How To Integrate Using U-Substitution

Summary

This instructional video demonstrates the technique of u-substitution for integrating definite integrals. It guides viewers through identifying the u variable and its derivative, du, to simplify complex integrals. The script covers several examples, including integrating functions with trigonometric and exponential components, and emphasizes the importance of isolating dx to avoid mistakes. The video aims to help viewers master u-substitution by practicing with various problems, showcasing how to transform and integrate expressions step by step.

Takeaways

• 📚 U-substitution is a method for integrating definite integrals by identifying a suitable 'u' variable and its derivative 'du'.
• 🔍 To apply u-substitution, choose 'u' to simplify the integral by making 'du' cancel out with part of the original integral's expression.
• 📝 The process involves setting 'u' equal to an expression, finding 'du', and then solving for 'dx' in terms of 'du'.
• 🧩 Replace the original variable with 'u' and 'dx' with 'du/dx' to transform the integral into a new form that is easier to integrate.
• 📉 In the example of integrating 4x * (x^2 + 5)^3, set 'u' to x^2 + 5 and 'du' to 2x dx, and then solve for 'dx'.
• 📈 For the integral of 8 cos(4x) dx, set 'u' to 4x and 'du' to 4 dx, simplifying the integral to involve 'u' and 'du/4'.
• 🌟 The antiderivative of cosine is sine, so the integral of 8 cos(4x) dx simplifies to 2 sin(4x) + C.
• 📌 When the integral involves a complex expression, like x^3 e^{x^4}, choose 'u' to be the most complex part, x^4 in this case, to simplify the integral.
• 🔢 In the script, several examples are provided to illustrate the process, including integrals with trigonometric functions and exponentials.
• 🎯 The final step in each example is to replace 'u' back with its original expression in terms of 'x' to find the final antiderivative.
• 📝 The video emphasizes the importance of isolating 'dx' during the substitution process to avoid mistakes and simplify calculations.

Q & A

What is the main topic of the video?

The main topic of the video is teaching how to integrate using u-substitution with a focus on definite integrals.

What is the first integral problem presented in the video?

The first integral problem presented is to find the anti-derivative of 4x * (x^2 + 5)^3.

How does the video suggest to choose the u variable for u-substitution?

The video suggests choosing the u variable to be something that when differentiated will cancel out terms in the integral, such as setting u = x^2 + 5 in the first example.

What is the purpose of finding du in u-substitution?

The purpose of finding du is to replace dx with du/dx, which allows you to change the integral in terms of u and simplify the expression.

How do you solve for dx in the context of u-substitution?

You solve for dx by isolating dx in the equation du = f(x)dx, for example, if du = 2x dx, then dx = du / 2x.

What is the antiderivative of u to the third power using the power rule?

Using the power rule, the antiderivative of u^3 is u^4/4.

What is the second integral problem presented in the video?

The second integral problem is to find the integration of 8 cos(4x) dx.

What is the antiderivative of cosine function?

The antiderivative of the cosine function is sine, because the derivative of sine is cosine.

Can you give an example of a more complex integral problem from the video?

An example of a more complex integral problem from the video is integrating x^3 e^{x^4}.

What is the final step in solving an integral using u-substitution?

The final step in solving an integral using u-substitution is to replace the u variable back with its original expression in terms of x.

How does the video handle the case when the u variable and dx cannot be directly canceled out?

The video demonstrates solving for x in terms of u when the u variable and dx cannot be directly canceled out, as shown in the examples involving 3x + 2 and 4x - 5.

What is the importance of isolating dx during the u-substitution process?

Isolating dx is important to avoid mistakes and to clearly see how to replace dx with du/dx in the integral.

How does the video illustrate the process of u-substitution with trigonometric functions?

The video illustrates the process with an example of integrating sin^4(x) cos(x) dx by setting u to the function with the higher exponent, which is sin(x) in this case.

What is the final answer for the integral of the square root of 5x + 4 as presented in the video?

The final answer for the integral of the square root of 5x + 4 is (2/15)(5x + 4)^(3/2) + C.

Outlines

00:00

📚 Introduction to U-Substitution for Definite Integrals

This paragraph introduces the concept of u-substitution for solving definite integrals. The focus is on finding the anti-derivative of a function involving x squared and a polynomial raised to a power. The speaker illustrates the process by defining a u variable and its derivative (du), which is used to transform the integral into a more manageable form. The example given involves integrating 4x times (x squared + 5) cubed, and the process includes changing variables, solving for dx, and simplifying the integral to find the antiderivative. The paragraph concludes with the substitution of u back into the original variable to obtain the final answer.

