Trigonometric Substitution - By The Organic Chemistry Tutor
00:0-1 | in this video , we're going to talk about how | |
00:02 | to find the indefinite integral . Using trigger metric substitution | |
00:07 | . Now there's three forms that you need to be | |
00:09 | familiar with . The first one is the square root | |
00:13 | of a squared minus X squared . The 2nd 1 | |
00:18 | you want to look for is a squared plus X | |
00:23 | squared inside a square root function . And the last | |
00:26 | one the square root of X squared minus a square | |
00:33 | . Now for the 1st 1 you need to substitute | |
00:36 | X with a sign data where a is a constant | |
00:42 | . And the reason for that is because one minus | |
00:44 | sine squared is close line squared . For the 2nd | |
00:47 | 1 , replace X with a tangent data and the | |
00:52 | reason for that is one plus tan squared is second | |
00:55 | squared And for the last one replace X with a | |
00:59 | sick and data because one 2nd squared -1 is tangent | |
01:06 | squared . Now these are the three forms that you | |
01:10 | need to look out for when using trigger metric substitution | |
01:14 | . So let's work on an example problem . Let's | |
01:17 | say if we have the square root of four minus | |
01:20 | X squared divided by X squared , how can we | |
01:25 | integrate this function ? So notice that we have the | |
01:29 | form square root a square minus X squared . So | |
01:33 | we can clearly see that A squared Is four which | |
01:36 | means that a . Is equal to the square root | |
01:39 | of four or 2 . Therefore we need to replace | |
01:43 | X with a signed data . In this case X | |
01:47 | has to be to sign better . So D . | |
01:50 | X . is going to be the derivative of two | |
01:52 | signed data . So that's to co sign data . | |
01:56 | T theater . And so now we're going to have | |
02:01 | the integral of four minus . Let's replace acts with | |
02:06 | to sign data . So this is gonna be too | |
02:11 | signed peter squared and then divided by X squared . | |
02:15 | Which is also to sign data squared . Now let's | |
02:20 | replace the DX with two co signed data data . | |
02:26 | We can get rid of this for now . So | |
02:37 | now we've got to do some math . So let's | |
02:40 | perform some algebra techniques to simplify this expression two squared | |
02:46 | is four . So two sine squared is going to | |
02:49 | be four sine squared theta . And on the bottom | |
02:52 | we're also going to have four signs square data . | |
03:00 | Now what do you think we need to do at | |
03:01 | this point ? Mhm . At this point We need | |
03:06 | to take out a four inside the square root so | |
03:10 | we can get one minus sine square . So we're | |
03:12 | gonna have the square root of four times one minus | |
03:16 | sine squared theta . Yeah . Yeah . Now something | |
03:23 | else that we can do Is that we can cancel | |
03:28 | to to over four Reduces to 1/2 . So there's | |
03:31 | gonna be two left over on the bottom and we | |
03:34 | still have coast Sine theta . D . Theta . | |
03:38 | Yeah . Now we can take the square root of | |
03:42 | four . The square root of four . It's too | |
03:47 | and then we can replace one minus sine squared with | |
03:50 | co sine squared . So the coastline square part is | |
03:54 | still inside the square root symbol . So now at | |
04:01 | this point we can cancel 22 divided by two is | |
04:04 | one and the square root of coastline square is cool | |
04:08 | . Saint Boehner . So we have co sign and | |
04:11 | this is supposed to be sine squared . Co sign | |
04:15 | over sine squared times . Cosine theta . D . | |
04:18 | Theta and co sign times co sign is Coulson square | |
04:32 | . Now what do you think we should do at | |
04:33 | this point ? The best thing I recommend doing at | |
04:38 | this point is to replace coastline square away from one | |
04:41 | minus sine square Because Science Square plus coastline squared is | |
04:45 | one . Now at this point we can split the | |
