How To Integrate Using U-Substitution - Free Educational videos for Students in K-12

How To Integrate Using U-Substitution - By The Organic Chemistry Tutor

Transcript


00:0-1	in this video , I'm going to show you how
00:02	to integrate using use substitution . So we're going to
00:07	focus on the definite integral seals . How can we
00:10	find the anti derivative of four x times X squared
00:15	plus five Raised to the 3rd power . So what
00:20	do we need to do ? We need to define
00:22	two things . We need to identify the U .
00:25	Variable . And do you now whatever you select you
00:32	to be , do you have to be the derivative
00:34	? If we select you to be four X ,
00:37	the derivative will be four and that's not going to
00:39	get rid of X squared plus five . We need
00:42	to change all of the X variables into you variables
00:45	. Now if we make U equal to X squared
00:49	plus five D'you is going to be two X .
00:51	Which can cancel the X and four X . And
00:54	that's what we want to do . So let's set
00:57	u equal to X square plus five d'you is going
01:01	to be two X but times dx . So what
01:08	I'm gonna do now is solve for dx in this
01:10	equation . So if I divide both sides by two
01:13	X . Dx is equal to do . You Divided
01:20	by two x . Now what you need to do
01:23	is replace this with you and replace the dx with
01:28	D . You over to X . And it will
01:30	all work out . So let's go ahead and do
01:32	that . So we have four X . And then
01:35	you raised to the 3rd power And then the U
01:39	divided by two x . So here we can cancel
01:42	X . four x divided by two X . is
01:45	too . So it's two times the anti derivative of
01:50	You to the 3rd . Using the power rule is
01:53	going to be two times U . To the fourth
01:56	over four plus C . Now to over four is
01:59	one half . So it's one half used the fourth
02:02	plus C . The last thing you need to do
02:05	is replace you with what equals and that's X .
02:11	Square plus five . So this is the answer .
02:15	Mhm . Let's try another problem . Go ahead and
02:27	find the integration of eight . Co sign for X
02:30	. Dx . So what should we make U .
02:34	Equal to ? We need to make U equal to
02:38	four x . D . You Will Equal four DX
02:43	. And if we divide by four do you over
02:45	four is dx . So let's replace four x .
02:50	With you . And let's replace detox With D .
02:55	U . divided by four . So this is going
02:58	to be eight co sign of the U . Variable
03:02	. And we placed the acts with D . You
03:04	over four . So now we can divide eight by
03:07	48 divided by four is two . Now what is
03:11	the anti derivative of co side ? The anti derivative
03:15	of co sign is signed because the derivative of science
03:18	co sign . So we have to sign you plus
03:23	the constant of integration . See now let's replace you
03:27	with four X . So the final answer is to
03:29	sign four X plus C . So hopefully you see
03:34	a pattern emerging when integrated by use substitution . The
03:39	key is to identify what you and do you is
03:42	going to be once you figure that out , you
03:45	just got to follow the process and it's not gonna
03:46	be that bad . So let's work on a few
03:48	more examples so you can master this technique . Let's
03:53	try X cube E . Race to the X .
03:57	to the 4th . So what should we make U
04:00	. Equal to X . Cuba X to the 4th
04:04	. If you is existent third D . You will
04:07	be three X squared . And that will not completely
04:10	get rid of excess the fourth . But if we
04:12	make U equal to X to the 4th . D
04:15	. You will be equal to four x cubed .
04:17	And that can get rid of the X . Cube
04:20	that we see here . So let's do that .
04:22	Let's make U equal to X to the 4th .
04:25	So do you is going to be four X cubed
04:29	dx . And then as always solve for dx if
04:34	you don't do that you can easily make a mistake
04:37	. So I recommend in this step isolate detox .
04:41	It'll save you a lot of trouble later on .
04:44	So do you is going to be I mean DX
04:46	is going to be do you over four x cube
04:50	Now let's replace X to the 4th with you and
04:55	let's replace detox With D . You over four x
04:59	Cube . So this is gonna be E . Mhm
05:05	. Race to you times T you over for X
05:08	cubed . So execute will cancel which is good and
