Problem 15 - Calculus Single Variable

Question

Calculate the indefinite integral. $$ \int(3 \cos (4 x)+2 x) d x $$

Step-by-Step Solution

Answer

\( \int(3 \cos (4 x) + 2x) \ dx = \frac{3}{4} \sin(4x) + x^2 + C \)

Step 1: Analyze the Integral

Observe the given integral \( \int (3 \cos (4x) + 2x) \, dx \). It is a sum of two functions, which allows us to integrate each part separately.

Step 2: Integrate the First Term

The first term is \( 3 \cos(4x) \). The integral of \( \cos(kx) \) is given by \( \frac{1}{k} \sin(kx) \), where \( k \) is a constant. Therefore, the integral of \( 3 \cos(4x) \) is \( \frac{3}{4} \sin(4x) \).

Step 3: Integrate the Second Term

The second term is \( 2x \). The integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} \). Here, \( n = 1 \), so the integral of \( 2x \) is \( x^2 \).

Step 4: Combine the Results with the Constant of Integration

Add the results from the integrations of both terms together: \( \int (3 \cos(4x) + 2x) \, dx = \frac{3}{4} \sin(4x) + x^2 + C \), where \( C \) is the constant of integration that appears because the integral is indefinite.

Key Concepts

Integration by Parts

Integration by parts is an essential technique used in calculus for transforming the integral of a product of functions into potentially easier integrals. The formula for integration by parts comes from the product rule for differentiation, and it is expressed as:\[ \int u \, dv = uv - \int v \, du \]Where,\( u \) is a function of \( x \), and \( dv \) is the differential of another function. The choice of \( u \) and \( dv \) is critical to simplify the problem.

Typically, one chooses \( u \) to be a function that becomes simpler when differentiated, while \( dv \) is chosen such that it is easy to integrate. This technique becomes particularly useful when dealing with integrals involving polynomial and logarithmic functions, or products involving sine and cosine. In the context of this exercise, it is more straightforward to integrate each separate component of the sum than to apply integration by parts, thanks to simpler basic integration rules involved.

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental components in calculus, often appearing in various forms of integrals. When dealing with integrals involving cosine, as seen in this exercise, it helps to recall some standard integral results.

For example, the integral of \( \cos(kx) \) is \( \frac{1}{k} \sin(kx) \). This comes from reversing the differentiation process of the sine function. In the given exercise, the term \( 3 \cos(4x) \) is integrated using this rule, yielding \( \frac{3}{4} \sin(4x) \).

Understanding these basic trigonometric integral formulas is crucial, as they not only simplify calculations but also enhance comprehension of more complex calculus problems. Remember, when integrating trigonometric functions, the coefficient of \( x \) in cosine or sine directly affects the integral—a common oversight for many students.

Integration Rules

Integration rules are the core tools for handling a variety of integral problems, enabling students to tackle complex expressions by breaking them into manageable parts. For the problem at hand, rules such as integrating sums term by term and using known integrals are applied.

- **Sum Rule**: The integral of a sum is the sum of the integrals. Thus, \( \int (3 \cos(4x) + 2x) \, dx \) can be split into two separate integrals.

- **Power Rule**: For polynomials like \( x^n \), the integral is \( \frac{x^{n+1}}{n+1} \), provided \( n eq -1 \). For this exercise, \( 2x \) fits this rule since \( n = 1 \), leading to \( x^2 \) after integration.

Integrating each term independently often simplifies the process and helps avoid common missteps. Always remember the constant of integration, \( C \), when dealing with indefinite integrals, as it is essential for capturing all potential solutions.