Problem 22
Question
Evaluate the integrals. \(\int e^{-y} \cos y d y\)
Step-by-Step Solution
Verified Answer
The integral is \( \frac{-e^{-y} (\cos y - \sin y)}{2} + C \).
1Step 1: Identify the Integration Method
The integral \( \int e^{-y} \cos y \, dy \) involves the product of an exponential function and a trigonometric function. This suggests using Integration by Parts, which is suitable for integrals of this form.
2Step 2: Apply Integration by Parts Formula
Recall the integration by parts formula: \( \int u \, dv = uv - \int v \, du \). We need to choose \( u \) and \( dv \) from the integral \( \int e^{-y} \cos y \, dy \). Let \( u = \cos y \) and \( dv = e^{-y} \, dy \). Then, \( du = -\sin y \, dy \) and we need to compute \( v \).
3Step 3: Compute the Derivatives and Integrals
Integrate \( dv = e^{-y} \, dy \) to find \( v \): \[ v = \int e^{-y} \, dy = -e^{-y} \]Using these components, we have: - \( u = \cos y \) - \( dv = e^{-y} \, dy \) - \( du = -\sin y \, dy \) - \( v = -e^{-y} \).
4Step 4: Substitute into Integration by Parts Formula
Substitute \( u, v, du, \) and \( dv \) into the integration by parts formula:\[ \int e^{-y} \cos y \, dy = uv - \int v \, du \]\[ = -e^{-y} \cos y - \int (-e^{-y})(-\sin y) \, dy \]\[ = -e^{-y} \cos y - \int e^{-y} \sin y \, dy \].
5Step 5: Use Integration by Parts Again
The integral \( \int e^{-y} \sin y \, dy \) also requires integration by parts. Let \( u = \sin y \) and \( dv = e^{-y} \, dy \). Then, \( du = \cos y \, dy \) and \( v = -e^{-y} \) (from previous computation). Substitute:\[ \int e^{-y} \sin y \, dy = -e^{-y} \sin y - \int -e^{-y} \cos y \, dy \]\[ = -e^{-y} \sin y + \int e^{-y} \cos y \, dy \].
6Step 6: Solve for the Original Integral
Notice that the original integral appears again in \( \int e^{-y} \sin y \, dy = -e^{-y} \sin y + \int e^{-y} \cos y \, dy \). Let \( I = \int e^{-y} \cos y \, dy \), then:\[ I = -e^{-y} \cos y + e^{-y} \sin y - I \]\[ 2I = -e^{-y} (\cos y - \sin y) \]\[ I = \frac{-e^{-y} (\cos y - \sin y)}{2} + C \], where \( C \) is the constant of integration.
Key Concepts
Exponential FunctionsTrigonometric FunctionsIndefinite Integrals
Exponential Functions
Exponential functions are a crucial component of calculus and broader mathematics. They have the general form of \( f(x) = e^x \), where \( e \) is the base of the natural logarithm, approximately equal to 2.71828. This function is unique due to its key property: the rate of change (derivative) of an exponential function is proportional to the value of the function itself.
\[ \frac{d}{dx}e^x = e^x \]
This property simplifies many calculus operations, particularly differentiation and integration. When faced with an integral involving exponential functions, recognizing this property can often guide the steps to solution.
In our problem, the function \( e^{-y} \) is an exponential function with a negative exponent. This converts it into \( \frac{1}{e^y} \), which decreases as \( y \) increases. This behavior is essential when integrating combined functions like in this exercise.
To integrate \( e^{-y} \), recall the antiderivative formula:
\[ \int e^{-y} \, dy = -e^{-y} + C \]
Understanding exponential functions is essential for tackling advanced integration techniques like Integration by Parts when they appear alongside other types of functions.
\[ \frac{d}{dx}e^x = e^x \]
This property simplifies many calculus operations, particularly differentiation and integration. When faced with an integral involving exponential functions, recognizing this property can often guide the steps to solution.