05:13	before we can move it to the front now because
05:16	it's in the bottom of the fraction it's running before
05:20	now the anti derivative of E . To the U
05:23	. Is simply easy to you . So the final
05:25	answer , we're not the final answer . But the
05:28	anti derivative is 1/4 E . T . U .
05:30	Plus C . And now let's replace you with exit
05:33	the fourth . So this is the final answer .
05:36	1/4 E . Raised to the X . to the
05:38	4th plus c . And that's it . Here's another
05:47	one . Find the indefinite integral of eight x times
05:53	the square root of 40 minus two X . Squared
05:56	detox . So typically you wanna make U . Equal
06:00	to the stuff that's more complicated and that is the
06:03	stuff on the inside of the square root . If
06:06	we make u equal to 40 -2 x squared .
06:10	Do you ? It's going to be The derivative of
06:14	40 so we can ignore that . And the derivative
06:17	of negative two X . Squared . That's gonna be
06:19	negative four X . Dx . So isolated dx we
06:25	need to divide both sides by negative four X .
06:27	So it's d you over negative four X . So
06:31	let's replace this with you And this part dx with
06:38	D . U . Over negative four X . So
06:41	it's going to be eight x times the square root
06:44	of you Times . Do you Divided by -4 X
06:49	. Eight X . Divided by negative four X .
06:52	Is negative two . I'm going to write that in
06:54	front . The square root of you is the same
06:56	as you to the one half . So now we
06:59	can use the powerball one half plus one is 3/2
07:05	. And then we could divide by three of the
07:07	two or multiply by 2/3 which is the better option
07:13	. So negative two times two thirds . That's negative
07:15	4/3 . Now the last thing we need to do
07:21	is replaced the U . Variable with 40 minus two
07:24	X squared . So the final answer is negative 4/3
07:28	40 -2 x squared . Race to the three of
07:32	the two Plus C . And that's all we need
07:36	to do . Let's work on some more problems .
07:43	Feel free to pause the video and try this one
07:45	, integrate X cubed divided by two plus X to
07:52	the 4th . Race to the 2nd power . Now
07:58	, typically it's better to make U equal to the
08:00	stuff that has the higher exponent four is higher than
08:03	three . So let's make U equal to two plus
08:09	X to the 4th . Do you ? It's going
08:12	to be the derivative of X to the fourth .
08:14	That's four X to the third power times dx .
08:17	And as always I recommend that you solve for dx
08:21	Just to avoid mistakes . So let's replace two plus
08:26	X to the 4th of you . Unless replace dx
08:31	with this thing that we have here . So we're
08:36	gonna have X to the third on top , U
08:39	squared on the bottom . And dx is do you
08:42	Over four X . Cubed . And if you do
08:45	it this way as you can see the remaining X
08:47	variables will cancel out nicely . So this floor is
08:51	in the bottom . Let's move it to the front
08:52	so it becomes 1/4 anti derivative one over U .
08:57	Squared do you ? And so let's move the youth
09:00	square from the bottom to the top . So this
09:04	is gonna be 1 4th integration of U . To
09:07	the negative too do you ? And now we can
09:10	use the power room . So if we add 12
09:13	negative two , That's gonna be negative one . And
09:16	then we need to divide by -1 . So now
09:19	let's bring this variable back to the bottom to make
09:22	the negative export and positive . So it's negative one
09:26	over for you Plus C . Now let's replace you
09:34	with what we said is equal to the beginning .
09:38	So I'm gonna need a bigger fraction so you is
09:42	two plus X to the fourth and then plus thing
09:47	. So this is the final answer . Let's integrate
09:54	sign To the 4th of X . Times . Co
09:59	sign of accident tax . So go ahead and find
10:02	the anti derivative of this trigger metric function . So
10:08	if we make U equal to co sign D .
10:11	You will be negative sign dx . That will only
10:13	cancel one of the sign variables . And it's best
10:17	to make U equal to the trick function that you
10:20	have more . We have four signs and only one
10:23	co sign . So it's best to make U equal
10:25	to syntax and D . You will be equal to
10:29	cosine X . And since there's only one co sign