05:01

🔍 Applying U-Substitution to Various Integration Problems

The speaker continues to demonstrate the application of u-substitution to different integral problems. The examples include integrating 8 cosine 4x dx, x cube e to the x to the fourth, and 8x times the square root of (40 - 2x) squared dx. For each problem, the speaker identifies the appropriate u substitution, simplifies the integral, and applies the power rule to find the antiderivative. The process involves isolating dx, simplifying the expression, and substituting u back with its original expression to get the final answer. The paragraph emphasizes the importance of recognizing patterns and mastering the technique through practice.

10:02

🎯 Mastering U-Substitution with More Examples

This paragraph presents additional examples to further illustrate the mastery of u-substitution. The problems involve integrating x cubed divided by (two plus x to the fourth) squared and sine to the fourth of x times cosine of x dx. The speaker explains the process of selecting the u variable, solving for dx, and integrating using the power rule. The examples show how to handle more complex expressions and the importance of isolating dx to avoid mistakes. The paragraph encourages viewers to pause the video and attempt the problems to reinforce their understanding of u-substitution.

15:03

🛠 Complex U-Substitution Scenarios and Techniques

The speaker tackles more complex scenarios where the degree of x variables is the same, requiring solving for x in terms of u. The examples include integrating the square root of (3x + 2) and (4x - 5) with respect to x. The paragraph explains the process of isolating x, performing u-substitution, and integrating the resulting expression. The speaker emphasizes the need to eliminate every x variable and provides a step-by-step guide to solving these more challenging problems, including the manipulation of algebraic expressions and the application of the power rule.

20:03

📘 Final Thoughts on U-Substitution Mastery

The final paragraph wraps up the discussion on u-substitution by summarizing the method and encouraging further practice. The speaker highlights that u-substitution becomes easier with practice and familiarity with the technique. The paragraph also presents a final example, integrating (4x - 5) with respect to x, to demonstrate the process one more time. The speaker provides a detailed walkthrough of this example, including solving for x, performing u-substitution, and integrating to find the antiderivative. The paragraph concludes with the final answer and an invitation for viewers to continue practicing to master the u-substitution method.

Keywords

💡U-substitution

U-substitution is a method used in calculus to evaluate integrals, particularly when the integrand can be expressed in terms of a new variable 'u'. In the video, it is the central technique for finding antiderivatives of complex functions. For example, when integrating '4x * (x^2 + 5)^3', the video uses u-substitution by setting 'u = x^2 + 5', which simplifies the integral.

💡Definite Integrals

Definite integrals are a fundamental concept in calculus that represent the signed area under a curve between two points on the x-axis. While the video focuses on indefinite integrals, the process of u-substitution is equally applicable to definite integrals. The video script does not explicitly mention definite integrals, but the technique discussed is foundational for both types.

💡Antiderivatives

Antiderivatives, also known as indefinite integrals, are the reverse process of differentiation. They represent a family of functions that, when differentiated, yield the original function. In the video, finding antiderivatives is the main goal, and the script provides multiple examples of how to use u-substitution to find them.

💡Variable

In the context of calculus and the video, a variable represents a symbol that stands for a number that can change. The script identifies 'u' as a new variable used in u-substitution, which is essential for transforming the integral into a more manageable form.

💡Derivative

The derivative is a measure of how a function changes as its input changes. In u-substitution, identifying the derivative of the new variable 'u' is crucial, as it allows for the transformation of the integral. For instance, when setting 'u = x^2 + 5', the derivative 'du = 2x dx' is found and used to express 'dx' in terms of 'du'.

💡Power Rule

The power rule is a basic principle in calculus used to find the derivative of a function raised to a power. In the video, the power rule is also applied when finding antiderivatives, as seen when integrating 'u^3' where the antiderivative is 'u^4/4 + C'.

💡Trigonometric Functions

Trigonometric functions, such as sine and cosine, are periodic functions that relate the angles of a triangle to the lengths of its sides. The video demonstrates u-substitution with trigonometric functions, as in the example where 'u = sin(x)' is used to integrate 'sin^4(x)cos(x)'.

💡Integration by Parts

Although not explicitly mentioned in the script, integration by parts is a technique related to u-substitution and is used for integrating products of functions. The script implicitly uses this concept when dealing with products like 'x^3e^(x^4)', where one part of the product is treated as 'u' and the other as 'dv'.

