Integration By Parts

Summary

This educational video script covers the technique of integration by parts to solve various integral problems. It explains the process step by step, starting with the integral of x times e^x and progressing to more complex examples such as the integral of x^2 ln(x) and combinations of exponential and trigonometric functions. The script also tackles integrals involving logarithmic functions and provides a method to solve definite integrals, exemplified by e^(3x)cos(4x). Each example builds on the previous, reinforcing the concept of integration by parts for students to master calculus integration.

Takeaways

• 📚 Integration by parts is a fundamental technique for solving integrals of products of functions, expressed as ∫udv = uv - ∫vdu.
• 🔍 Choosing 'u' and 'dv' wisely is crucial; typically, 'u' is chosen to simplify when differentiated, and 'dv' is chosen to be easily integrable.
• 📈 The integral of x * e^x is found by setting x as 'u' and e^x dx as 'dv', resulting in x * e^x - e^x + C.
• 🌀 For integrals involving trigonometric functions, such as x * sin(x), setting x as 'u' and sin(x) dx as 'dv' leads to -x * cos(x) + sin(x) + C.
• 📘 When integrating a product of a polynomial and a logarithmic function, like x^2 * ln(x), it's often necessary to apply integration by parts multiple times.
• ⚡ The integral of ln(x) is x * ln(x) - x + C, which is derived using integration by parts and recognizing the integral of 1 dx as x.
• 🔗 The integral of x^2 * sin(x) involves using integration by parts twice and results in a complex expression involving both sine and cosine functions.
• 🌟 The integral of x^2 * e^x requires applying integration by parts twice, leading to a solution involving powers of e^x.
• 📉 The integral of ln(x) squared involves rewriting the expression and applying integration by parts, resulting in x * ln(x) squared - 2x * ln(x) + 2x + C.
• 📊 The integral of ln(x) to the seventh power is simplified by recognizing it as 7 * ln(x), leading to 7 * (x * ln(x) - x) + C.
• 🌌 The integral of e^x * sin(x) is solved by recognizing the pattern in the integral and using a clever manipulation to find the solution as (1/2) * e^x * (sin(x) - cos(x)) + C.

Q & A

What is the integral of x times e^x?

The integral of x times e^x is found using integration by parts, setting u=x and dv=e^x dx, which gives du=dx and v=e^x. The result is x * e^x - integral of e^x dx, which simplifies to x * e^x - e^x + C, where C is the constant of integration.

How do you integrate x times sine x dx?

For x times sine x, set u=x and dv=sine x dx, giving du=dx and v=-cosine x. Using integration by parts, the integral becomes -x * cosine x + integral of cosine x dx, which simplifies to -x * cosine x + sine x + C.

What is the antiderivative of x squared times ln x?

To find the antiderivative of x squared times ln x, set dv=x squared dx and u=ln x, giving v=x cubed / 3 and du=1/x dx. The integral by parts results in (1/3) * x cubed * ln x - (1/3) * integral of x squared dx, which simplifies to (1/3) * x cubed * ln x - (1/9) * x cubed + C.

How can you find the integral of the natural log of x using integration by parts?

For the natural log of x, set u=ln x and dv=dx, giving du=1/x dx and v=x. The integral by parts formula yields x * ln x - integral of x * (1/x) dx, which simplifies to x * ln x - x + C.

What is the indefinite integral of x squared times sine x dx?

For x squared times sine x, you need to use integration by parts twice. Initially, set u=x squared and dv=sine x dx, giving du=2x dx and v=-cosine x. The process involves further integration by parts on the resulting integral of 2x * cosine x dx, leading to a final answer involving x squared * cosine x, 2x * sine x, and 2 * cosine x, plus the constant C.

How do you integrate x squared times e^x dx?

For x squared times e^x, use integration by parts twice. Start by setting u=x squared and dv=e^x dx, giving du=2x dx and v=e^x. The process involves another round of integration by parts with u=2x and dv=e^x dx, leading to a final expression involving e^x terms and a constant C.

What is the integral of ln x squared dx?

