Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

use the method of substitution to find each of the following indefinite integrals. $$ \int \sin (2 x-4) d x $$

Step-by-Step Solution

Answer

The indefinite integral is \( -\frac{1}{2} \cos(2x - 4) + C. \)

Step 1: Identify the Inner Function

The integrand is \( \sin(2x - 4) \). Here, the inner function is \( u = 2x - 4 \). This substitution will help us simplify the integration process.

Step 2: Differentiate the Inner Function

Differentiate the chosen inner function \( u = 2x - 4 \) with respect to \( x \). This yields \( \frac{du}{dx} = 2 \) or equivalently \( du = 2dx \).

Step 3: Solve for dx

From \( du = 2dx \), solve for \( dx \). This gives \( dx = \frac{1}{2} du \).

Step 4: Substitute u and dx into the Integral

Substitute \( u = 2x - 4 \) and \( dx = \frac{1}{2} du \) into the integral: \[ \int \sin(2x - 4) \; dx = \int \sin u \cdot \frac{1}{2} du = \frac{1}{2} \int \sin u \; du. \]

Step 5: Integrate with Respect to u

Integrate \( \frac{1}{2} \int \sin u \; du \) using standard integration formulas. The integral of \( \sin u \) is \( -\cos u \). Therefore, \[ \frac{1}{2} \int \sin u \; du = \frac{1}{2} \left(-\cos u\right) = -\frac{1}{2} \cos u + C. \]

Step 6: Substitute Back to x

Replace \( u \) with the original expression in terms of \( x \): \( u = 2x - 4 \). So, the solution in terms of \( x \) is: \[ -\frac{1}{2} \cos(2x - 4) + C. \]

Key Concepts

Substitution Method

The substitution method is a powerful technique in calculus used to simplify the process of finding indefinite integrals. It involves identifying a portion of the integrand that can be represented as a separate function, often referred to as the inner function. This helps in transforming the integral into a simpler form. For example, in the integral \( \int \sin (2x - 4) \, dx \), we recognize the inner function as \( u = 2x - 4 \). Once identified, this inner function is differentiated with respect to \( x \) to determine \( du \), leading to \( \frac{du}{dx} = 2 \). This differentiation facilitates substitution, making the integration more manageable. Key aspects of the substitution method include:

• Identifying the inner function (often, the argument of a trigonometric or exponential function).
• Differentiating this function to relate \( du \) and \( dx \).
• Rewriting the integral in terms of \( u \), simplify, and integrate with respect to \( u \).

In summary, the substitution method redefines complicated expressions into simpler ones, easing the integration process.

Integral Calculus

Integral calculus is one of the two main branches of calculus, dealing primarily with the concept of integration. It enables us to determine areas, volumes, central points, and many useful things. The fundamental aspect of integral calculus involves finding the function whose derivative is the given function, known as the antiderivative or the indefinite integral. Understanding the following scenarios is crucial:

• Indefinite integrals represent a family of functions, typically including a constant of integration \( C \).
• Definite integrals compute a numerical value based on the limits of integration and often represent areas under a curve.

In the given exercise, we are asked to find an indefinite integral, \( \int \sin(2x - 4) \, dx \), which uses substitution to simplify and solve the problem. The objective is to determine an antiderivative, in this case, \( -\frac{1}{2} \cos(2x - 4) + C \), thereby reflecting the essence of integral calculus.

Trigonometric Integrals

Trigonometric integrals involve functions that contain trigonometric expressions, such as sine, cosine, tangent, and others. They frequently arise in calculus when solving problems related to periodic functions or waveforms. These integrals often require specific strategies, like the substitution method, for simplification. In this exercise, the function \( \sin(2x - 4) \) is a classic example of a trigonometric integral. For successful integration:

• Know the basic antiderivatives of trigonometric functions, like \( \int \sin(u) \, du = -\cos(u) + C \).
• Apply suitable substitutions to transform complex trigonometric terms into simpler ones.
• Utilize trigonometric identities if needed, to further simplify the expressions.

By understanding the properties and rules of trigonometric integrals, one can approach solving them with confidence. The integration of \( \int \sin u \, du \) turned into \( -\cos u + C \) demonstrates the facility offered by utilizing these methods.