Problem 78 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that you want to evaluate the integral $$ \int e^{5 x}(4 \cos 7 x+6 \sin 7 x) d x $$ and you know from experience that the result will be of the form \(e^{5 x}\left(C_{1} \cos 7 x+C_{2} \sin 7 x\right)+C_{3}\). Compute \(C_{1}\) and \(C_{2}\) by differentiating the result and setting it equal to the integrand. Many surprising theoretical results can be derived through the use of integration by parts. In all cases, one starts with an integral. We explore two of these results here.

Step-by-Step Solution

Answer

\(C_1 = \frac{148}{185}\), \(C_2 = \frac{17}{37}\).

Step 1: Assume the Form of the Solution

Assume the solution to the integral will be in the form \[ e^{5x} (C_1 \cos 7x + C_2 \sin 7x) + C_3 \]where \( C_1 \), \( C_2 \), and \( C_3 \) are constants to be determined.

Step 2: Differentiate the Assumption

Differentiate the assumed solution \[ e^{5x}(C_1 \cos 7x + C_2 \sin 7x) + C_3 \] with respect to \( x \). This requires using the product rule and differentiating \[ e^{5x} \] and \[ C_1 \cos 7x + C_2 \sin 7x \].

Step 3: Apply the Product Rule

The product rule is \((uv)' = u'v + uv'\). Let \( u = e^{5x} \) and \( v = C_1 \cos 7x + C_2 \sin 7x\).Differentiating, we get:\[ u' = 5e^{5x}, \quad v' = -7C_1 \sin 7x + 7C_2 \cos 7x \].Thus, \[ \frac{d}{dx} [e^{5x} (C_1 \cos 7x + C_2 \sin 7x)] = 5e^{5x}(C_1 \cos 7x + C_2 \sin 7x) + e^{5x} (-7C_1 \sin 7x + 7C_2 \cos 7x). \]

Step 4: Simplify the Differentiated Expression

Combine terms from the differentiated expression:\[ 5e^{5x}C_1 \cos 7x + 5e^{5x}C_2 \sin 7x - 7e^{5x}C_1 \sin 7x + 7e^{5x}C_2 \cos 7x. \]Group like terms:\[ e^{5x} ((5C_1 + 7C_2) \cos 7x + (5C_2 - 7C_1) \sin 7x). \]

Step 5: Equate to the Original Integrand

Set the expression obtained after differentiation equal to the original integrand:\[ e^{5x} ((5C_1 + 7C_2) \cos 7x + (5C_2 - 7C_1) \sin 7x) = e^{5x} (4 \cos 7x + 6 \sin 7x). \]This results in the system of equations:1. \( 5C_1 + 7C_2 = 4 \)2. \( 5C_2 - 7C_1 = 6 \).

Step 6: Solve the System of Equations

Solve the system:1. \( 5C_1 + 7C_2 = 4 \)2. \( 5C_2 - 7C_1 = 6 \)From equation 1, express \( C_1 \) as \( C_1 = \frac{4 - 7C_2}{5} \).Substitute \( C_1 \) into equation 2:\[ 5C_2 - 7\left(\frac{4 - 7C_2}{5}\right) = 6 \]Solve for \( C_2 \):\[ 5C_2 = 6 + 28 - 49C_2/5 \]Multiply through by 5:\[ 25C_2 = 34 - 49C_2 \]\[ 74C_2 = 34 \]\[ C_2 = \frac{34}{74} = \frac{17}{37} \]Substitute \( C_2 \) back into \( C_1 = \frac{4 - 7C_2}{5} \) to solve for \( C_1 \):\[ C_1 = \frac{4 - 7(\frac{17}{37})}{5} = \frac{4 - \frac{119}{37}}{5} = \frac{148}{185} \]

Step 7: Verify the Solution

Double-check the computations and ensure that both values satisfy the original system of equations.1. Check with \( 5C_1 + 7C_2 = 4 \)2. Check with \( 5C_2 - 7C_1 = 6 \)Both equations should be satisfied with \( C_1 = \frac{148}{185} \) and \( C_2 = \frac{17}{37} \).

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used to integrate products of functions. It's closely related to the product rule for differentiation. Imagine you need to integrate a product of two functions, like \(u \cdot v'\). The integration by parts formula helps tackle this:

• \( \int u \cdot dv = u \cdot v - \int v \cdot du \)

To choose \(u\) and \(dv\), follow an acronym like "LIATE": Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential.
When using integration by parts multiple times, it can sometimes result in recursive patterns, or result in simpler integrals.
The key idea is to transform a difficult integral into simpler parts. If you choose \(u\) as \(e^{5x}\) and \(dv\) as \(\cos 7x + \sin 7x\), be prepared for repeated applications. Always look for simpler results in practice.

Differentiation

Differentiation is all about finding the rate at which a quantity changes. It's fundamental in calculus and tells us how a function varies.
For instance, the derivative of a function \(f(x)\) measures the slope of a tangent line to the function at any point \(x\).
In the exercise, differentiation helps us verify the integral form by breaking down the expression \(e^{5x}(C_1 \cos 7x + C_2 \sin 7x)\). Here's why it's crucial:

• The product rule allows us to find the derivative of products of functions, \((uv)' = u'v + uv'\).
• Knowing derivatives facilitates solving equations and finding constants in integrals.

Differentiation helps reverse-engineer the integrand, offering a path to validate the constants \(C_1\) and \(C_2\). It helps confirm that we have matched the form needed for integration.

System of Equations

Systems of equations are key in finding solutions where multiple relationships are at play.
When working with integrals or differentiating solutions, like finding constants \(C_1\) and \(C_2\), you often end up with a system of equations.
A system like:

• \(5C_1 + 7C_2 = 4\)
• \(5C_2 - 7C_1 = 6\)

can be solved using substitution or elimination methods.
Start by isolating one variable, either \(C_1\) or \(C_2\), then substitute into the other equation to find the numeric values:

• If \(5C_1 + 7C_2 = 4\), express \(C_1\) to substitute into the second equation.

After solving the system, verify the results by plugging values back into the original equations. This ensures accuracy and aligns the constants with the initial integral framework.