💡Constant of Integration

The constant of integration, typically denoted by 'C', is added to the antiderivative to account for the infinite family of functions that could result from differentiation. In the video, after finding the antiderivative using u-substitution, 'C' is added to indicate the general solution.

💡Exponential Functions

Exponential functions are mathematical functions of the form e^x, where 'e' is the base of the natural logarithm. The video includes examples of integrating exponential functions using u-substitution, such as 'e^(x^4)', showcasing how to handle the exponentiation within the integral.

Highlights

Introduction to u-substitution method for integrating definite integrals.

Identifying the u variable and du for the integral of 4x times (x^2 + 5)^3.

Transforming the integral by changing x variables into u variables.

Solving for dx to facilitate the substitution process.

Simplifying the integral to 2 * (u^4/4) + C by canceling terms.

Substituting u back with x^2 + 5 to find the anti-derivative.

Demonstration of integrating 8 * cos(4x) dx using u-substitution.

Using the derivative of cosine to find the anti-derivative of the integral.

Pattern recognition in u-substitution for integrating functions.

Integrating x^3 * e^(x^4) by setting u = x^4 and simplifying.

Dealing with the integral of 8x * sqrt(40 - 2x^2)^2 dx.

Isolating dx and simplifying the integral using u-substitution.

Integrating x^3 / (2 + x^4)^2 by setting u = 2 + x^4.

Solving for x when the degree of x variables is the same.

Integrating sin^4(x) * cos(x) dx by setting u = sin(x).

Integrating sqrt(5x + 4) by setting u = 5x + 4.

Complex example of integrating (3x + 2) with x variables of the same degree.

Final example integrating (4x - 5) with a similar approach to the previous complex example.

Conclusion emphasizing the ease of u-substitution once the method is understood.

Transcripts

in this video i'm going to show you how

to integrate using u-substitution

so we're going to focus on the definite

integrals

how can we find the anti-derivative of

4x

times x squared plus five

raised to the third power

so what do we need to do

we need to define two things we need to

identify the u variable

and d u

now whatever you select u to be

du has to be the derivative if we select

u to be 4x the derivative will be 4 and

that's not going to get rid of x squared

plus 5.

we need to change all of the x variables

into u variables

now

if we make u equal to x squared plus 5

d u is going to be 2x which can cancel

the x and 4x and that's what we want to

do

so let's set u equal to x squared plus

five

d u is going to be 2x but times dx

so what i'm going to do now is solve for

dx in this equation so if i divide both

sides by 2x

dx

is equal to du

divided by 2x

now what you need to do is replace this

with u

and replace the dx

with du over 2x and it will all work out

so let's go ahead and do that so we have

4x

and then u

raised to the third power and then u

divided by two x

so here we can cancel x

four x divided by two x is two

so it's two times the antiderivative of

u to the third

using the power rule is going to be two

times

u to the fourth over four

plus c

now two over four is one half so it's

one half u to the fourth plus c

the last thing you need to do

is replace u with what it equals

and that's

x squared plus five

so this is the answer

let's try another problem

go ahead and find the integration of 8

cosine

4x dx

so what should we make u equal to

we need to make u equal to 4x

d u

will equal 4 dx

and if we divide by 4 du over 4 is dx

so let's replace 4x

with u

and let's replace dx

with du divided by 4.

so this is going to be 8

cosine

of the u variable

and replace dx with du over four

so now we could divide eight by four

eight divided by four is two

now what is the anti-derivative of

cosine

the anti-derivative of cosine is sine

because the derivative of sine is cosine

so we have 2

sine u

plus the constant of integration c

now let's replace u with 4x

so the final answer is 2 sine 4x plus c

so hopefully you see a pattern

emerging

when integrating by u substitution

the key is to identify

what u and d u is going to be

once you figure that out

you just got to follow the process and

it's not going to be that bad so let's

work on a few more examples so you can

master this technique

let's try

x cube e

raised to the x to the fourth

so what should we make u equal to

x cube or x to the fourth

if u is x to the third

d u will be three x squared and that

will not completely get rid of x to the

fourth

but if we make u equal to x to the

fourth

d u will be equal to four x cubed and

that can get rid of the x cubed that we

see here

so let's do that let's make u equal to

x to the fourth

so d u

is going to be 4x cubed dx

and then as always solve for dx

if you don't do that you can easily make

a mistake so i recommend in a step

isolate dx

it'll save you a lot of

trouble later on so du is going to be i

mean dx is going to be du over 4x cubed

now let's replace

x to the fourth with

u and let's replace dx

with d u over four x cubed

so this is going to be e

raised to the u

times d u over four x cubed

so x cubed will cancel which is good

and the 4 we can move it to the front

now because it's in the bottom of the

fraction

it's 1 over 4.