To integrate ln x squared, set u=ln x squared and dv=dx, giving du=2 * ln x * (1/x) dx and v=x. The integral by parts results in x * ln x squared - 2 integral of ln x * (1/x) dx. Solving this requires recognizing the integral of ln x and simplifying to get the final answer involving x * ln x squared, -2x * ln x, and 2x, plus C.

How can you rewrite the integral of ln x to the seventh power dx?

The integral of ln x to the seventh power can be rewritten using the property of logs to move the exponent to the front, resulting in 7 * integral of ln x dx. Since the integral of ln x is x * ln x - x, the final answer is 7 * (x * ln x - x) + C.

What is the indefinite integral of e^x times sine x dx?

For e^x times sine x, set u=sine x and dv=e^x dx, giving du=cosine x dx and v=e^x. Using integration by parts, the process involves recognizing the similarity of the resulting terms to the original integral and solving for the integral, leading to a final answer involving e^x * (sine x - cosine x) / 2 plus C.

How do you find the antiderivative of ln x squared divided by x dx?

To find the antiderivative of ln x squared / x, rewrite it as ln x squared * 1/x dx and use integration by parts with u=ln x squared and dv=1/x dx, giving du=2 * ln x * (1/x) dx and v=ln x. The process involves recognizing the similarity of terms and solving for the integral, resulting in ln x cubed / 3 plus C.

What is the indefinite integral of e^(3x) times cosine 4x dx?

For e^(3x) times cosine 4x, set u=cosine 4x and dv=e^(3x) dx, giving du=-sine 4x * 4 dx and v=(1/3)e^(3x). Using integration by parts, the process involves moving constants to the front and recognizing the similarity of terms, leading to a final answer involving e^(3x) * (cosine 4x / 3 + 4/3 * sine 4x) plus C.

Outlines

00:00

📚 Introduction to Integration by Parts

This paragraph introduces the concept of integration by parts, a technique used to find the integral of a product of two functions. It explains the formula for integration by parts, \( \int u \, dv = uv - \int v \, du \), and illustrates the process by setting up the functions 'u' and 'dv' for the integral of \( x \cdot e^x \). The explanation includes choosing 'u' and 'dv', finding 'du' and 'v', and applying the formula to arrive at the integral of \( x \cdot e^x \). It also briefly touches on the integral of \( x \cdot \sin(x) \) and \( x^2 \ln(x) \) as further examples.

05:01

🔍 Detailed Walkthrough of Integration by Parts

The second paragraph delves deeper into the application of integration by parts with the integral of \( x \cdot \sin(x) \). It outlines the selection of 'u' and 'dv', calculates 'du' and 'v', and applies the formula to find the integral. The explanation continues with the integral of \( x^2 \ln(x) \), emphasizing the importance of choosing 'u' and 'dv' wisely to simplify the integration process. The paragraph concludes with the final answer for the integral of \( x^2 \ln(x) \) and introduces additional examples to practice the technique.

10:04

📘 Advanced Integration by Parts Examples

This paragraph presents more complex examples of integration by parts, starting with the integral of \( x^2 e^x \). It discusses the strategy of choosing 'u' and 'dv' and the need for applying integration by parts multiple times due to the complexity of the function. The explanation proceeds with the integral of \( \ln(x) \) and \( x^2 \sin(x) \), demonstrating the process of repeatedly using integration by parts and simplifying the expressions to reach the final answers.

15:09

📙 Integration of Logarithmic and Exponential Functions

The fourth paragraph focuses on integrating functions involving logarithms and exponentials. It starts with the integral of \( \ln(x) \) using integration by parts and then moves on to the integral of \( x^2 e^x \), which requires a strategic approach to selecting 'u' and 'dv'. The explanation includes the step-by-step process of integrating these functions and concludes with the final expressions for each integral.

20:10

📒 Dealing with Higher Power Logarithms

This paragraph addresses the integration of higher power logarithmic functions, specifically \( \ln(x)^7 \). It explains how to rewrite the integral using logarithmic properties to simplify the process. The explanation demonstrates the integration of \( 7 \ln(x) \) and arrives at the final answer by applying the previously discussed anti-derivative of \( \ln(x) \).