04:49 | fraction into two fractions . So we could divide one | |
04:55 | by sine squared and we can divide sine squared by | |
05:00 | itself . Yeah . Now you need to be familiar | |
05:06 | with the reciprocal identities intrigue . one over sine is | |
05:09 | Corsica , so one over sine squared is cosecha squared | |
05:15 | and Science squared divided by sine squared is one . | |
05:18 | So this is what we now have . Now . | |
05:20 | What is the anti derivative of co Seacon square ? | |
05:23 | The derivative of co tangent is a negative cosine squared | |
05:27 | . So the anti derivative of negative cosine squared is | |
05:30 | co tangent . So the anti derivative of positive Corsican | |
05:34 | squared is negative co tangent And the anti derivative of | |
05:40 | -1 d . Theta is going to be negative data | |
05:43 | . And then plus scene . Now this is the | |
05:46 | answer . It's the integral but not with the appropriate | |
05:49 | variables because we started with an X . Variable and | |
05:53 | now we have to change data back into an X | |
05:56 | . Variable . So how can we do that ? | |
06:03 | Now recall ? We said that X Is equal to | |
06:07 | two signed data . So if we divide both sides | |
06:10 | by tune , signed data Is X over two . | |
06:15 | So we can make a right triangle . Now you | |
06:18 | need to be familiar with the principles of Socotra , | |
06:27 | the soul part of sackatoga tells us that sine theta | |
06:30 | is equal to the opposite side , divided by the | |
06:33 | adjacent side . So let's place the angle theta here | |
06:37 | , so opposite to theater is X . That's on | |
06:40 | top divided by the ipod owners . So the hypothesis | |
06:43 | two . Now we've got to find the missing side | |
06:47 | . So whenever you have a right triangle , you | |
06:48 | can use the pythagorean theorem . C squared is equal | |
06:50 | to a squared plus B squared C . Is the | |
06:53 | hypotenuse , which is to we could say is X | |
06:57 | and B is the miss inside that we're looking for | |
07:00 | . So two squared is four . And if we | |
07:02 | subtract both sides by X squared , we're gonna have | |
07:04 | four minus X squared is equal to B squared . | |
07:07 | So the missing side B is going to be for | |
07:10 | the square root of four minus X squared . So | |
07:13 | we can put that here . Now if science data | |
07:24 | is X divided by two , what is tangent data | |
07:27 | based on circle ? Tour , tangent is opposite over | |
07:31 | adjacent , this is opposite , this is adjacent . | |
07:34 | So tangent will be X over the square root of | |
07:39 | four minus X squared . Yeah co tangent is the | |
07:45 | reciprocal of tangent . It's one over tangent . So | |
07:50 | if tangent is X divided by the square root of | |
07:53 | four minus x square . Co tangent , it's gonna | |
07:56 | be the reciprocal of that fraction . So it's gonna | |
07:58 | be the square of four minus X squared over X | |
08:02 | . So now what about feta , what can we | |
08:04 | replace theater with ? Now recall that sine theta is | |
08:13 | X divided by two . So if we take the | |
08:15 | arc sine of both sides , what's going to happen | |
08:19 | ? What is the arc sine of signed data ? | |
08:25 | Well these two expressions will cancel . And so we | |
08:28 | can say that fada is the arc sine of X | |
08:33 | over two . So we have negative data . So | |
08:35 | it's gonna be negative arc sine X divided by two | |
08:39 | and then plus C . So this is the final | |
08:42 | answer . So that's how you can find the indefinite | |
08:47 | integral . Using trig substitution . Now let's work on | |
08:52 | finding the integral of X cube divided by the square | |
08:59 | root of X squared plus nine . So we have | |
09:07 | the form X squared plus a square . Or you | |
09:10 | can write it as a square plus X squared five | |
09:14 | plus three and three plus five is the same . | |
09:17 | So what should we replace acts with If we see | |
09:21 | this particular form in this case we need to use | |
