Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-28, perform the indicated integrations. \(\int \sin 4 y \cos 5 y d y\)

Step-by-Step Solution

Answer

The integral is \(-\frac{1}{18} \cos 9y + \frac{1}{2} \cos y + C.\)

Step 1: Use a Trigonometric Identity

To integrate \( \sin 4y \cos 5y \), we can use the product-to-sum identities in trigonometry. The identity states: \[ \sin A \cos B = \frac{1}{2}(\sin(A+B) + \sin(A-B)). \] For our problem, \( A = 4y \) and \( B = 5y \). Apply this identity to express the integrand as a sum of simpler sine functions.

Step 2: Substitute Values into the Identity

Substitute \( A = 4y \) and \( B = 5y \) into the identity: \[ \sin 4y \cos 5y = \frac{1}{2}(\sin(4y + 5y) + \sin(4y - 5y)). \] This simplifies to: \[ \sin 4y \cos 5y = \frac{1}{2}(\sin 9y + \sin(-y)). \] Since \( \sin(-y) = -\sin(y) \), the expression becomes: \[ \sin 4y \cos 5y = \frac{1}{2}(\sin 9y - \sin y). \]

Step 3: Rewrite the Integral

Rewrite the integral using the result from Step 2: \[ \int \sin 4y \cos 5y \, dy = \int \frac{1}{2}(\sin 9y - \sin y) \, dy. \] This can be split into two separate integrals: \[ \frac{1}{2} \left( \int \sin 9y \, dy - \int \sin y \, dy \right). \]

Step 4: Integrate Each Term

Integrate \( \int \sin 9y \, dy \) and \( \int \sin y \, dy \): - The antiderivative of \( \sin ky \) is \( -\frac{1}{k} \cos ky \). - Thus, \( \int \sin 9y \, dy = -\frac{1}{9} \cos 9y \). - Likewise, \( \int \sin y \, dy = -\cos y \). Substitute these antiderivatives back into the expression from Step 3: \[ \frac{1}{2} \left( -\frac{1}{9} \cos 9y + \cos y \right). \]

Step 5: Simplify the Expression

Simplify the expression obtained in Step 4: \[ -\frac{1}{18} \cos 9y + \frac{1}{2} \cos y. \] This is the integrated result. Finally, add the constant of integration \( C \): \[ -\frac{1}{18} \cos 9y + \frac{1}{2} \cos y + C. \]

Key Concepts

Trigonometric Identities

Trigonometric identities are powerful tools in calculus that help simplify complex expressions involving trigonometric functions. For example, when dealing with products of sine and cosine functions, you can use the product-to-sum identities to transform them into sums or differences of simpler trigonometric functions.
This simplification is helpful because sums of sine and cosine are straightforward to integrate or differentiate.
Let's break down the product-to-sum identity used in this exercise:

• Identity: \( \sin A \cos B = \frac{1}{2}(\sin(A + B) + \sin(A - B)) \)

Using this identity, you reduce the complexity from a product to a series of addition or subtraction, which is easier to handle.
This identity not only simplifies the integration process but also deepens your understanding of how different trigonometric forms relate within a problem.

Calculus

Calculus is the mathematical study of continuous change. It comprises two main branches: differentiation and integration.
Integration is used to calculate areas under curves and solve differential equations, among other applications.
In this exercise, integration is applied using trigonometric identities to aid in simplifying the problem before integrating.

• Integration of Trigonometric Functions: When you integrate functions like \( \sin x \) and \( \cos x \), you need to remember their antiderivatives.
• Antiderivatives: The antiderivative of \( \sin x \) is \( -\cos x \), and of \( \cos x \) is \( \sin x \).

By knowing these basic antiderivatives and using trigonometric identities, you can manage otherwise complex calculus problems with greater ease. This exercise shows how combining calculus with trigonometric identities can simplify the integration process efficiently.

Definite and Indeterminate Integrals

Understanding the distinction between definite and indefinite integrals is crucial in calculus. Definite integrals evaluate the area between a curve and the x-axis within a specific interval, returning a numerical value. Indefinite integrals, like the one in this exercise, find the general form of antiderivatives, resulting in a function plus a constant of integration.

• Indeterminate Integral: Expressed as \( \int f(x) \, dx \), represents a family of functions without specific bounds or limits.
• Constant of Integration: In indefinite integrals, the constant \( C \) accounts for all possible vertical shifts of the antiderivative function.

In this integrative process, you focus on rewriting complex trigonometric forms into simpler expressions using identities, then systematically integrating each part. This insight into both forms of integrals equips students with versatile approaches to solving a wide range of calculus problems.