10:33	, this will be completely canceled , solving for Dx
10:36	is going to be D . You divided by co
10:38	sign X . And don't forget that last step always
10:42	isolate dx . So let's replace this with you .
10:52	So this is going to be U . to the
10:53	4th times Cosine X . And dx is D .
10:57	U divided by co sign . So we could cancel
11:01	cosign X . And so we left with the indefinite
11:06	integral of U . To the fourth . D .
11:08	You using the power rule four plus one is five
11:11	And then divide by five . So we have 1
11:17	5th You to the five Plus C . And the
11:21	last thing we need to do is replace you with
11:22	syntax . So it's 1/5 Sign race to the 5th
11:27	power of X plus E . And that concludes this
11:32	problem . Now , what would you do if you
11:40	have to integrate the square root of five X plus
11:43	four ? Now in this problem , all we could
11:47	do is set U equal to five X plus four
11:50	. There's nothing else that we can do . So
11:52	let's go ahead and do this . The derivative of
11:55	five x is going to be five and then times
11:58	dx so isolated detox , it's going to be d'you
12:02	over five . So just like before we're going to
12:05	replace five X plus four with you and dx with
12:10	D you over five . So then this becomes the
12:14	square root of you and DX is d'You divided by
12:18	five And move the 5 to the front . So
12:21	this is 1 5th and then instead of the square
12:24	root of you were gonna right as you to the
12:26	one half . So now let's use of horrible one
12:30	half plus one . That's going to be 3/2 .
12:34	And If you divide it by 3/2 , What I
12:38	would recommend is multiply the top and the bottom by
12:40	2/3 . So the threes in the bottom will cancel
12:46	and the tools will cancel as well . So in
12:48	the end you get 1 5th , You to the
12:51	3/2 times to over three , which becomes to over
12:57	15 , You to the 3/2 . And let's not
13:01	forget the constant of integration plus C . So now
13:06	to write the final answer , all we need to
13:08	do is replace you with five X plus four .
13:11	So it's to over 15 five x plus four .
13:16	Race to the three of the two plus C .
13:18	And so that's the solution . So as you can
13:22	see use substitution . It's not very difficult once you
13:26	get the hang of it , as long as you
13:28	do a few problems and get used to the method
13:32	and the techniques employed here , it's a piece of
13:34	cake . Now this problem is a little bit different
13:40	from the others . Go ahead and try it .
13:43	I recommend that you pause the video and Give this
13:48	1 to go . So let's set you equal to
13:52	the stuff . That's more complicated . Do we express
13:55	to now , do you ? That's gonna be the
14:00	derivative of three acts , which is three times DX
14:03	. So isolating detox , it's going to be do
14:06	you over three and this time this X variable will
14:10	not cancel . So notice that the X variables are
14:14	of the same degree and when you see the situation
14:18	it indicates that in this expression you need to solve
14:21	for X . So isolated X . I need to
14:25	move the to to the other side . So I
14:27	have U minus two Is equal to three acts and
14:32	then divided by three . U -2/3 is X .
14:39	Which you can write it as you can say X
14:42	is one third u minus two . Which I think
14:46	looks a lot better now . Keep in mind in
14:50	order to perform use substitution . We need to eliminate
14:53	every X variable in this expression . If we replaced
14:58	reacts plus two with you And then DX with D
15:02	You over three . This act will still be here
15:05	and so that's why we need to solve for X
15:07	in this expression . So make sure to do that
15:12	if these two are of the same degree . So
15:18	this is gonna be 1/3 . I'm going to have
15:21	to rewrite it because I can't fit it in here
15:25	. So let's replace experts with 1/3 u minus two
15:30	and then we have the square root of you and
15:34	DX is d'You divided by three so one third times
15:40	do you over three ? That's going to be Do
15:42	you over nine . So I'm going to take the
15:44	one knife and move it to the front And then
15:47	they have U -2 . And the square root of
15:51	you is you to the one half . And then
15:53	do you . So now we only have the you