25:12

📕 Complex Exponential and Trigonometric Integrals

The sixth paragraph tackles the integration of complex functions involving exponentials and trigonometric functions, such as \( e^x \sin(x) \). It describes the process of using integration by parts and the need for careful selection of 'u' and 'dv'. The explanation includes the iterative application of integration by parts and the algebraic manipulation required to simplify the integral to its final form.

30:14

📔 Integration of Products of Logarithms and Trigonometric Functions

The final paragraph discusses the integration of products of logarithmic and trigonometric functions, such as \( \ln(x)^2 / x \). It outlines the process of rewriting the integral and applying integration by parts, including the selection of 'u' and 'dv' and the calculation of 'du' and 'v'. The explanation concludes with the final answer for the integral and provides a method for solving similar problems.

📓 Definite Integrals with Exponential and Trigonometric Functions

This paragraph explores the calculation of a definite integral involving \( e^{3x} \cos(4x) \). It suggests making the trigonometric function the 'u' in the integration by parts process and outlines the steps to find the integral. The explanation includes the manipulation of constants and the application of integration by parts multiple times to arrive at the final expression for the integral.

Keywords

💡Indefinite Integral

An indefinite integral represents the family of all antiderivatives of a given function. It is the reverse process of differentiation and is denoted by the integral symbol ∫ without limits of integration. In the video, the indefinite integral is the main focus, with various functions being integrated to find their antiderivatives, such as x times e^x and x squared sine x.

💡Integration by Parts

Integration by parts is a specialized method for integrating products of functions, based on the product rule for differentiation. It is expressed as ∫u dv = uv - ∫v du. The video script uses this technique multiple times to find the integrals of products like x times e^x and x squared times e^x, by choosing appropriate u and dv to simplify the integration process.

💡Natural Logarithm (ln x)

The natural logarithm, denoted as ln x, is the logarithm of a number to the base e, where e is the Euler's number, an important mathematical constant. In the video, ln x is used in integrals such as the integral of ln x and ln x squared, which are solved using integration by parts and other integration techniques.

💡Exponential Function (e^x)

An exponential function is a mathematical function of the form e^x, where e is the base of the natural logarithm. The video discusses the integration of exponential functions, especially when they are multiplied by other functions like x or sine x, and how to handle these using integration by parts.

💡Trigonometric Functions

Trigonometric functions, such as sine and cosine, are periodic functions that relate angles to the ratios of two sides of a right triangle. The script covers the integration of products involving trigonometric functions, like x times sine x and e^x times sine x, demonstrating how to apply integration by parts to these cases.

💡Derivative

A derivative represents the rate at which a function changes with respect to its variable. In the context of integration by parts, the derivative of a chosen function (u) is needed to find du. The script mentions finding derivatives of functions like x squared and ln x squared to apply the integration by parts formula.

💡Antiderivatives

An antiderivative is a function that represents the reverse process of differentiation, essentially 'undoing' what the derivative does. The video's theme revolves around finding antiderivatives of various functions, such as x times e^x and ln x squared divided by x, using different integration techniques.

💡Chain Rule

The chain rule is a method for finding the derivative of a composite function. In the script, the chain rule is implied when dealing with the derivative of ln x squared, which requires multiplying the derivative of the inner function (1/x) by the derivative of the outer function (2ln x).

💡Constant of Integration (C)

The constant of integration, typically denoted as C, is added to the result of an indefinite integral to account for the infinite family of antiderivatives that differ by a constant. The script includes C in the final answers of integrals to indicate that the solution is an antiderivative plus an arbitrary constant.

💡Power Rule

The power rule is a basic principle in calculus for finding the derivative of a function of the form x^n, where n is a constant. In the script, the power rule is applied when integrating functions like x squared, resulting in x to the power of (n+1) divided by (n+1).

💡Product Rule

The product rule is a fundamental rule in calculus for differentiating the product of two functions. Although the video focuses on integration, the product rule is the basis for the integration by parts formula, which is derived from the product rule in reverse. The script uses integration by parts, which is essentially the product rule applied to integration.

