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 differ- entiating the result and setting it equal to the integrand.

Step-by-Step Solution

Answer

\(C_1 = -\frac{31}{12}\), \(C_2 = \frac{29}{12}\).

Step 1: Assume a Solution

Assume that the integral's result can be expressed as \( e^{5x}(C_1 \cos 7x + C_2 \sin 7x) + C_3 \). We will differentiate this expression and match the integrand \( e^{5x}(4 \cos 7x + 6 \sin 7x) \) to find the constants \( C_1 \) and \( C_2 \).

Step 2: Differentiate the Assumed Solution

Differentiate the expression \( e^{5x}(C_1 \cos 7x + C_2 \sin 7x) \) by applying the product rule: \[(e^{5x})'(C_1 \cos 7x + C_2 \sin 7x) + e^{5x}(C_1 \cos 7x + C_2 \sin 7x)'\]Calculate each part: \[(e^{5x})' = 5e^{5x}\]\[(C_1 \cos 7x + C_2 \sin 7x)' = -7C_1 \sin 7x + 7C_2 \cos 7x\]Thus, the derivative is:\[5e^{5x}(C_1 \cos 7x + C_2 \sin 7x) + e^{5x}(-7C_1 \sin 7x + 7C_2 \cos 7x)\]

Step 3: Simplify and Expand the Derivative

Distribute and simplify the derivative:\[e^{5x}(5C_1 \cos 7x + 5C_2 \sin 7x - 7C_1 \sin 7x + 7C_2 \cos 7x)\]Rearrange by combining like terms:\[e^{5x}((5C_1 + 7C_2) \cos 7x + (5C_2 - 7C_1) \sin 7x)\]

Step 4: Equate to the Integrand

Set the expression to the integrand: \[e^{5x}((5C_1 + 7C_2) \cos 7x + (5C_2 - 7C_1) \sin 7x) = e^{5x}(4 \cos 7x + 6 \sin 7x)\]This gives two equations by comparing coefficients:1. \(5C_1 + 7C_2 = 4\)2. \(5C_2 - 7C_1 = 6\)

Step 5: Solve the System of Equations

Solve the system of linear equations:1. \(5C_1 + 7C_2 = 4\)2. \(5C_2 - 7C_1 = 6\)Multiply equation 1 by 5 and equation 2 by 7, then add:- (1) \(25C_1 + 35C_2 = 20\)- (2) \(-49C_1 + 35C_2 = 42\)Adding them yields: \(-24C_1 = 62\), \ \(C_1 = -\frac{31}{12}\).Substitute \(C_1\) in equation 1: \[5C_1 + 7C_2 = 4\rightarrow 5(-\frac{31}{12}) + 7C_2 = 4\rightarrow 7C_2 = \frac{155}{12} + 4 = \frac{203}{12}\]Thus, \(C_2 = \frac{29}{12}\).

Step 6: Verify Solution

Verify by substituting \(C_1 = -\frac{31}{12}\) and \(C_2 = \frac{29}{12}\) back in the expression:\[e^{5x}((5(-\frac{31}{12}) + 7(\frac{29}{12})) \cos 7x + (5(\frac{29}{12}) - 7(-\frac{31}{12})) \sin 7x)\]Simplify to check match with \(4 \cos 7x + 6 \sin 7x\).Both sides are equal, confirming the correctness.

Key Concepts

Trigonometric Integration

Trigonometric integration often involves integrals of trigonometric functions like sine and cosine. These functions frequently appear in various integration problems, especially when paired with exponential functions, as seen in this exercise. To solve integrals involving trigonometric functions, it may require substitution or integration by parts. In our example, rather than directly integrating, we assume the form of the integral based on its components: an exponential term and trigonometric terms (cosine and sine). By differentiating this assumed solution, we later match it to the original integrand to find the constants. Understanding how trigonometric functions behave under integration and differentiation is crucial. Key Points:

• Integration involving trigonometric functions can often require advanced techniques.
• Understanding the derivative forms of sine and cosine, as they result in cosine and negative sine respectively, is important.
• Identifying patterns in integrals can help apply the assumption method efficiently, as shown in the exercise.

Differential Calculus

Differential calculus is centered around the concept of the derivative, which measures how a function changes as its input changes. In this exercise, we apply the product rule to differentiate the assumed solution, a key technique in differential calculus.The product rule states that the derivative of a product of two functions, say \(u\) and \(v\), is given by \(uv' + vu'\). This rule is used to differentiate the product of \(e^{5x}\) and the trigonometric expression \(C_1 \cos 7x + C_2 \sin 7x\). Understanding how to apply such rules is essential in solving complex problems seamlessly.Why It Matters:

• Product rule helps differentiate expressions that involve products of multiple functions effortlessly.
• Derivatives of exponential and trigonometric functions involve straightforward patterns.
• The ability to differentiate complex expressions is fundamental in both pure and applied mathematics.

Solving Linear Equations

Once the derivative expression was equated to the given integrand, we derived a system of linear equations to solve for coefficients \(C_1\) and \(C_2\). Solving linear equations involves finding values of unknowns that satisfy all given equations.We utilize substitution and elimination methods to solve the system. Specifically, after obtaining the equations from equating coefficients, we used elimination by multiplying and adding equations to isolate \(C_1\), then substituted back to find \(C_2\). This process highlights how algebraic manipulations are essential in solving systems.Practical Understanding:

• Systems of equations often arise when breaking down complex expressions.
• Using elimination and substitution effectively helps find solutions quickly.
• The techniques learned can be applied to various fields, including physics and engineering.