Problem 66 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int \frac{\sec ^{2} y d y}{\sqrt{1-\tan ^{2} y}}$$

Step-by-Step Solution

Answer

The integral is \( \ln | \tan y + \sec y | + C \).

Step 1: Identify the Function Forms

The given integral is \( \int \frac{\sec^2 y \, dy}{\sqrt{1 - \tan^2 y}} \). Notice that it involves \( \sec^2 y \) and \( \tan^2 y \), which are trigonometric functions related through derivatives: the derivative of \( \tan y \) is \( \sec^2 y \). Also, recall the identity \( \sin^2 y + \cos^2 y = 1 \), which can be rewritten using \( \tan y \) and \( \sec y \).

Step 2: Simplify the Denominator

The term \( \sqrt{1 - \tan^2 y} \) can be simplified using the identity \( \sec^2 y = 1 + \tan^2 y \). Therefore, \( 1 - \tan^2 y = \frac{1}{\sec^2 y} \), and the square root of this term is \( \frac{1}{\sec y} \).

Step 3: Simplify and Substitute

Substituting the simplified expression into the integral, we get: \( \int \frac{\sec^2 y}{\frac{1}{\sec y}} \, dy = \int \sec^3 y \, dy = \int \sec y \cdot \sec^2 y \, dy \). Let \( u = \tan y \); then \( du = \sec^2 y \, dy \).

Step 4: Perform the Integration

The integral becomes \( \int \sec y \, du = \int \frac{1}{\cos y} \, du \). This integral can be evaluated by recognizing an easy substitution: let \( v = \sin y \), then \( dv = \cos y \, dy \), and \( y = \arcsin v \). The integral updates: \( \int \frac{1}{\sqrt{1-v^2}} \, du \), which integrates directly to \( \ln |\tan y + \sec y| + C \).

Step 5: Final Result

Thus, the evaluated integral \( \int \frac{\sec^2 y \, dy}{\sqrt{1 - \tan^2 y}} \) simplifies and evaluates to \( \ln | \tan y + \sec y | + C \), where \( C \) is the constant of integration.

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in solving integration problems involving trigonometric functions. These identities simplify expressions and make it easier to evaluate integrals. In the integral \( \int \frac{\sec^2 y \, dy}{\sqrt{1-\tan^2 y}} \), two crucial identities are used. The first is the Pythagorean identity, \( \sin^2 y + \cos^2 y = 1 \), which can be rewritten using \( \tan y \) and \( \sec y \) as \( \sec^2 y = 1 + \tan^2 y \). This identity helps express \( 1 - \tan^2 y \) in terms of \( \sec y \). By manipulating the identity, \( 1 - \tan^2 y \) becomes \( \frac{1}{\sec^2 y} \), allowing further simplification. Understanding and applying these identities is key to solving integrals involving trigonometric functions effectively.

Integration Techniques

Integration techniques are methods used to evaluate integrals that are not easily integrable at first glance. One popular technique is simplifying the integral using trigonometric identities. In this exercise, simplifying \( \sqrt{1 - \tan^2 y} \) to \( \frac{1}{\sec y} \) removes complexity in the denominator. This transforms the integral into a more workable form: \( \int \sec^3 y \, dy \). After simplifying, another technique involves addressing this form by breaking the integral into simpler parts: \( \int \sec y \cdot \sec^2 y \, dy \). This step paves the way for using substitution, a powerful strategy that changes the variable, simplifying the integration process by transforming it into a standard form. These techniques, when applied judiciously, make seemingly complex integrals more manageable.

Substitution Method

The substitution method is a straightforward yet powerful technique for solving integrals. It involves changing the variable to simplify both the integrand and the process of integration. In the exercise, the substitution \( u = \tan y \) is proposed, with the derivative \( du = \sec^2 y \, dy \). This choice arises naturally from noticing that the derivative matches part of the integrand, leading to \( \int \sec y \, du \). This substitution essentially reduces the integral to a simpler form, allowing us to recognize that the integral is a standard one, known to yield \( \ln |\tan y + \sec y| + C \). Applying substitution can greatly simplify solving integrals by transforming them into more familiar forms, highlighting the method’s value in integration.