Highlights

Introduction to finding the indefinite integral of x times e^x using integration by parts.

Setting u=x and dv=e^x dx for the integral of x times e^x.

Deriving the integral of e^x as e^x and applying integration by parts formula.

Solving the integral of x times sine x dx using integration by parts.

Choosing u=x and dv=sin(x)dx for the integral of x times sine x.

Finding the antiderivative of sine x as -cosine x and applying it to the integral.

Integrating x squared ln x using a strategic choice of u and dv.

Deciding on u=ln(x) and dv=x^2 dx for the integral of x squared ln x.

Using the derivative of ln(x) as 1/x to simplify the integral.

Deriving the integral of ln(x) as x ln(x) - x and applying integration by parts.

Integrating x squared sine x dx using a two-step integration by parts process.

Choosing u=x^2 and dv=sin(x)dx for the integral of x squared sine x, and applying integration by parts twice.

Solving the integral of x squared e^x dx using a two-step integration by parts approach.

Deciding on u=x^2 and dv=e^x dx for the integral of x squared e^x, and applying integration by parts twice.

Finding the indefinite integral of ln(x) squared dx using integration by parts.

Rewriting ln(x) squared as 2ln(x) and applying integration by parts.

Solving the integral of ln(x) to the power of seven using properties of logarithms.

Integrating e^x sin(x) dx using a recursive integration by parts strategy.

Rewriting the integral of ln(x) squared over x as a product of two functions.

Solving the definite integral of e^(3x)cos(4x) dx using integration by parts.

Applying integration by parts to the integral of e^(3x)sin(4x) dx and simplifying the result.

Transcripts

so how can we find the indefinite

integral of x

times e raised to the x

to do this

we need to use something called

integration by parts

the integral of u

dv

is equal to u times v

minus the integral of v

d u

so we need to determine u

dv

v

and u and d u

let me put it in a different order

so which one should we set equal to u

and which one

should we set equal to dv

it's best to set u

equal to x

because if we do so

d u will simply be equal to one times dx

and when dealing with constants

they're a lot easier to integrate than

variables

so we're going to make dv equal to e to

the x dx

now v

is the integral of dv

the integral of e to the x is simply e

to the x

so now using the formula u times v

that's going to be x

times e raised to the x

minus the integral of v

d u

v

is e to the x d u is 1 dx

and the integral of

e to the x

is e to the x plus some constant c

so this is the final answer that's the

integral

of

x raised to the or x times e raised to

the x

let's try another problem

let's integrate x times sine x dx

so what should we make u

and

dv equal to

so just like before i'm going to make u

equal to x

so that d u is going to be 1

dx

now dv

has to be what remains

sine x

dx

now what is the integral of sine

the derivative of what function will

give us sign

the integral of sine is negative cosine

the derivative of negative cosine is

positive sine

so using the formula the integral of u

dv

is equal to u times v

minus the integral of v

d u

so u is x

v is negative cosine x

and then minus v

and d u is

1 dx

so far we have negative x

cosine x

and then these two negative signs will

become positive so plus

the integral of cosine x

dx

now the antiderivative of positive

cosine

is positive sine

so the final answer

is what you see here negative cosine i

mean negative x cosine x

plus sine x plus c

now here's another problem to work on

let's integrate x squared ln x

so what should we make u

and

dv equal to in this problem

if we make u equal to x squared

and we make dv equal to ln x

dx the integral of ln x

will be quite difficult to do

unless you know what it is so we're not

going to do that

so it's best to make

dv

x squared dx

and u is going to be ln x because we

know the derivative of l and x

the derivative of l and x

is 1 over x

and the antiderivative of x squared

the antiderivative of dv will give us v

so the antiderivative of x squared dx

is x cubed over 3.

