Problem 81 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

\( e^{\sin^{-1} x} + C \)

Step 1: Recognize the Inverse Trigonometric Function

The function inside the integral, \( \sin^{-1} x \), is the inverse sine function, also known as arcsin. Recall that its derivative is \( \frac{1}{\sqrt{1-x^2}} \). This indicates that substitution involving this function might simplify the integral.

Step 2: Use a Substitution for Simplification

Let \( u = \sin^{-1} x \), which implies \( x = \sin u \). Compute the derivative \( \frac{du}{dx} = \frac{1}{\sqrt{1 - x^2}} \), so \( dx = \sqrt{1-x^2} \, du \). Substituting these transformations into the integral simplifies it.

Step 3: Perform the Substitution

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

Step 4: Integrate the Simplified Expression

Now, integrate \( e^u \) with respect to \( u \): \[ \int e^u \, du = e^u + C \]. This antiderivative is straightforward as the derivative of \( e^u \) with respect to \( u \) is \( e^u \).

Step 5: Substitute Back to Original Variable

Replace \( u \) with \( \sin^{-1} x \) to revert back to the original variable: \[ e^{\sin^{-1} x} + C \]. The integral evaluates to this expression with the constant of integration \( C \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential elements in calculus, helping in various integrations and transformations. One of the most commonly encountered functions is the inverse sine function, noted as \( \sin^{-1} x \) or \( \text{arcsin} \,x \). This function is used to find the angle whose sine is \( x \). Thus, when you see an integral that involves \( \sin^{-1} x \), it's beneficial to remember its derivative formula.

• The derivative of \( \sin^{-1} x \) is \( \frac{1}{\sqrt{1-x^2}} \).
• This derivative can assist in recognizing when a substitution method might simplify an integral.

Inverse trigonometric functions extend our toolbox in solving integrals that would otherwise be challenging. They also appear in various pre-calculus topics, making them foundational for further studies in mathematics.

Substitution Method

The substitution method in calculus is like changing the clothes of integral problems, making them easier to handle. It's a tactical approach where you replace a part of the integral with another variable, simplifying the process of finding the solution. In solving the integral \( \int \frac{e^{\sin^{-1} x} \ dx}{\sqrt{1-x^{2}}} \), substitution plays a critical role:

• Identify the substitution with \( u = \sin^{-1} x \), leading to \( x = \sin u \).
• Find \( dx \) using \( \frac{du}{dx} = \frac{1}{\sqrt{1 - x^2}} \), hence \( dx = \sqrt{1-x^2} \, du \).
• Replace \( dx \) and simplify the integral to \( \int e^u \, du \).

With the proper substitution, you convert the complex integral into a simpler form like \( \int e^u \, du \), which is easy to evaluate. Substitution is an indispensable technique that leverages algebraic manipulation to make calculus more manageable.

Calculus

Calculus, the magical land of derivatives and integrals, is about change and accumulation. In this exercise, you've walked through an integral that combines trigonometric components with exponential functions. Calculus helps in understanding how to manipulate and solve such integrals.

• The exercise illustrates using differentiation, which is understanding rates of change, to solve antiderivatives (integration).
• Integrating \( e^u \) becomes straightforward once substitution transforms the complex function into something simpler.
• The constant of integration, \( C \), always appears in indefinite integrals, reminding us of the family of functions sharing the same derivative.

As you venture through calculus, these concepts of differentiation, integration, and strategy selection like substitution will become your constant companions. They equip you with tools to unravel the intricacies of functions and their behaviors.