15:56	variable in this form . We can integrate it .
16:00	Yeah . Yeah . Now the next thing we need
16:09	to do Is distribute you to the 1/2 to U
16:12	-2 . So you to the first power Times .
16:17	Used to 1/2 we need to add one and one
16:21	half . That's going to be you to the three
16:23	of a tune . And then if we multiply negative
16:25	to buy , used to 1/2 that's negative too .
16:28	You to the 1/2 and then times do you .
16:32	So now we could find the anti derivative of each
16:34	one . So for you to the three halfs it's
16:37	gonna be 3/2 plus one Which is 5/2 . And
16:41	instead of dividing it by five of the two we're
16:42	gonna multiply by 2/5 . And for you to the
16:46	1/2 . One half plus one is three of the
16:49	two . And then we're gonna multiply it by two
16:50	thirds . And then we need to add plus C
16:59	. So now let's distribute 1/9 to everything on the
17:02	inside . So one of the nine times to over
17:05	five . That's gonna be too over 45 Times .
17:10	You raise to the 5/2 and then we have one
17:13	knife times . This is basically negative 4/3 . So
17:19	that's gonna be negative four over 27 . And that's
17:24	you to the three of the two and then plus
17:26	C . So now let's replace you with three X
17:29	. Plus two . So the final answer It's going
17:36	to be to over 45 . three x plus two
17:41	. Race to the 5/2 -4/27 . Time Street Acts
17:46	Plus two . Race at a 3/2 Plus C .
17:52	And that is it . Yeah . Now let's work
18:00	on this example . It's very similar to the last
18:03	one . So you can try if you want more
18:05	practice . So let's set U equal to four X
18:10	-5 . Which means do you . It's going to
18:13	be the derivative of four X . That's four and
18:16	in times dx . So solving for DX It's d'You
18:20	divided by four . Now we need to isolate X
18:23	. And this expression . So if we add five
18:27	, you plus five is equal to four X .
18:29	And then if we divide by four X . is
18:34	equal to this which we can write it as 1/4
18:38	. You plus five . So let's replace acts with
18:45	this expression And then four x -5 with you .
18:50	And then dx With D . You over four .
18:56	So what we have is two times 1/4 . You
19:01	plus five And then the square root of you or
19:05	you to them 1/2 And then de acciones de you
19:09	over four . So two times 1/4 is one half
19:15	and one half times 1/4 is one . If So
19:19	we can move the 1/8 to the front and then
19:22	we have you to the half Times U . Plus
19:25	five . Now let's distribute you to the one half
19:29	. So you 2 1/2 times . Used the first
19:31	power 1/2 plus one . That's going to be three
19:34	of the two And then plus 2 5 You two
19:38	and 1/2 . So now we need to integrate the
19:44	expression that we now have . So this is gonna
19:52	be 1/8 you three of the two plus one .
19:57	That's 5/2 . And then times 2/5 and then one
20:02	half plus one is three of the two Times to
20:05	over three . And then plus C . Yeah 1/8
20:20	times 2/5 That's gonna be to over 40 . And
20:26	then times you raised to the five or 2 and
20:30	then we have five times 2/3 That's 10/3 Times 1/8
20:37	. So that's going to be 10 over eight times
20:42	3 is 24 . And this is going to be
20:47	used at 3/2 plus C . Now to over 40
20:50	can be reduced to 1/20 . And let's replace you
20:54	with four X -5 . So this is going to
20:57	be four X -5 . Race to the five of
21:01	the two And then 10/24 . You can reduce that
21:04	to 5/12 , And then it's going to be four
21:08	X -5 today , three halfs and then plus C
00:0-1	.

Summarizer

DESCRIPTION:

This calculus video tutorial provides a basic introduction into u-substitution. It explains how to integrate using u-substitution. You need to determine which part of the function to set equal to the u variable and you to find the derivative of u to get du and solve for dx. After replacing all x variables with u variables, find the antiderivative of f(u) and substitute u in the new function with x variables. This video contains plenty of examples and practice problems of finding the indefinite integral using u-substitution. Examples include polynomial functions, trigonometric functions, exponential functions, square root functions, and rational functions.

OVERVIEW:

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