09:24 | the expression X . Is equal to a tangent data | |
09:29 | . So a squared is the same as a nine | |
09:35 | . So if a squared is equal to nine , | |
09:37 | that means A is equal to three . Which means | |
09:41 | we should replace X with three tangent . Theta . | |
09:47 | So let's go ahead and do that . Now let's | |
09:54 | calculate detox . The derivative of tangent is second square | |
09:59 | . So dx is going to be three seconds squared | |
10:01 | theta D fader . So on top we can replace | |
10:05 | XQ with three tangent beta Raised to the 3rd power | |
10:14 | and then X squared . It's going to be three | |
10:18 | tangent beta squared plus nine . And so d access | |
10:23 | three seek and square data detailer . Now three to | |
10:30 | the third power is 27 . So on top we | |
10:32 | have 27 tangent cube . And on the bottom three | |
10:37 | squared is nine . So we're gonna have the square | |
10:39 | root of nine , tangent squared . Theta plus nine | |
10:44 | . And then we still have three seconds squared fed | |
10:47 | to the theater . Now in the denominator . Inside | |
10:58 | the square root Let's take out a nine . So | |
11:01 | we're gonna have the square root of nine . And | |
11:03 | then after we factor out the G C . F | |
11:05 | we're going to be left over with tangent squared plus | |
11:08 | one and everything else . I'm just going to leave | |
11:12 | it the way it is for now . Yeah . | |
11:26 | Now if you recall one plus tangent squared is seeking | |
11:31 | square And the square root of nine is 3 . | |
11:35 | So we now have this expression . So now at | |
11:49 | this point we can cancel three and a square root | |
11:52 | of Seacon square Jessica . So this is what we | |
11:56 | now have 27 tangent cube times . Seek and square | |
12:02 | fattah D . Theater divided by seeking . So now | |
12:07 | at this point we could cancel a second . And | |
12:15 | so now we're left with the integral of 27 tangent | |
12:24 | cube seeking to . So what can we do to | |
12:28 | integrate this expression ? What we have here is a | |
12:33 | trick biometric instagram . And instead of writing tangent cube | |
12:38 | I'm going to replace it with tangent data times . | |
12:42 | Actually tangent squared data times tangent data . Let's write | |
12:47 | it like that . Now we need to perform another | |
12:51 | substitution particularly use substitution at this point . So I'm | |
12:56 | gonna make you equal to tangent beta . And the | |
13:01 | reason why I want to do that is so that | |
13:02 | D . U . Will be actually no that's not | |
13:06 | gonna work . D . You will be seeking squared | |
13:10 | . Mhm . I need to change it up a | |
13:14 | bit . Let's replace tan squared With 2nd squared -1 | |
13:21 | . Because one plus tangent squared is seeking squared . | |
13:27 | Now in this format we can replace you with sick | |
13:30 | and theater so that D . U . The derivative | |
13:34 | of C . Can and will be C . Can't | |
13:37 | tangent . So it's gonna be seeking tangent theta . | |
13:41 | D . Theta so I can replace second with you | |
13:50 | . So let me get rid of this first . | |
13:55 | And so now this is going to be 27 integral | |
13:58 | of U squared minus one . And then tangent C | |
14:05 | can't defeat to is the same as do you ? | |
14:10 | So this becomes to you . You can see that | |
14:12 | here these two expressions are exactly the same . Now | |
14:25 | the anti derivative of U squared , That's going to | |
14:28 | be used to the 3rd power divided by three . | |
14:31 | And the anti derivative of one is simply you . | |
14:35 | And keep in mind we said U . Is equal | |
14:36 | to seek him . So we now have is 27 | |
14:40 | Times seeking to the third power divided by three minus | |
14:44 | C . Can't data . And let's not forget plus | |
14:47 | C . So we could simplify that 27 divided by | |
14:52 | three is nine . So we have nine . Seek | |
14:54 | out to the third power And then if we distribute | |
14:57 | to 27 that's gonna be minus 27 times seek and | |