In our problem, the function \( e^{-y} \) is an exponential function with a negative exponent. This converts it into \( \frac{1}{e^y} \), which decreases as \( y \) increases. This behavior is essential when integrating combined functions like in this exercise.
To integrate \( e^{-y} \), recall the antiderivative formula:
\[ \int e^{-y} \, dy = -e^{-y} + C \]
Understanding exponential functions is essential for tackling advanced integration techniques like Integration by Parts when they appear alongside other types of functions.
Trigonometric Functions
Trigonometric functions are another key area in calculus and represent periodic behaviors. In this exercise, they appear as \( \cos y \) and \( \sin y \), which are fundamental trigonometric functions.
These functions relate to angles and sides in right-angled triangles and extend to the unit circle for all angle measurements. Their periodic nature means that they repeat values in a predictable pattern, critical in solving integrals.
When integrating expressions involving trigonometric and other functions, especially using methods like Integration by Parts, it's important to remember the basic derivatives and integrals. Some of these include:
These functions relate to angles and sides in right-angled triangles and extend to the unit circle for all angle measurements. Their periodic nature means that they repeat values in a predictable pattern, critical in solving integrals.
When integrating expressions involving trigonometric and other functions, especially using methods like Integration by Parts, it's important to remember the basic derivatives and integrals. Some of these include:
- The derivative of \( \cos y \) is \( -\sin y \)
- The antiderivative of \( \sin y \) is \( -\cos y + C \)
Indefinite Integrals
Indefinite integrals, or antiderivatives, extend the idea of integration into finding functions whose derivative is the given function.
They are represented with the integral symbol \( \int \) followed by the function and include a constant of integration, \( C \), since derivatives of constants are zero, leaving room for multiple solutions.
For any function \( f(x) \), an indefinite integral is:
\[ \int f(x) \, dx = F(x) + C \]
where \( F(x) \) is the antiderivative of \( f(x) \).
In the integration by parts method, indefinite integrals help break down the product of functions — such as the exponential \( e^{-y} \) and trigonometric \( \cos y \) in our problem. Finding separate antiderivatives for involved functions and then using the integration by parts formula allows us to solve otherwise challenging integrals.
They are represented with the integral symbol \( \int \) followed by the function and include a constant of integration, \( C \), since derivatives of constants are zero, leaving room for multiple solutions.
For any function \( f(x) \), an indefinite integral is:
\[ \int f(x) \, dx = F(x) + C \]
where \( F(x) \) is the antiderivative of \( f(x) \).
In the integration by parts method, indefinite integrals help break down the product of functions — such as the exponential \( e^{-y} \) and trigonometric \( \cos y \) in our problem. Finding separate antiderivatives for involved functions and then using the integration by parts formula allows us to solve otherwise challenging integrals.
- The formula used was:
\[ \int u \, dv = uv - \int v \, du \] - In the context of our problem, choosing \( u = \cos y \) and \( dv = e^{-y} \, dy \) simplifies evaluating \( \int e^{-y} \cos y \, dy \).
Other exercises in this chapter
Problem 22
Evaluate the integrals in Exercises \(1-34\) without using tables. $$ \int_{0}^{\infty} 2 e^{-\theta} \sin \theta d \theta $$
View solution Problem 22
Evaluate the integrals in Exercises \(15-22\). $$ \int_{-\pi}^{\pi}\left(1-\cos ^{2} t\right)^{3 / 2} d t $$
View solution Problem 22
In Exercises \(21-28,\) express the integrands as a sum of partial fractions and evaluate the integrals. $$ \int_{1}^{\sqrt{3}} \frac{3 t^{2}+t+4}{t^{3}+t} d t
View solution Problem 22
Evaluate each integral in Exercises \(1-36\) by using a substitution to reduce it to standard form. $$ \int \frac{2^{\ln x}}{x} d x $$
View solution