Problem 86 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int \frac{d y}{\left(\sin ^{-1} y\right) \sqrt{1-y^{2}}}$$

Step-by-Step Solution

Answer

\(\ln|\sin^{-1}(y)| + C\)

Step 1: Identify the Form

The integral \( \int \frac{dy}{(\sin^{-1} y) \sqrt{1-y^2}} \) suggests a trigonometric substitution. Recognize that the term \( \sqrt{1-y^2} \) can be related to trigonometric identities.

Step 2: Use Trigonometric Substitution

Let \( y = \sin(\theta) \). Then \( dy = \cos(\theta) d\theta \) and \( \sqrt{1 - y^2} = \cos(\theta) \). Thus the integral becomes \( \int \frac{\cos(\theta) d\theta}{(\theta) \cos(\theta)} \).

Step 3: Simplify the Integral

The simplification results in \( \int \frac{d\theta}{\theta} \). This is a standard integral that can be solved directly.

Step 4: Integrate

The integral \( \int \frac{d\theta}{\theta} \) is equal to \( \ln|\theta| + C \), where \( C \) is the constant of integration.

Step 5: Substitute Back to Original Variable

Recall that \( \theta = \sin^{-1}(y) \), so substitute back to get \( \ln|\sin^{-1}(y)| + C \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique in integral calculus used to simplify integrals involving expressions like \( \sqrt{1-y^2} \). This substitution is particularly useful when the integrals resemble the Pythagorean identity \( 1 - \sin^2(\theta) = \cos^2(\theta) \). The purpose is to transform a complex algebraic expression into a more manageable trigonometric one.

In the original exercise, we make use of the substitution \( y = \sin(\theta) \). This transforms \( dy \) to \( \cos(\theta) d\theta \) and \( \sqrt{1 - y^2} \) into \( \cos(\theta) \). The trigonometric substitution leads us from a complex integral to a simpler form where standard integration techniques can be applied without hassle. A key aspect is choosing the right substitution that aligns perfectly with the identity, enabling the cancellation of complex terms.

Definite Integral

The definite integral is a concept in calculus that calculates the net area under a curve over a specific interval. Unlike indefinite integrals, which include a constant of integration, definite integrals solve for an actual number, providing a specific solution or value.

While the original exercise provided did not specify bounds, the idea of a definite integral is still crucial to understanding how integrals can be used practically. By integrating from one boundary to another, you can find areas under curves, volumes of solids, and more.

• Definite integrals have limits of integration which denote the range of the area being covered.
• They can also be solved using trigonometric substitutions, especially if the bounds make the substitution straightforward.

For students, mastering definite integrals provides a foundation for further exploration into how calculus is applied in physical sciences and engineering.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for any value of the variable. They are essential tools in simplifying complex trigonometric expressions and solving equations.

In the case of the integral \( \int \frac{dy}{(\sin^{-1} y) \sqrt{1-y^2}} \), understanding identities like \( \cos^2(\theta) = 1 - \sin^2(\theta) \) allows us to substitute \( y = \sin(\theta) \), simplifying the \( \sqrt{1-y^2} \) to \( \cos(\theta) \). This transformation is pivotal as it breaks down complicated expressions into easily integrable forms.

Some important trigonometric identities include:

• Pythagorean identities, such as \( \sin^2(x) + \cos^2(x) = 1 \).
• Angle addition formulas, like \( \sin(a + b) = \sin a \cos b + \cos a \sin b \).

Firm grasp of these identities will simplify your work with integrals and enable the use of trigonometric substitution more effectively.

Inverse Trigonometric Functions

Inverse trigonometric functions are functions that reverse the operation of the standard trigonometric functions. They are crucial in finding angles corresponding to given trigonometric ratios.

In our exercise, the notable inverse function is \( \sin^{-1}(y) \), which serves as the angle whose sine is \( y \). Applying this inverse function within the integral allows for the back substitution after integration.

Inverse trigonometric functions appear frequently in integration, particularly when evaluating definite integrals or when solving differential equations. They are the tools we use to switch back to the original variable after making a trigonometric substitution that involves an inverse operation.

• Common inverse trigonometric functions include \( \sin^{-1}(x), \cos^{-1}(x), \tan^{-1}(x) \).
• These functions are used to resolve angles and are employed in calculus to convert trigonometric expressions back to their algebraic forms.

Understanding how these functions work will help you effectively complete integrations and make proper substitutions.