Problem 60 - Applied Calculus

Question

Find the indefinite integrals. $$ \int 6 \cos (3 x) d x $$

Step-by-Step Solution

Answer

The indefinite integral is \( 2 \sin(3x) + C \).

Step 1: Identify the Integral Type

The problem involves the integral \( \int 6 \cos(3x) \, dx \). This is a basic integral of the form \( \int a \cos(bx) \, dx \), where \(a = 6\) and \(b = 3\). Calculating this involves using the formula for the integral of \( \cos(bx) \):\[ \int \cos(bx) \, dx = \frac{1}{b} \sin(bx) + C \] where \(C\) is the constant of integration.

Step 2: Apply the Integral Formula

Replace the values of \(a\) and \(b\) in the integral formula \( \int \cos(bx) \, dx = \frac{1}{b} \sin(bx) + C \). We have:\[ \int 6 \cos(3x) \, dx = 6 \int \cos(3x) \, dx \]Now apply the formula:\[ \int \cos(3x) \, dx = \frac{1}{3} \sin(3x) + C \]Multiply by 6:\[ 6 \times \frac{1}{3} \sin(3x) + 6C \]

Step 3: Simplify the Expression

Simplify the expression obtained:\[ 6 \times \frac{1}{3} \sin(3x) + 6C \] becomes \[ 2 \sin(3x) + C \].So, the indefinite integral of \( 6 \cos(3x) \, dx \) is \( 2 \sin(3x) + C \), where \(C\) is a constant.

Key Concepts

Integration Techniques

Integration techniques are key in finding indefinites integrals. A variety of methods exist, each suiting different types of functions. When dealing with basic functions, formulas can be applied directly.
For more complex integrals, techniques like substitution, integration by parts, or partial fractions might be necessary.

• Substitution: Often used when an integral contains an expression whose derivative is also present.
• Integration by Parts: Useful for products of different functions such as polynomials and exponentials.
• Partial Fractions: Breaks down complex rational expressions into simpler parts that can be integrated individually.

For the given problem, the direct formula for integrating \( \cos(bx) \) is effective because it eliminates the need for more complex methods. Recognizing the correct technique saves time and simplifies the problem.

Trigonometric Integrals

Trigonometric integrals are an essential part of calculus that involve integrating trigonometric functions like sine and cosine. Understanding how to work with these integrals is crucial for solving a wide range of problems.When dealing with a trigonometric integral, it often involves:

• Simplifying using trigonometric identities.
• Applying standard integration formulas.
• Sometimes requiring substitution to simplify or resolve the integral into known forms.

In our example, we have \( \int 6 \cos(3x) \, dx \). This fits the formula for integrating \( \cos(bx) \), resulting in \( \frac{1}{b} \sin(bx) + C \). Simplifying the initial integral using this formula leads to \( 2 \sin(3x) + C \). Recognizing these patterns helps in efficiently calculating the results for trigonometric integrals.

Constant of Integration

The constant of integration \( C \) appears in indefinite integrals because when integrating, we reverse differentiation—a process that can eliminate the constant term originally present in the function.Why include a constant of integration?
The differentiation of a constant is zero, hence, when we integrate and find an antiderivative, we need to introduce \( C \) to account for any lost constant.
Every antiderivative is part of a family of functions that differ by a constant.

• Two functions can have the same derivative, but differ by a constant value.
• Including \( C \) ensures all possible original functions are considered.

In our integral \( \int 6 \cos(3x) \, dx \), the resulting function \( 2 \sin(3x) + C \) reflects the general solution to the antiderivative without any specific constraints. Thus, \( C \) is crucial to demonstrating all possible solutions.