0:00:01.010,0:00:03.854
Sometimes and apparently[br]complicated integral can

0:00:03.854,0:00:08.120
be evaluated by first of[br]all making a substitution.

0:00:09.140,0:00:13.573
The effect of this is to change[br]the variable, say from X to

0:00:13.573,0:00:14.596
another variable U.

0:00:15.350,0:00:18.150
It changes the integrand.[br]That's the quantity that

0:00:18.150,0:00:19.550
we're trying to integrate.

0:00:20.620,0:00:23.076
And if we're dealing with[br]definite integrals, those

0:00:23.076,0:00:25.532
with limits than the limits[br]may change too.

0:00:26.730,0:00:29.824
Before we look at our first[br]example, I want to give you a

0:00:29.824,0:00:32.442
preliminary result which will be[br]very useful in all of the

0:00:32.442,0:00:33.632
examples that we look at.

0:00:34.570,0:00:36.790
Suppose we have a variable U.

0:00:37.860,0:00:42.852
Which is a function of X, so you[br]is a function of X.

0:00:43.860,0:00:46.060
Suppose we can[br]differentiate this

0:00:46.060,0:00:49.140
function of X and[br]workout du DX.

0:00:53.590,0:00:56.090
Then a quantity[br]called the

0:00:56.090,0:00:59.590
differential du is[br]given by du DX.

0:01:02.780,0:01:07.114
Multiply by DX and this will be[br]a particularly important result

0:01:07.114,0:01:12.630
in all of the examples that we[br]do. So, for example, if I had

0:01:12.630,0:01:17.358
you equals 1 - 2 X, so there's[br]my function of X.

0:01:18.280,0:01:21.040
We can differentiate this du DX.

0:01:24.660,0:01:27.810
Well, in this case du[br]DX will be minus 2.

0:01:30.350,0:01:34.715
Then the differential du is[br]given by du DX.

0:01:35.350,0:01:39.360
Minus 2 multiplied by[br]DX.

0:01:41.800,0:01:46.300
So this formula will be very[br]useful in all the examples that

0:01:46.300,0:01:51.175
we do. Let's have a look at our[br]first example of integration by

0:01:51.175,0:01:54.175
substitution. Suppose we want to[br]perform this integration.

0:01:54.175,0:01:57.925
Suppose we want to find the[br]integral of X +4.

0:02:01.120,0:02:05.878
All raised to the power 5 and we[br]want to integrate this with

0:02:05.878,0:02:06.976
respect to X.

0:02:10.100,0:02:13.532
Now the problem here in this[br]example is the X +4.

0:02:14.270,0:02:17.240
This is the complicated bit[br]and the reason why it's

0:02:17.240,0:02:20.210
complicated is because if this[br]was just a single variable,

0:02:20.210,0:02:24.071
say X would be integrating X[br]to the power 5, and we all

0:02:24.071,0:02:27.635
know how to do that already.[br]So this is the problem. We'd

0:02:27.635,0:02:31.199
like this to be a single[br]variable, and so we make a

0:02:31.199,0:02:33.872
substitution and let you be[br]this quantity X +4.

0:02:39.080,0:02:42.995
Let's see what happens when we[br]make this substitution.

0:02:44.110,0:02:49.076
This integral is going to become[br]the integral X +4 is going to

0:02:49.076,0:02:50.604
become just simply you.

0:02:52.510,0:02:54.365
We've got you to the power 5.

0:02:55.880,0:03:00.374
And then we have to be a little[br]bit careful because we've got to

0:03:00.374,0:03:04.226
take care of this quantity an[br]appropriate way as well. Now we

0:03:04.226,0:03:08.078
do that using the results I've[br]just given you. We workout the

0:03:08.078,0:03:10.967
differential du. Now remember[br]that du is du DX.

0:03:12.700,0:03:14.338
Multiplied by DX.

0:03:16.550,0:03:20.390
In this case, du DX is[br]just one. If you equals X

0:03:20.390,0:03:23.910
+4 du DX, the derivative[br]of you will be just one.

0:03:25.960,0:03:29.652
So in this first example with[br]a nice simple case in which du

0:03:29.652,0:03:33.912
is just equal to 1, DX or du[br]is equal to DX. So when we

0:03:33.912,0:03:36.468
make the substitution of, the[br]DX can become simply.

0:03:38.030,0:03:38.450
See you.

0:03:39.640,0:03:43.012
Now this is a much simpler[br]integral than the one we started

0:03:43.012,0:03:47.227
with and will be able to finish[br]it off in the usual way just by

0:03:47.227,0:03:50.880
increasing the power by one, so[br]will get you to the Palace 6.

0:03:51.760,0:03:53.608
We divide by the new power.

0:03:54.300,0:03:56.720
And we don't forget to add the[br]constant of integration.

