Problem 52 - Calculus Single Variable

Question

Evaluate each of the integrals. $$ \int 9 x \sin (3 x+4) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(- (3x + 4) \cos(3x + 4) + \sin(3x + 4) + 4 \cos(3x + 4) + C\)."

Step 1: Choose a Substitution

Let's denote the substitution variable as \( u = 3x + 4 \). This simplifies the integrand by changing the composite function \( 3x + 4 \) into something simpler.

Step 2: Differentiate the Substitution

Differentiate the substitution: \( u = 3x + 4 \). This gives us \( \frac{du}{dx} = 3 \), or equivalently \( du = 3 \, dx \).

Step 3: Express dx in terms of du

Rearrange \( du = 3 \, dx \) to solve for \( dx \): \( dx = \frac{du}{3} \).

Step 4: Substitute in the Integral

Substitute into the integral: \( \int 9x \sin(3x + 4) \, dx \) becomes \( \int \frac{9x \sin u}{3} \, du \).

Step 5: Express x in terms of u

From the substitution \( u = 3x + 4 \), express \( x \) as \( x = \frac{u - 4}{3} \) and substitute it back in: \( \int 3\left(\frac{u-4}{3}\right) \sin u \, du \).

Step 6: Simplify the Integral

Simplify \( \int (u-4) \sin u \, du \), which separates into \( \int u \sin u \, du - 4 \int \sin u \, du \).

Step 7: Integrate \( \int u \sin u \, du \) using Integration by Parts

Let \( v = u \), \( dv = du \), \( dw = \sin u \, du \), so \( w = -\cos u \). The first integral becomes \( -u \cos u + \int \cos u \, du \).

Step 8: Solve Each Integral

The integral \( \int \cos u \, du = \sin u \). Therefore, \( \int u \sin u \, du = -u \cos u + \sin u \). The second integral \( \int \sin u \, du = -\cos u \).

Step 9: Combine Results

Combine the results: \( \int u \sin u \, du - 4 \int \sin u \, du = [-u \cos u + \sin u] + 4\cos u \).

Step 10: Substitute Back

Replace \( u \) back with \( 3x + 4 \): \([- (3x + 4) \cos(3x + 4) + \sin(3x + 4) + 4 \cos(3x + 4)] + C \) where \( C \) is the constant of integration.

Key Concepts

Integration by Parts

One of the more versatile techniques in calculus, integration by parts, helps solve integrals involving product functions like polynomial and trigonometric functions. It’s based on the product rule for differentiation and formulated as:

• \[ \int u \, dv = uv - \int v \, du \]

The difficult part is often choosing the right parts for \( u \) and \( dv \). A general tip is to choose \( u \) so it becomes simpler when differentiated.
For example, in our integral problem, \( u \) corresponds to the polynomial \( x \) and \( dv \) is derived from the \( \sin(3x + 4) \) term. By differentiating \( u \) and integrating \( dv \), you progress through the process of simplifying the integral until it’s manageable. It is a powerful method for dealing with composite functions where substitution alone doesn’t immediately simplify them.

Trigonometric Integration

Trigonometric functions, such as sine and cosine, present unique challenges in integration, especially when paired with algebraic expressions. In the exercise, you encounter an integral with \( \sin(3x + 4) \), which must be carefully managed with trigonometric techniques.

• Recognizing patterns within repeating integrals.
• Utilizing identities like \( \sin^2 x + \cos^2 x = 1 \) if needed.

This problem specifically uses trigonometric substitution within a broader substitution method, where \( 3x + 4 = u \). Understanding how to manipulate these functions is fundamental to solving more complex expressions, leveraging identities and alterations to reach a calculable endpoint. Often, breaking down composite angles and converting triggers like \( \sin(3x+4) \) into more familiar terms simplifiers the expression into manageable parts.

Definite and Indefinite Integrals

An understanding of integrals is key in both calculus and real-world applications. There are two types of integrals: definite, which provide a numerical value, and indefinite, which include a constant of integration \( C \). The problem given requires finding an indefinite integral, indicated by its resulting expression containing \( + C \), which accounts for any constant that was lost during differentiation.

• Definite integrals are calculated between two bounds \( a \) and \( b \), offering the area under the curve in this interval.
• Indefinite integrals encompass a family of functions, represented typically as \( F(x) + C \).

In the solution, once the integral is solved using substitution and then integration by parts, we recall to include \( C \) to account for this general solution aspect, ensuring it represents not just one curve but a family of possible curves that satisfy the integral equation.