Indefinite Integral - Choice Of First And Second Function

Indefinite Integral - Choice of first and second function

In this topic, we will learn about integrating functions where the integrand is a product of two functions.
This technique is known as “Integration by parts”.
The main idea behind integration by parts is to differentiate one function and integrate another, in order to simplify the integral.
We will also focus on understanding the concept of the choice of the first and second function.
Let’s begin exploring this technique further in the upcoming slides.

Integration by Parts Formula

The integration by parts formula is given by: ∫(u * dv) = u * v - ∫(v * du) where u and v are functions of x and du and dv are their respective derivatives.
This formula allows us to rewrite the integral of a product of two functions in a different form, making it easier to evaluate.

Choosing the First Function

Choosing the first function is a crucial step in using integration by parts.
It is generally recommended to choose the function that will simplify after differentiating.
The choice of the first function is often guided by the acronym “LIATE”.
L: Logarithmic functions
I: Inverse trigonometric functions
A: Algebraic functions (polynomials)
T: Trigonometric functions
E: Exponential functions
The functions mentioned in the acronym are ranked in order of how easy they become after differentiation.

Choosing the Second Function

Once the first function is chosen, the second function is determined by integration.
We integrate the second function until it becomes simpler or a known function.
Often, the second function is chosen in such a way that it becomes simpler after integrating.

Example 1

Let’s consider the integral: ∫ (x * ln(x)) dx

We will choose u = ln(x) (logarithmic function) as our first function.
Then, dv = x dx, which implies v = (1/2) * x^2.
Now, we can apply the integration by parts formula: ∫ (x * ln(x)) dx = u * v - ∫ (v * du) = ln(x) * (1/2)x^2 - ∫ (1/2x^2 * 1) dx = (1/2) * x^2 * ln(x) - (1/4) * x^2 + C where C is the constant of integration.

Example 2

Let’s consider the integral: ∫ (x^2 * cos(x)) dx

We will choose u = x^2 (algebraic function) as our first function.
Then, dv = cos(x) dx, which implies v = sin(x).
Now, we can apply the integration by parts formula: ∫ (x^2 * cos(x)) dx = u * v - ∫ (v * du) = x^2 * sin(x) - ∫ (2x * sin(x)) dx
To evaluate the remaining integral, we can apply integration by parts again with u = x and dv = sin(x) dx.

Example 2 (continued)

After applying integration by parts a second time, we have: ∫ (x^2 * cos(x)) dx = x^2 * sin(x) - [x * (-cos(x)) - ∫ (-cos(x) dx)] = x^2 * sin(x) + x * cos(x) - ∫ cos(x) dx

The remaining integral is straightforward to evaluate: ∫ cos(x) dx = sin(x) + C where C is the constant of integration.

Example 2 (continued)

Finally, we substitute the value of ∫ cos(x) dx back into the previous expression: ∫ (x^2 * cos(x)) dx = x^2 * sin(x) + x * cos(x) - ∫ cos(x) dx = x^2 * sin(x) + x * cos(x) - sin(x) + C where C is the constant of integration.

Summary

To summarize the process of integrating by parts:

Choose the first function based on the LIATE acronym.
Differentiate the first function to obtain du.
Integrate the second function to obtain v.
Apply the integration by parts formula: ∫(u * dv) = u * v - ∫(v * du).
Simplify the resulting integral and evaluate if possible.

Key Points to Remember

Integration by parts is a technique used to integrate products of functions.
It involves choosing a first function and a second function and applying the formula: ∫(u * dv) = u * v - ∫(v * du).
The choice of the first function is guided by the LIATE acronym.
The second function is determined by integration, with the goal of simplifying the integral.
Examples help us understand the application of integration by parts in solving different types of integrals.

Example 3

Let’s consider the integral: ∫ (e^x * sin(x)) dx

We will choose u = e^x (exponential function) as our first function.
Then, dv = sin(x) dx, which implies v = -cos(x).
Now, we can apply the integration by parts formula: ∫ (e^x * sin(x)) dx = -e^x * cos(x) - ∫ (-cos(x) * e^x) dx
The remaining integral can be evaluated using integration by parts again.