so let's use the formula now

u times v that's going to be ln x times

x cubed over three

minus the integral of v

times d u

and

d u is

one over x

it's supposed to be dx

so this is equal to 1 3

x cubed times ln x

and then we could cancel an x and so we

have minus one third the integral of x

squared

dx

the integral of x squared is going to be

x to the third over three

and then plus some constant c

so the final answer is one third

x cubed ln x minus one over nine

x to the third plus c

and that concludes this problem

so let's work on some more examples

let's try this problem what is the

anti-derivative

of the natural log of x

how can we use integration by parts to

get this answer

now we can't make u

equal well we can't make a dv equal to l

and x

because we don't know the integral of l

and x we're trying to figure that out so

therefore we have to make u equal to

ln x the derivative of that is one over

x dx

dv

has to be

just we can make dv dx if we want to

or one dx

and so v the integral of one dx is just

x

so now let's start with the formula

and so it's going to be u times v

u is ln x

and v

is x

minus the integral of v

and then d u

is one over x

dx

so far we have x

ln x

x times one over x the x variables

cancel

so we have the integral of one

dx now the integral of one dx is x

so this is the final answer

x ln x

minus x plus c

that is the integral

of the natural log of x

let's find the indefinite integral of x

squared

sine x

dx

so which part should be u

and which part

is going to be

dv

we need to make u equal to x squared

and we need to differentiate it twice to

bring it down to a constant so this is

one of those problems where you have to

use integration

of you know integration by parts two

times

d u is going to be 2x dx

dv

is sine x dx

so this is dv

and this is going to be u

the integral of sine

is negative cosine

so let's start with the formula

so u times v that's x squared

times negative cosine x

minus

the integral of v which is negative

cosine x

times d u

that's

2x dx

so let's organize what we have negative

x squared cosine x

and then these two negative signs will

become positive so positive the integral

of

2x cosine x

we could take the 2 and move it to the

front so it's just going to be x cosine

x

now we need to use the integration by

parts formula

on the integral of x cosine x

we're going to make u equal to x

and

dv

is going to be cosine x

dx

the integral of cosine is sine and the

derivative of x is 1

dx

so using the formula

it's going to be u times v

which is

x times sine x

minus the integral

of v d u

so v is sine x

d u is just dx

and so the integral of sine

is

negative cosine

and then plus c

so now what we need to do is replace

this part

with what we have here

so it's going to be plus we need to

distribute 2 to everything in here

so it's going to be 2x

sine x

plus

this would be plus cosine x but times

two so plus two cosine x

and we don't need to multiply c by two

because c is just

some generic constant

and

this whole thing is equal to the

original integral and so this is the

answer

negative x squared cosine x

plus

2x sine x

plus 2 cosine x plus c

now let's work on this problem

let's say we have the integral of x

squared

e to the x dx

go ahead and try that problem

so let's begin by writing the formula

the integral of u

dv is equal to

u times v minus the integral of v

d u

now what should we make u equal to

and what should be

dv

i prefer making dv

e c x dx

because once we integrate it

it's going to stay the same

and i want x square to eventually become

a constant so this is one of those

problems where you need to do

use the integration by parts formula

twice

so we're going to make u

equal to x squared

du is going to be 2x dx

so using the formula it's going to be uv

that's going to be

x squared

that's you v is e to the x

minus the integral of v which is e to

the x

times d u that's 2x

dx

so now this becomes

well before we uh do that let's use

integration by parts one more time

so i'm going to make u

equal to 2x

du is going to be the derivative of 2x

which is 2 but times dx

dv

we're going to make that equal to

e x dx

just like we did before

and the integral of that is simply e x

so we're going to have

x squared

e x

and then minus

now let's use the formula again for this

part so it's uv

so 2x

e to the x

and then minus the integral

of vdu

so that's 2

e to the x

dx

or multiplying those two

now let's begin by distributing the

negative signs

so this is going to be negative 2x

e to the x and then these two negatives

will cancel

giving us

positive

well let's take out the two so positive

two integral

e to the x dx