0:03:59.100,0:04:01.908
Now, for all intents and[br]purposes, that's the problem

0:04:01.908,0:04:05.028
solved. However, it's normal to[br]go back to our original

0:04:05.028,0:04:07.212
variables. Remember, we[br]introduced this new variable

0:04:07.212,0:04:10.644
you, so we go back to the[br]variable of the original

0:04:10.644,0:04:12.204
problem, which was X. Remember

0:04:12.204,0:04:14.970
that you. Is equal to X +4.

0:04:16.330,0:04:18.760
So replacing you by X +4.

0:04:21.080,0:04:26.428
We have X +4 to the power[br]six or divided by 6 plus a

0:04:26.428,0:04:27.574
constant of integration.

0:04:29.340,0:04:33.309
And that's our first example of[br]integration by substitution.

0:04:37.110,0:04:39.740
Let's have a look at another[br]example. Suppose this time

0:04:39.740,0:04:42.107
that we want to integrate a[br]trigonometric function and

0:04:42.107,0:04:44.737
the one I'm going to look at[br]is the cosine.

0:04:46.240,0:04:48.048
Of three X +4.

0:04:50.800,0:04:52.276
And we want to integrate this.

0:04:53.820,0:04:54.968
With respect to X.

0:04:55.930,0:04:59.450
I'm going to assume that we[br]already know how to integrate

0:04:59.450,0:05:03.290
cosine X Sign X, the standard[br]trig functions. This is a bit

0:05:03.290,0:05:06.810
more complicated, and the reason[br]it's more complicated is the 3X

0:05:06.810,0:05:08.090
plus four in here.

0:05:09.030,0:05:13.430
So as before, we make a[br]substitution to change this

0:05:13.430,0:05:18.270
into just a single variable,[br]so I'm going to write you

0:05:18.270,0:05:20.030
equals 3X plus 4.

0:05:22.230,0:05:26.000
The effect of doing that is to[br]change the integral to this one.

0:05:26.000,0:05:27.450
The integral of the cosine.

0:05:28.980,0:05:32.130
And instead of 3X plus[br]four, we're now going to

0:05:32.130,0:05:34.965
have just simply. You are[br]much simpler problem than

0:05:34.965,0:05:36.540
the one we started with.

0:05:38.220,0:05:41.542
As before, we need to take[br]particular care with this term.

0:05:41.542,0:05:46.380
Here the DX. And we do that with[br]our standard results that du.

0:05:48.710,0:05:50.680
Is equal to du DX.

0:05:51.490,0:05:52.180
DX

0:05:53.870,0:05:57.379
now what's du DX? In this[br]example? Well, you equals 3X

0:05:57.379,0:06:00.250
plus four, and if we[br]differentiate this function with

0:06:00.250,0:06:03.759
respect to X, will get that du[br]DX is just three.

0:06:05.000,0:06:10.656
So our differential du is simply[br]3D XDU is 3D X if we just

0:06:10.656,0:06:15.504
rearrange this will be able to[br]replace the DX by something in

0:06:15.504,0:06:21.564
terms of du over here. So if du[br]is 3 DX then dividing both sides

0:06:21.564,0:06:25.604
by three, we can write down that[br]DX is 1/3.

0:06:27.170,0:06:27.960
Of du

0:06:29.660,0:06:34.524
so the DX in here is replaced by[br]1/3 of du. Now I'll put the du

0:06:34.524,0:06:38.476
there and the factor of 1/3 can[br]be put outside the integral sign

0:06:38.476,0:06:41.820
like that. A constant factor can[br]immediately be moved outside the

0:06:41.820,0:06:45.906
integral. Now this is very[br]straightforward to finish. We

0:06:45.906,0:06:50.370
all know that the integral of[br]the cosine of you with respect

0:06:50.370,0:06:52.230
to you is sign you.

0:06:54.040,0:06:55.440
Plus a constant of integration.

0:06:57.620,0:07:02.317
And as before, we revert to our[br]original variables using the

0:07:02.317,0:07:07.014
given substitution. You was 3X[br]plus four, so this becomes 1/3

0:07:07.014,0:07:10.003
of the sign of three X +4.

0:07:11.630,0:07:13.230
Plus a constant of integration.

0:07:15.220,0:07:19.170
And that is our second example[br]of integration by substitution.

0:07:20.630,0:07:24.582
Now before I go on, I want to[br]generalize this a little bit.

0:07:24.582,0:07:27.926
This particular example we had[br]the cosine of a constant times

0:07:27.926,0:07:30.966
X plus another constant and[br]let's just generalize that so

0:07:30.966,0:07:34.006
we can deal with any[br]situations where we have a

0:07:34.006,0:07:37.350
cosine of a constant times X[br]plus a constant. So let's

0:07:37.350,0:07:39.174
suppose we look at this[br]example.