Example 3 (continued)

After applying integration by parts a second time, we have: ∫ (e^x * sin(x)) dx = -e^x * cos(x) - ∫ (-cos(x) * e^x) dx = -e^x * cos(x) + ∫ (cos(x) * e^x) dx

Notice that the integral on the right side is the same as the original integral.
We can rewrite the equation as:

2∫ (e^x * sin(x)) dx = -e^x * cos(x)

Finally, solving for the original integral, we get: ∫ (e^x * sin(x)) dx = -0.5 * e^x * cos(x) + C

Choosing the First Function - Logarithmic Functions

When the integrand contains a logarithmic function, it is often a good choice for the first function.
Logarithmic functions have the property that their derivative involves the reciprocal of the argument.
For example:
- The derivative of ln(x) is 1/x.
- The derivative of ln(x^2) is (2/x) = 2ln(x).
Logarithmic functions can simplify after differentiation, making the integral easier to evaluate.

Choosing the First Function - Inverse Trigonometric Functions

Inverse trigonometric functions are another good choice for the first function when they are present in the integrand.
The derivatives of inverse trigonometric functions often involve algebraic expressions.
For example:
- The derivative of arcsin(x) is 1/√(1 - x^2).
- The derivative of arctan(x) is 1/(1 + x^2).
Inverse trigonometric functions can simplify after differentiation, making the integral easier to evaluate.

Choosing the First Function - Algebraic Functions

Algebraic functions, such as polynomials, are a common choice for the first function.
The derivatives of algebraic functions are usually straightforward to calculate.
For example:
- The derivative of x^2 is 2x.
- The derivative of 3x^3 - 2x^2 + x is 9x^2 - 4x + 1.
Choosing an algebraic function as the first function can simplify the integral, especially when combined with other functions in the integrand.

Choosing the First Function - Trigonometric Functions

Trigonometric functions, such as sin(x) and cos(x), are commonly used as the first function in integration by parts.
The derivatives of trigonometric functions involve other trigonometric functions.
For example:
- The derivative of sin(x) is cos(x).
- The derivative of cos(x) is -sin(x).
Trigonometric functions can simplify or eliminate themselves after differentiation, making the integral easier to evaluate.

Choosing the First Function - Exponential Functions

Exponential functions, such as e^x, are also suitable choices for the first function.
The derivative of an exponential function is simply the function itself.
For example:
- The derivative of e^x is e^x.
- The derivative of 2e^x is 2e^x.
Exponential functions often remain unchanged or get simplified after differentiation, making the integral easier to evaluate.

Recap: Choosing the First Function

The choice of the first function plays a crucial role in integration by parts.
It is essential to consider the properties of different types of functions to make an appropriate choice.
Logarithmic functions, inverse trigonometric functions, algebraic functions, trigonometric functions, and exponential functions are commonly used as the first function.
The first function should simplify or eliminate itself after differentiation, making the integral easier to evaluate.

Recap: Choosing the Second Function

Once the first function is chosen, the second function is determined by integration.
The goal is to choose the second function such that integrating it makes the integral simpler or a known function.
The choice of the second function depends on the specific problem and requires practice and experience.
Often, integrating the second function repeatedly simplifies the integral and leads to the desired solution.

Summary

Integration by parts is a powerful technique for integrating products of functions.
Choosing the first function based on the LIATE acronym helps simplify the integral.
Logarithmic functions, inverse trigonometric functions, algebraic functions, trigonometric functions, and exponential functions are commonly used as the first function.
The second function is determined through integration, aiming to simplify the integral or obtain a known function.
Examples and practice are essential to understand the application of integration by parts in solving different types of integrals.

Integration by Parts - Example 4

Consider the integral: ∫ (x * e^x) dx

We will choose u = x (algebraic function) as our first function.
Then, dv = e^x dx, which implies v = e^x.
Now, we can apply the integration by parts formula: ∫ (x * e^x) dx = x * e^x - ∫ (1 * e^x) dx = x * e^x - e^x + C where C is the constant of integration.