so our final answer is going to be x

squared

e c x minus

2 x e to the x

anti derivative of e to the x is just e

to the x and then plus c

now if we want to we can factor out

e to the x

so we're going to have x squared minus

2x

plus 2 and then plus the constant c

so this right here is our final answer

that is the anti-derivative

of

x squared e to the x

what about this problem

what is the integral of

ln x

squared

dx

try that

so for this problem what we need to do

is we need to make u

equal

to ln x squared

dv is going to be equal to the other

part

we're going to make dv

equal to dx

so v

is going to be the integral of dx which

is x

d u i'm going to write that here i need

more space

we need to use the chain rule

so first let's use the power rule by

moving the 2 to the front

so it's going to be 2 and then we'll

keep everything on the inside the same 2

times ln x

and then we're going to subtract that by

1

and then take the derivative of the

inside the derivative of l and x is one

over x

so using the formula this is going to be

u times v

so v is x u is

ln x squared

minus the integral of v d u

v is x

d u is

2

ln x

times 1 of x and then

let's not forget the dx part

so we can cancel x and one of x

so this becomes x

ln x squared

and then we could take out the two we

can move it to the front so minus two

integral

ln x dx

earlier in this video we determined that

the integral of l and x is x ln x minus

x

so this is going to be -2

and then we can replace this with

x ln x minus x

now let's go ahead and distribute the

two so our final answer is going to be x

times ln x squared

minus 2x

ln x

and then negative 2 times negative x

that's going to be positive

2x and then plus c

so this

is

the indefinite integral

of ln x squared

now what would you do if you were to see

a problem like this

ln x to the seventh power dx

go ahead and try that

well first you need to realize that we

can rewrite this problem

a property of logs allows us to move the

exponent to the front

so make sure you understand this this is

not ln x

to the seventh power like this

if it was we can't move the seven

the seven is only affecting the x

variable as a result we can move the 7

to the front

so this becomes 7

ln x

so this is what we're integrating

this is equivalent to 7

integral

ln x dx

and we know the anti-derivative of l and

x it's x

ln x minus x

so this is our final answer it's just 7

x ln x

minus 7 x plus c

now let's try this one let's find the

indefinite integral

of e raised to the x sine x dx

go ahead and work on that

so for this problem i want to make

u equal to sine x

and then dv i'm going to set that equal

to

e to the x dx

so v is simply going to be e to the x

d u

is the derivative of sine x which is

cosine x

so using the formula it's going to be

uv

that's going to be e

sine x

minus the integral of v d u

so v is e to the x d u is cosine x dx

so we need to use integration by parts

one more time

but this time we're going to make

u equal to

cosine x instead of sine

and dv is going to be the same

dv is going to be e to the x

dx

if v is the same

d u is going to be the derivative of

cosine which is negative

sine

and then times dx

so now we're going to have

e x

sine x

minus and then in brackets

u v so that's going to be e x

cosine x

minus the integral of v to u

so v is e to the x d u is

negative

sine x and then dx

so let's go ahead and distribute the

negative sign i'm going to rewrite the

original integral

so the integral of e x sine x

that's going to be e to the x

sine x

these two negative signs

will become positive

and then we need to distribute this

negative so it's negative e raised to

the x cosine x

and then we need to apply the negative

to this

which is going to be negative

e to the x sine x

now the key to solving this problem is

to realize that this term is very

similar to

the original problem

so what we want to do is we want to

add this to both sides

if we move it to the other side we're

going to have 2

e x sine x

which is equal to this

now we want to find the value of just

one

integral e to the x sine x

so what we need to do is divide both

sides

by two

so that's we can say that our final

answer

is one half

we also factor out e to the x so it's

one half e to the x

and then times

sine x

minus cosine x

plus c

so that's the anti-derivative

of

e raised to the x times sine x

now let's move on to our next example

let's find the antiderivative of ln

x squared

divided by

x dx

so go ahead and try that

now the first thing that i want to do is

i'm going to rewrite this as

ln x squared times 1 over x

dx

i want to multiply as a product of two

functions

