Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

use the method of substitution to find each of the following indefinite integrals. $$ \int \cos (3 x+2) d x $$

Step-by-Step Solution

Answer

The indefinite integral is \( \frac{1}{3} \sin(3x + 2) + C \).

Step 1: Choose a Substitution Variable

First, we need to make a substitution to simplify the integration process. Let's set the inside of the cosine function as a new variable. We choose \( u = 3x + 2 \).

Step 2: Differentiate the Substitution Variable

Differentiate \( u \) to find \( du \). Given \( u = 3x + 2 \), the derivative \( \frac{du}{dx} = 3 \). Thus, \( du = 3 \, dx \).

Step 3: Solve for dx

Using the derivative relationship, solve for \( dx \): \[ dx = \frac{du}{3} \]

Step 4: Substitute into the Integral

Replace \( 3x + 2 \) with \( u \) and \( dx \) with \( \frac{du}{3} \) in the integral: \[ \int \cos(3x + 2) \, dx = \int \cos(u) \cdot \frac{du}{3} = \frac{1}{3} \int \cos(u) \, du \]

Step 5: Integrate with Respect to u

Now, integrate \( \int \cos(u) \, du \): \( \int \cos(u) \, du = \sin(u) + C \), where \( C \) is the constant of integration. So, \[ \frac{1}{3} \int \cos(u) \, du = \frac{1}{3} \sin(u) + C \]

Step 6: Substitute Back to x

Replace \( u \) back with \( 3x + 2 \): \[ \frac{1}{3} \sin(u) + C = \frac{1}{3} \sin(3x + 2) + C \]

Key Concepts

Method of Substitution

The method of substitution is a powerful technique in calculus that helps us evaluate integrals by simplifying the integrand. It’s like undoing the chain rule of differentiation. The idea is to replace a part of the integral with a single variable, usually noted as \( u \), making the problem easier to solve.

Here's how it works:

• Identify a part of the integrand that can be substituted with \( u \).
• Differentiate \( u \) to express \( dx \) in terms of \( du \).
• Rewrite the integral in terms of \( u \).
• Integrate with respect to \( u \).
• Substitute back to the original variable to get the final answer.

In the exercise, we set \( u = 3x + 2 \), differentiated to find \( du = 3dx \), and solved for \( dx \). After substituting, our task was simplified to integrating \( \cos(u) \), which is much more manageable. The last step was returning to the variable \( x \) to complete the integration.

Trigonometric Integration

Trigonometric integration involves finding indefinite integrals where the integrand contains trigonometric functions like sine, cosine, tangent, etc. These kinds of integrals can often be tricky because trigonometric identities can be applied in inventive ways.

Understanding certain basic trigonometric integrals is crucial. For example, the integral of \( \cos(x) \) is \( \sin(x) + C \), and working knowledge of these can save significant time.

In our particular problem, the substitution turned the original integrand into a simple trigonometric function \( \cos(u) \). Knowing that \( \int \cos(u) \, du = \sin(u) + C \) directly provided the solution. Recognizing these basic forms is essential for quick resolution of such integrals.

Calculus Integration Techniques

Apart from substitution and dealing with trigonometric forms, calculus offers other integration techniques. Techniques like integration by parts, partial fraction decomposition, and using symmetry each have their place depending on the nature of the integrand.

Some scenarios call for specific methods:

• Use substitution when you can identify a derivative within the integrand.
• Integration by parts is helpful when dealing with products of functions.
• Partial fractions are typically used for rational functions where the degree of the numerator is less than the degree of the denominator.

These techniques often work together. A complex problem might need more than one approach for a complete solution. Using the correct method speeds up finding the integral and deepens understanding of calculus operations.