Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify Functions for Integration by Parts

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). To apply this to \( \int e^{at} \sin t \, dt \), we need to choose \( u \) and \( dv \). Let \( u = \sin t \) and \( dv = e^{at} \, dt \). This way, \( du = \cos t \, dt \) and for \( v \), integrate \( dv \) to get \( v = \frac{1}{a} e^{at} \), assuming \( a eq 0 \).

Step 2: Apply Integration by Parts First Time

Apply the integration by parts formula: \[ \int e^{at} \sin t \, dt = \sin t \cdot \frac{1}{a} e^{at} - \int \frac{1}{a} e^{at} \cdot \cos t \, dt. \] Simplify the equation to: \[ \int e^{at} \sin t \, dt = \frac{1}{a} e^{at} \sin t - \frac{1}{a} \int e^{at} \cos t \, dt. \] We will need to evaluate \( \int e^{at} \cos t \, dt \) using integration by parts again.

Step 3: Choose Functions for Integration by Parts Again

For \( \int e^{at} \cos t \, dt \), let \( u = \cos t \) and \( dv = e^{at} \, dt \). Therefore, \( du = -\sin t \, dt \) and \( v = \frac{1}{a} e^{at} \).

Step 4: Apply Integration by Parts Second Time

Use the formula again on \( \int e^{at} \cos t \, dt \): \[ \int e^{at} \cos t \, dt = \cos t \cdot \frac{1}{a} e^{at} + \int \frac{1}{a} e^{at} \sin t \, dt. \] Simplify the expression: \[ \int e^{at} \cos t \, dt = \frac{1}{a} e^{at} \cos t + \frac{1}{a} \int e^{at} \sin t \, dt. \] Here, notice that we have arrived back to our original integral, \( \int e^{at} \sin t \, dt \).

Step 5: Solve for Original Integral

Substitute the expression for \( \int e^{at} \cos t \, dt \) into the equation from Step 2: \[ \int e^{at} \sin t \, dt = \frac{1}{a} e^{at} \sin t - \frac{1}{a} \left( \frac{1}{a} e^{at} \cos t + \frac{1}{a} \int e^{at} \sin t \, dt \right). \] Simplify this equation: \[ \int e^{at} \sin t \, dt = \frac{1}{a} e^{at} \sin t - \frac{1}{a^2} e^{at} \cos t - \frac{1}{a^2} \int e^{at} \sin t \, dt. \] Move the integral terms together: \[ \int e^{at} \sin t \, dt + \frac{1}{a^2} \int e^{at} \sin t \, dt = \frac{1}{a} e^{at} \sin t - \frac{1}{a^2} e^{at} \cos t. \] Factor out \( \int e^{at} \sin t \, dt \) from the left side: \[ \left( 1 + \frac{1}{a^2} \right) \int e^{at} \sin t \, dt = \frac{1}{a} e^{at} \sin t - \frac{1}{a^2} e^{at} \cos t. \] Divide both sides by \( 1 + \frac{1}{a^2} \): \[ \int e^{at} \sin t \, dt = \frac{1}{1 + \frac{1}{a^2}} \left( \frac{1}{a} e^{at} \sin t - \frac{1}{a^2} e^{at} \cos t \right). \] Simplify the expression: \[ \int e^{at} \sin t \, dt = \frac{a}{a^2 + 1} e^{at} \sin t - \frac{1}{a^2 + 1} e^{at} \cos t + C, \] where \( C \) is the constant of integration.

Key Concepts

Exponential Functions

Exponential functions are functions of the form \( e^{x} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.718. In our exercise, we deal with \( e^{at} \), an exponential function showing how the value changes with respect to the variable \( t \). This function is critical in modeling growth and decay processes.
Key characteristics of exponential functions include:

• They are continuous and smooth, meaning they have no gaps or sharp corners.
• The rate of growth of \( e^{x} \) functions is proportional to their current value.
• They have a constant relative growth rate.

In calculus, exponential functions frequently appear because their derivatives and integrals yield the same exponential form, which simplifies calculations.

Trigonometric Integration

Trigonometric integration involves integrating functions that contain trigonometric functions like \( \sin x \) or \( \cos x \). This exercise involves finding the integral of a product of a trigonometric function \( \sin t \) and an exponential function \( e^{at} \).
Trigonometric functions have distinct characteristics:

• They are periodic, repeating their values in regular intervals.
• \( \sin \) and \( \cos \) functions are continuous and have smooth wave-like graphs.

Understanding these properties is crucial when performing integration as the integration of products often necessitates advanced techniques such as Integration by Parts, especially when combined with non-trigonometric functions, as in our given problem.

Integral Calculus

Integral calculus focuses on finding the area under curves. This is done by calculating the integral of a function, which essentially sums up an infinite number of infinite small quantities. In this exercise, we are tasked with solving an integral using the integration by parts technique.
Key points in integral calculus:

• The integral symbol \( \int \) represents the sum of continuous elements over a defined range.
• Definite integrals compute the area under a curve over a specific interval, whereas indefinite integrals find antiderivatives, not limited to a specific range.
• The inclusion of the constant of integration \( C \) in indefinite integrals accounts for all antiderivatives.

Integral calculus is fundamental in understanding areas, volumes, central points, among other continuous aspects in mathematical functions.

Integration Techniques

Integration by Parts is one of the prominent techniques used to solve integrals that appear as the product of two functions. This method can be quite helpful, especially when dealing with integrals involving both exponential and trigonometric functions as seen in this task.
The general formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). The technique works by breaking down a complicated integral into simpler parts.
Steps in Integration by Parts:

• Select \( u \) and \( dv \), such that differentiating \( u \) and integrating \( dv \) leads to simpler computations.
• Compute \( du \) by differentiating \( u \) and find \( v \) by integrating \( dv \).
• Apply the formula and solve the resulting integrals.

This approach can require repeated application, as seen in our solution where it is applied twice to simplify the process, eventually enabling us to solve for the original integral with persistent terms.