Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

apply integration by parts twice to evaluate each integral. $$ \int e^{t} \cos t d t $$

Step-by-Step Solution

Answer

\(\int e^{t} \cos t \ dt = \frac{1}{2} (e^{t} \cos t + e^{t} \sin t) + C\)

Step 1: Identify Parts for Integration by Parts

Integration by parts follows the formula \( \int u \ dv = uv - \int v \ du \). For \( \int e^{t} \cos t \ dt \), we can choose \( u = \cos t \) and \( dv = e^t \ dt \). Then, compute \( du = -\sin t \ dt \) and \( v = e^t \).

Step 2: Apply Integration by Parts Formula

Apply the integration by parts formula: \( \int e^{t} \cos t \ dt = e^{t} \cos t - \int e^{t} (-\sin t) \ dt \). Simplify it to: \( e^t \cos t + \int e^{t} \sin t \ dt \).

Step 3: Apply Integration by Parts for the Second Integral

For \( \int e^{t} \sin t \ dt \), use integration by parts again. Choose \( u = \sin t \) and \( dv = e^t \ dt \). Then \( du = \cos t \ dt \) and \( v = e^t \).

Step 4: Apply Integration by Parts Formula Again

Using integration by parts on the second integral: \( \int e^{t} \sin t \ dt = e^{t} \sin t - \int e^{t} \cos t \ dt \), which simplifies back to \( e^{t} \sin t - \int e^{t} \cos t \ dt \).

Step 5: Solve the System of Equations

Notice that the integral \( \int e^{t} \cos t \ dt \) appears on both sides of the equation after substituting back: \( \int e^{t} \cos t \ dt = e^t \cos t + e^t \sin t - \int e^{t} \cos t \ dt \). Add \( \int e^{t} \cos t \ dt \) to both sides and simplify to get \( 2\int e^{t} \cos t \ dt = e^{t} \cos t + e^{t} \sin t \).

Step 6: Solve for the Integral

Divide both sides by 2 to find the integral: \( \int e^{t} \cos t \ dt = \frac{1}{2} (e^{t} \cos t + e^{t} \sin t) + C \), where \( C \) is the constant of integration.

Key Concepts

Integration Techniques in Calculus

Integration is a fundamental concept in calculus, providing tools to calculate areas under curves and solve complex mathematical problems. A key technique within integration is "integration by parts," which is especially useful when dealing with products of functions. This method requires us to identify two parts of the integrand, usually labeled as \( u \) and \( dv \), then apply the integration by parts formula: \[ \int u \ dv = uv - \int v \ du \].

This technique effectively transforms and simplifies the integral into a more manageable form, often making it easier to solve. It's particularly beneficial for integrals involving polynomial and exponential functions, as well as trigonometric expressions.

Despite its efficiency, integration by parts can sometimes require multiple applications to fully solve an integral, as seen in the exercise involving \( \int e^{t} \cos t \ dt \). By carefully choosing the right parts and methodically applying the formula, integration by parts proves to be a versatile tool in the calculus toolkit.

Exponential and Trigonometric Integrals

Exponential and trigonometric functions frequently appear in calculus problems, often requiring specific strategies for integration. These functions, like \( e^t \) and \( \cos t \), have unique properties that can complicate the integration process. The combination of both in a single integral necessitates advanced techniques, such as integration by parts.

When we integrate \( e^t \) and \( \cos t \) together, as in the problem \( \int e^{t} \cos t \ dt \), one efficient approach is to apply integration by parts. This involves expressing the integral as a product of these functions, and iteratively simplifying using the integration by parts formula.

The nature of \( e^t \) as a function with the property of being its own derivative, and trigonometric functions like \( \sin t \) and \( \cos t \), whose derivatives oscillate between each other, indicates they can sometimes cyclically substitute each other in repeated integration processes. This interaction between exponential and trigonometric functions is effectively handled by integration by parts, helping to reduce complex expressions.

Calculus Problem Solving Steps

Problem solving in calculus often follows a structured approach, allowing for consistent and logical solutions. For integration by parts, this involves several systematic steps:

• Identifying Parts: First, choose the parts of the integrand that will be \( u \) and \( dv \). The choice impacts the complexity of the resulting integrals. Often, choosing \( u \) to be a part that simplifies when differentiated is key.
• Applying the Integration by Parts Formula: Use the formula \( \int u \ dv = uv - \int v \ du \) to transform the integral.
• Addressing Remaining Integrals: If a new integral remains complex, apply the technique again. It might involve iteration or other methods to resolve it completely.
• Simplifying through Algebra: Sometimes, simplification involves solving for the integral if it appears on both sides of the equation, as seen in the given exercise.
• Including the Constant of Integration: Don't forget to add \( C \), the constant of integration, as this represents the family of solutions within indefinite integrals.

By following these steps, calculus problems become more approachable, ensuring a thorough understanding and application of integration techniques.