Integration by Parts - Example 5

Consider the integral: ∫ (x^4 * ln(x)) dx

We will choose u = ln(x) (logarithmic function) as our first function.
Then, dv = x^4 dx, which implies v = (1/5) * x^5.
Now, we can apply the integration by parts formula: ∫ (x^4 * ln(x)) dx = ln(x) * (1/5) * x^5 - ∫ [(1/5) * x^5 * (1/x)] dx = (1/5) * x^5 * ln(x) - (1/5) * ∫ x^4 dx = (1/5) * x^5 * ln(x) - (1/25) * x^5 + C where C is the constant of integration.

Integration by Parts - Example 6

Consider the integral: ∫ (arcsin(x) * √(1 - x^2)) dx

We will choose u = arcsin(x) (inverse trigonometric function) as our first function.
Then, dv = √(1 - x^2) dx, which implies v = (1/2) * (x * √(1 - x^2) + arcsin(x)).
Now, we can apply the integration by parts formula: ∫ (arcsin(x) * √(1 - x^2)) dx = (1/2) * [arcsin(x) * (x * √(1 - x^2) + arcsin(x)) - ∫ [(x * √(1 - x^2) + arcsin(x) * (1/√(1 - x^2))] dx = (1/2) * [arcsin(x) * x * √(1 - x^2) + (arcsin(x))^2] - ∫ [(x/√(1 - x^2)) + (arcsin(x)/√(1 - x^2))] dx The remaining integrals can be evaluated separately.

Integration by Parts - Example 7

Consider the integral: ∫ (x^2 * e^x) dx

We will choose u = x^2 (algebraic function) as our first function.
Then, dv = e^x dx, which implies v = e^x.
Now, we can apply the integration by parts formula: ∫ (x^2 * e^x) dx = x^2 * e^x - ∫ (2x * e^x) dx
To evaluate the remaining integral, we can apply integration by parts again with u = 2x and dv = e^x dx.

Integration by Parts - Example 7 (continued)

After applying integration by parts a second time, we have: ∫ (x^2 * e^x) dx = x^2 * e^x - [2x * e^x - ∫ (2 * e^x) dx] = x^2 * e^x - 2x * e^x + 2∫ (e^x) dx

The remaining integral is straightforward to evaluate: ∫ (e^x) dx = e^x + C where C is the constant of integration.

Integration by Parts - Example 7 (continued)

Finally, we substitute the value of ∫ (e^x) dx back into the previous expression: ∫ (x^2 * e^x) dx = x^2 * e^x - 2x * e^x + 2∫ (e^x) dx = x^2 * e^x - 2x * e^x + 2e^x + C where C is the constant of integration.

Integration by Parts - Example 8

Consider the integral: ∫ (ln(x) * e^x) dx

We will choose u = ln(x) (logarithmic function) as our first function.
Then, dv = e^x dx, which implies v = e^x.
Now, we can apply the integration by parts formula: ∫ (ln(x) * e^x) dx = ln(x) * e^x - ∫ (1/x * e^x) dx
The remaining integral can be challenging to evaluate. This integral is commonly encountered in calculus and does not have a known elementary antiderivative.

Integration by Parts - Example 8 (continued)

We can rewrite the remaining integral as: ∫ (1/x * e^x) dx = e^x * ln(x) - ∫ (e^x/x) dx

The integral on the right side, ∫ (e^x/x) dx, does not have a known elementary antiderivative.
In some cases, the resulting integral might require special functions or numerical techniques for evaluation.

Integration by Parts - Strategy

When using integration by parts, remember:

Choose the first function based on the LIATE acronym.
Differentiate the first function to obtain du.
Integrate the second function to obtain v.
Apply the integration by parts formula: ∫(u * dv) = u * v - ∫(v * du).
Simplify the resulting integral and evaluate if possible.
If the resulting integral is difficult or impossible to evaluate, consider alternative methods or specialized techniques.

Summary

Integration by parts is a valuable technique for integrating products of functions.
The choice of the first function based on the LIATE acronym can simplify the integral.
Examples of integrals with algebraic, logarithmic, trigonometric, inverse trigonometric, and exponential functions were discussed.
Some integrals can be evaluated directly, while others might require specialized techniques or numerical methods.
Practice and experience are crucial for mastering integration by parts and applying it effectively to solve a variety of integration problems.