How To Integrate Using U-Substitution
Summary
TLDRThis 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
- π U-substitution is a method for integrating definite integrals by identifying a suitable 'u' variable and its derivative 'du'.
- π To apply u-substitution, choose 'u' to simplify the integral by making 'du' cancel out with part of the original integral's expression.
- π The process involves setting 'u' equal to an expression, finding 'du', and then solving for 'dx' in terms of 'du'.
- 𧩠Replace the original variable with 'u' and 'dx' with 'du/dx' to transform the integral into a new form that is easier to integrate.
- π 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'.
- π 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'.
- π The antiderivative of cosine is sine, so the integral of 8 cos(4x) dx simplifies to 2 sin(4x) + C.
- π 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.
- π’ In the script, several examples are provided to illustrate the process, including integrals with trigonometric functions and exponentials.
- π― The final step in each example is to replace 'u' back with its original expression in terms of 'x' to find the final antiderivative.
- π 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
π 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.
π 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.
π― 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.
π 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.
π 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.
Mindmap
Keywords
π‘U-substitution
π‘Definite Integrals
π‘Antiderivatives
π‘Variable
π‘Derivative
π‘Power Rule
π‘Trigonometric Functions
π‘Integration by Parts
π‘Constant of Integration
π‘Exponential Functions
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
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