now the antiderivative of e to the u

is simply e to the u

so the final answer well not the final

answer but

the antiderivative is one fourth e to

the u plus c

and now let's replace u with x to the

fourth

so this is the final answer

one fourth e raised to the x to the

fourth plus c

and that's it

here's another one

find the indefinite integral of 8x

times the square root of 40 minus 2x

squared

dx

so typically you want to make u equal to

the stuff that's more complicated

and that is the stuff on the inside of

the square root if we make u equal to 40

minus 2x squared

d u

is going to be

the derivative of 40 is zero so we can

ignore that

and the derivative of negative 2x

squared that's going to be negative 4x

dx

so isolating dx

we need to divide both sides by negative

4x so it's

du over negative 4x

so let's replace

this

with u

and this part dx

with du over negative 4x

so it's going to be 8x times the square

root of u

times du

divided by negative four x

eight x divided by negative four x is

negative two and i'm going to write that

in front

the square root of u is the same as u to

the one half

so now we can use the power rule

one half plus one is three over two

and then we could divide by three over

two or multiply by two over three

which is the better option

so negative two times two thirds that's

negative four over three

now the last thing we need to do

is replace the u variable with 40 minus

two x squared so the final answer is

negative four over three

40

minus 2x squared

raised to the 3 over 2 plus c

and that's all we need to do

let's work on some more problems

feel free to pause the video and try

this one

integrate x cubed divided by

two

plus

x to the fourth

raised to the second power

now typically it's better to make u

equal to the stuff that has the higher

exponent four is higher than three

so let's make u equal to

two plus

x to the fourth

d u

is going to be the derivative of x to

the fourth so that's 4x to the third

power

times dx and as always

i recommend that you solve for dx just

to avoid mistakes

so let's replace two plus x to the

fourth with u

and let's replace dx

with this thing that we have here

so we have x to the third on top

u squared on the bottom and dx is d u

over 4 x cubed

and if you do it this way

as you can see the remaining x variables

will cancel out nicely

so this 4 is in the bottom let's move it

to the front so it becomes 1 4

anti-derivative 1 over u squared d u

and so let's move the u squared from the

bottom to the top

so this is going to be 1 4

integration of u to the negative two

d u

and now we can use the power rule so if

we add one to negative two

that's going to be negative one and then

we need to divide by negative one

so now let's bring this variable back to

the bottom to make the negative exponent

positive

so it's negative one

over four u

plus c

now let's replace u with

what we set it equal to in the beginning

so i'm going to need a bigger fraction

so u is 2 plus x to the fourth and then

plus c

so this

is the final answer

let's integrate sine

to the fourth of x

times cosine of x dx

so go ahead and find the anti-derivative

of this trigonometric function

so if we make u equal to cosine

d u will be negative sine dx that will

only cancel one of the sine variables

and it's best to make u equal to

the trig function that you have more of

we have four sines and only one cosine

so it's best to make u equal to sine x

and d u

will be equal to cosine x

and since there's only one cosine this

will be completely cancelled

solving for dx is going to be du

divided by cosine x

and don't forget that last step always

isolate

dx

so let's replace this with you

so this is going to be u to the fourth

times cosine x and dx is du

divided by cosine

so we could cancel cosine x

and so we're left with

the indefinite integral of u to the

fourth d u

using the power rule four plus one is

five

and then divide by five

so we have one-fifth

u to the five plus c

and the last thing we need to do is

replace u with sine x

so it's one-fifth

sine raised to the fifth power of x

plus c

and that concludes this problem

now what would you do if you have to

integrate the square root of 5x plus 4.

now in this problem

all we could do is set u equal to 5x

plus 4. there's nothing else that we can

do

so let's go ahead and do this

the derivative of 5x is going to be 5

and then times dx

so isolating dx it's going to be

du over 5.