15:01 | data plus C . Now the last thing we need | |
15:06 | to do is convert that expression replace data with X | |
15:12 | . Somehow . Now the first substitution that we made | |
15:16 | was that acts is equal to three tangent beta Dividing | |
15:20 | both sides by three . Acts over three is tangent | |
15:24 | . So now we can make our right triangle . | |
15:29 | So this is going to be fatal . And here's | |
15:32 | the right angle . Now based on Socotra , tangent | |
15:35 | is opposite over adjacent so opposite to theater is X | |
15:44 | . And adjacent to it right next to his dream | |
15:47 | . So now we gotta find a missing side . | |
15:48 | So using the pythagorean theorem , C squared is a | |
15:51 | squared plus B squared . We could say a stream | |
15:54 | be his ex . So C squared is going to | |
15:58 | be nine plus X squared . And to solve for | |
16:01 | C we gotta take the square root of both sides | |
16:04 | . So the third side of the triangle is the | |
16:07 | square root of nine plus X squared . So now | |
16:11 | we can evaluate second there using a triangle . But | |
16:15 | let's evaluate co sign . The reciprocal of seeking . | |
16:19 | Now based on sackatoga . Co signed data is adjacent | |
16:23 | over hypotenuse adjacent history . So it's gonna be three | |
16:27 | over the square root of nine plus X . Square | |
16:31 | . So sick and theater , which is one divided | |
16:34 | by co sign , is the reciprocal of this fraction | |
16:37 | . That's gonna be the square root of nine plus | |
16:40 | X squared over three . So now what I'm gonna | |
16:51 | do is I'm going to take out 9 2nd Theatre | |
16:56 | . I'm going to factor out that expression . So | |
16:59 | it's nine seconds to times second squared minus stream plus | |
17:07 | scene . Mhm . Now let's replay 2nd with the | |
17:14 | square root of nine plus X . Square over three | |
17:19 | . So seek and squared . We're gonna have to | |
17:23 | square this expression And then that's going to be -1 | |
17:28 | plus E . So all we gotta do is simplify | |
17:32 | what we now have . So nine divided by three | |
17:38 | history . So we have three in front and then | |
17:40 | square root nine plus X squared . Now , once | |
17:45 | we square the square root of nine plus X squared | |
17:48 | , that will simply be nine plus X squared . | |
17:51 | On the bottom . We have three squared . So | |
17:53 | that's gonna be nine . And then Ministry now let's | |
18:03 | get common denominators . So negative three or negative 3/1 | |
18:08 | . I'm going to multiply the top and bottom by | |
18:10 | three . And so inside the bracket I'm gonna have | |
18:17 | nine plus X squared divided by nine . Actually need | |
18:21 | to multiply top and bottom by nine and not three | |
18:25 | . What was I thinking ? So it's going to | |
18:26 | be -3 times nine , which is negative 27 And | |
18:31 | then one times 9 . That's going to be nine | |
18:36 | . I want to turn this into a single fraction | |
18:47 | . Mhm . So I have nine plus X squared | |
18:54 | minus 27 Divided by nine plus scene . Now 3/9 | |
19:01 | that reduces to one third so I can get rid | |
19:04 | of the fraction inside the bracket . So I have | |
19:07 | a one third outside and then square root nine plus | |
19:10 | X squared and then nine minus 27 is negative 18 | |
19:15 | . So I have X squared minus 18 plus C | |
19:20 | . And so this is the final answer for this | |
19:24 | example . |
DESCRIPTION:
This calculus video tutorial provides a basic introduction into trigonometric substitution. It explains when to substitute x with sin, cos, or sec. It also explains how to perform a change of variables using u-substitution integration techniques and how to use right triangle trigonometry with sohcahtoa to convert back from angles in the form of theta to an x variable. There's plenty of examples and practice problems in this lesson.
OVERVIEW:
Trigonometric Substitution is a free educational video by The Organic Chemistry Tutor.
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