now what i'm going to do is i'm going to

make u

equal to ln x

squared

du will be 2

times ln x to the first power

times 1 over x

and then dx

so since i made u ln x squared

dv is going to be the other part

dv is going to be 1 over x

dx

so anti-derivative of 1 over x is ln x

now using the formula this is going to

be u times v

u is ln x

squared

v is just lnx

minus the integral of vdu

so v is ln x

d u is 2

ln x times 1 over x

dx

so let's put this all together so first

i'm going to rewrite the original

problem

that's equal to ln x squared times ln x

that's

ln x raised to the third power

in this 2 i'm going to move to the front

here we have ln x times ln x

that's ln x

squared

and it's multiplied by one over x which

we can move that to the bottom

and we need to realize that these two

terms are similar

so we need to add

this to both sides or if we move it to

the other side

we'll basically add in 1

plus 2 which will become 3. so on the

left side we're going to have 3

integral

ln x

squared

over x dx

and that's going to equal

ln x

to the third power

so if we divide both sides by 3

this 3 will disappear

and we're going to have it underneath

here

and then plus our constant c

so this

is the answer

to the original problem that's how we

can find the integral of ln x squared

over x

now for the sake of practice go ahead

and try this

finding a definite integral of e to the

3x cosine

4x dx

so whenever you have an e to the x times

the sine or cosine

it's good to make

the cosine part

equal to u

that's what i'd like to do personally

so in this one i'm going to make u equal

to cosine

4x

du

the derivative of cosine is going to be

negative

sine

4x and then times the derivative of the

angle the derivative of 4x is 4

and then we're going to have times dx

so dv in this problem

is going to be e

to the 3x to dx

the antiderivative of e to the 3x it's e

to the 3x over 3

which

we can write as one third e to the 3x

so using the formula

it's going to be uv

so that's 1 3

e to the 3x times cosine

4x

minus the integral

of v d u

so v

is

one third

e to the three x

d u

that's

negative

four

sine 4x

dx

so what we're going to do now is we're

going to move the constants to the front

the two negative signs

will cancel they will become positive

and we can move one third times four to

the front

so this is going to be plus

four over three

integral

e to the three x

sine four x dx

now we need to use integration by parts

one more time

so dv will be the same we don't need to

adjust this part of the formula

but u

is going to be sine 4x as opposed to

cosine for x

so du is going to change as well

the derivative of sine is cosine

4x

and then times 4 times dx

so using integration by parts we're

going to have u times v

so that's one-third

e to the 3x times sine

4x

and then minus

the integral of vdu

so v is one third

e to the three x

and then d u that's times four

cosine four x

d x

now let's go ahead and distribute four

over three

so multiply those together that's going

to be

4 over 9

e to the 3x

sine 4x

and then this 2 will also be 4 over 9

but it's going to be negative

four over nine

actually

there's a four here so this is four

thirds times four thirds

so that's 16 over nine with a negative

sign

and then we're going to have the

integral of e

to the 3x

cosine 4x

dx

now notice that these two are similar

so we're going to add

16 over 9 integral e to the 3x cosine

4xdx to both sides

so this is 1 but we can treat it as if

it's

9 over 9 just to get common denominators

so 9 over 9 plus

16 over 9

that's going to be

25

over 9

integral

e to the 3x

cosine 4x dx

and then that's going to be 1 3

e to the 3x cosine 4x

and then plus

4

over 9

e to the

3x sine 4x

now to get this to 1

what we're going to do is we're going to

multiply both sides by the reciprocal of

25 over nine

so that is by nine over twenty five

so those two fractions will cancel

on the left side we're just going to

have

e to the three x cosine four x

dx

and

what i'm going to do

i want to keep this in factored form

so we have 9 over twenty five

from one third and four over nine we can

factor out one third

and we can also factor out e to the

three x

so here we took out e to the three x

we're left with just cosine four x

here we took out e to the three x and we

took out one third from four over nine

so we're gonna have four thirds left

over times sine 4x

the last thing we could do is maybe

reduce this

so if we divide both numbers by 3 this

becomes

3

over 25

e to the 3x

and then times

cosine 4x plus 4 over 3

sine 4x plus c

so that's going to be the answer for

this problem

that's the indefinite integral of

e to the 3x cosine 4x

you