so just like before we're going to

replace 5x plus 4 with u

and

d x

with d u over five

so then this becomes the square root of

u

and dx is d u divided by five

and move the five to the front

so this is one fifth

and then instead of the square root of u

we're gonna write it as u to the

one-half

so now let's use the power rule

one-half

plus one that's gonna be three over two

and

if you divide it by three over two

what i would recommend is multiply the

top and the bottom by two thirds

so the threes in the bottom will cancel

and the twos will cancel as well

so in the end you get one-fifth

u to the three-halves

times two over three

which becomes

two over fifteen

u to the three-half

and let's not forget the constant of

integration plus c

so now to write the final answer

all we need to do is replace u with five

x plus four

so it's two over fifteen

five x plus four

raised to the three over two plus c and

so that's the solution

so as you can see u-substitution

is not very difficult once you get the

hang of it

as long as you do a few problems and get

used to the method

and the techniques employed here

it's a piece of cake

now this problem

is a little bit different from the

others

go ahead and try it i recommend that you

pause the video

and give this one a go

so let's set u

equal to the stuff that's more

complicated

three x plus two

now d u

that's gonna be the derivative of 3x

which is 3 times dx

so isolating dx

it's going to be du over 3 and this time

this x variable will not cancel

so notice that the x variables are of

the same degree and when you see this

situation

it indicates that

in this expression you need to solve for

x

so isolating x

i need to move the two to the other side

so i have

u minus two

is equal to three x

and then dividing by three

u minus two over three

is x

which you can write it as

you can say x is 1 3

u minus 2

which i think looks a lot better

now keep in mind in order to perform u

substitution we need to eliminate every

x variable in this expression

if we replace three x plus two with u

and then dx with

d u over three

this x will still be here and so that's

why we need to solve for x in this

expression

so make sure to do that

if

these two are of the same degree

so this is going to be one third

i'm gonna have to rewrite it because i

can't fit it in here

so let's replace x first with one third

u minus two

and then we have the square root of u

and dx

is d u divided by three

so 1 3 times du over 3 that's going to

be

du over 9.

so i'm going to take the 1 9 and move it

to the front

and then i have u minus 2

and the square root of u is u to the one

half

and then d u

so now we only have the u variable in

this form we can integrate it

now the next thing we need to do

is distribute u to the one half

to u minus two

so u

to the first power

times u to the one half

we need to add one and one half that's

going to be u

to the three over two and then if we

multiply negative two by u to the one

half

that's negative two

u to the one half

and then times d u so now we can find

the antiderivative of each one so for u

to the three halves is going to be three

over two plus one

which is five over two and instead of

dividing it by five over two we're gonna

multiply by two over five

and for u to the one-half

one-half plus one is three over two and

then we're going to multiply by two

thirds

and then we need to add plus c

so now let's distribute one over nine

to everything on the inside so 1 over 9

times 2 over 5 that's going to be 2

over 45

times u raised to the 5 over 2.

and then we have one knife times

this is basically negative four-thirds

so that's gonna be negative

four

over twenty-seven

and that's u to the three over two and

then plus c

so now let's replace u with three x plus

two

so the final answer

is going to be two over forty five

three x plus two

raised to the five over two

minus four over twenty seven times three

x plus two

raised to the three over two plus c

and that is it

now let's work on this example

it's very similar to the last one

so you can try if you want more practice

so let's set u equal to 4x minus 5

which means du

is going to be the derivative of 4x

that's 4 and then times dx

so solving for dx it's du divided by 4.

now we need to isolate x in this

expression so if we add 5

u plus five is equal to four x

and then if we divide by four

x is equal to this which we can write it

as

one fourth

u plus five

so let's replace x

with this expression

and then 4x minus 5

with u and then dx

with du

over 4.

so what we have is two times one fourth

u plus five

and then the square root of u or u to

the one half

and then dx is d u over four

so two times one fourth is one half

and one half times one over four

is one eighth

so we can move the one eighth to the

front and then we have u to the half

times u plus five

now let's distribute u to the one half

so u to the one half times u to the

first power

one half plus one that's going to be

three over two

and then plus 5

u to the one half

so now we need to integrate the

expression that we now have

so this is going to be one over eight

u

three over two plus one that's five over

two and then times two over five

and then

one half plus one is three over two

times two over three

and then plus c

one over eight times two over five

that's going to be two

over forty

and then times u raised to the five over

two

and then we have five

times two over three that's ten over

three

times one over eight

so that's going to be ten

over

eight times three is twenty four

and this is going to be u to the 3 over

2 plus c

now 2 over 40 can be reduced to 1 over

20.

and let's replace u with 4x minus 5.

so this is going to be 4x minus 5

raised to the 5 over 2.

and then 10 over 24 you could reduce

that to 5 over 12

and then it's going to be 4x minus 5

to the 3 halves

and then plus c

you