u = e^x \)) to substitute \( dx \).
• Re-write the entire integral in terms of this new variable, \( u \).

By substituting effectively, we turn an intricate integral into a known form, which paves the way for the straightforward application of integration techniques. This method is widely used when the integral includes compositions of functions.

Inverse Trigonometric Functions

Inverse trigonometric functions like \( \sin^{-1} \), \( \cos^{-1} \), and \( \tan^{-1} \) are essential in integration, especially when dealing with integrals involving roots and fractions. These functions can be tricky to integrate directly. In this exercise, the expression \( \sin^{-1} e^x \) is paired with \( \frac{1}{\sqrt{1-u^2}} \), a derivative form linked to inverse sine functions.

• The derivative of \( \sin^{-1} u \) is \( \frac{1}{\sqrt{1-u^2}} \), making these integral expressions recognizable.
• Inverse trigonometric integrals often appear in situations where the standard substitution method or partial fractions don't apply.
• Understanding these derivatives helps identify when an inverse trigonometric technique can simplify integration.

Mastering inverse trigonometric integrals adds a robust tool to your mathematical toolkit, crucial for tackling a wider range of problems.

Definite and Indefinite Integrals

The distinction between definite and indefinite integrals is fundamental to calculus. In our exercise, we dealt with an indefinite integral as indicated by the presence of the constant \( C \).With indefinite integrals:

• The objective is to find the general antiderivative of a function.
• They are represented as \( \int f(x) \, dx = F(x) + C \), where \( C \) is the constant of integration.
• This constant accounts for any vertical shifts in the family of antiderivative functions.

On the other hand, definite integrals compute the net area under a curve between two points. They provide a numerical value, indicating a specific solution.In practice, recognizing when an integral needs to end with a constant or numerical value helps clarify your approach and solution. Indefinite integrals are foundational for solving differential equations and understanding the behavior of functions.