0:07:41.140,0:07:45.716
Suppose we want to integrate the[br]cosine of AX plus B.

0:07:47.690,0:07:49.158
With respect to X.

0:07:50.160,0:07:52.868
When A&B are constants.

0:07:59.930,0:08:03.188
Again, the problem is the[br]quantity in the brackets

0:08:03.188,0:08:06.446
here, and we avoid the[br]problem by making a

0:08:06.446,0:08:10.428
substitution. We let you be[br]all this quantity X plus B.

0:08:14.140,0:08:19.780
That changes the integral to the[br]integral of the cosine and X

0:08:19.780,0:08:22.130
Plus B becomes simply you.

0:08:24.540,0:08:28.193
We've got to deal with the DX,[br]and we do that by calculating

0:08:28.193,0:08:29.036
the Differentials Du.

0:08:30.250,0:08:34.270
Du will be du DX, which in this[br]case is simply A.

0:08:35.860,0:08:36.590
DX

0:08:38.730,0:08:42.902
and this result will allow us to[br]replace the DX in here or DX

0:08:42.902,0:08:44.690
will be one over a du.

0:08:50.570,0:08:55.512
So the DX here will become one[br]over a DUI, right? The du there

0:08:55.512,0:08:59.042
under one over a being a[br]constant factor. I'll bring

0:08:59.042,0:09:00.101
straight outside here.

0:09:02.770,0:09:05.666
This is straightforward to[br]finish off because the

0:09:05.666,0:09:10.372
integral of the cosine you is[br]just the sign of you plus a

0:09:10.372,0:09:11.458
constant of integration.

0:09:14.310,0:09:17.970
In terms of our original[br]variables, will have sign and

0:09:17.970,0:09:20.166
you is a X plus B.

0:09:24.340,0:09:25.770
Plus a constant of integration.

0:09:27.810,0:09:31.038
And that's a general result.[br]We can use any time that we

0:09:31.038,0:09:33.997
have to integrate the cosine[br]of a quantity like this. This

0:09:33.997,0:09:37.225
is an example of a linear[br]function X plus be, so we've

0:09:37.225,0:09:40.184
got the cosine of a linear[br]linear function, and we can

0:09:40.184,0:09:43.143
use this results anytime we[br]want to integrate it. So for

0:09:43.143,0:09:45.833
example, if I write down,[br]what's the integral of the

0:09:45.833,0:09:47.178
cosine of Seven X +3?

0:09:49.460,0:09:52.583
Then immediately we recognize[br]that the A is 7.

0:09:53.830,0:09:57.964
The B is 3 and we use this[br]general result to state that

0:09:57.964,0:09:59.872
this integral is one over A.

0:10:00.110,0:10:01.538
Which will be one over 7.

0:10:02.980,0:10:04.080
The sign.

0:10:05.200,0:10:08.428
Of the original content in the[br]brackets, which was Seven X +3.

0:10:10.720,0:10:13.690
Plus a constant of[br]integration. So anytime we get

0:10:13.690,0:10:17.320
a function to integrate, which[br]is the cosine of a linear

0:10:17.320,0:10:19.300
function, we can use this[br]result.

0:10:20.690,0:10:22.898
There's a very similar[br]results for integrating sign,

0:10:22.898,0:10:26.486
and I won't prove it. I'll[br]leave it for you to to prove

0:10:26.486,0:10:29.522
for yourself, but the result[br]is this one that the integral

0:10:29.522,0:10:30.350
of the sign.

0:10:31.900,0:10:32.920
AX plus B.

0:10:35.990,0:10:39.630
With respect to X will be minus[br]one over A.

0:10:40.290,0:10:43.360
Cosine of X plus B.

0:10:44.680,0:10:46.556
Plus a constant of[br]integration, and that's

0:10:46.556,0:10:49.236
another standard result you[br]should be aware of, and you

0:10:49.236,0:10:51.380
should now be able to prove[br]for yourself.

0:10:52.920,0:10:53.450
OK.

0:10:58.040,0:10:59.324
Let's have a look at another

0:10:59.324,0:11:06.336
example. Suppose we want to find[br]the integral of 1 / 1 - 2 X.

0:11:07.040,0:11:08.248
With respect to X.

0:11:10.330,0:11:13.750
Now I'm going to assume that[br]if you had a single letter

0:11:13.750,0:11:17.170
down here, if they've just[br]been an X down here, so we

0:11:17.170,0:11:20.590
we're integrating one over X,[br]you'd know how to do that just

0:11:20.590,0:11:24.295
by the. By that the integral[br]of one over X with respect to

0:11:24.295,0:11:28.000
X is the natural logarithm of[br]the modulus of X. So I'm going

0:11:28.000,0:11:30.850
to assume that you know how to[br]do that already.

0:11:32.440,0:11:36.376
So if we can convert this[br]integral here into one where we

0:11:36.376,0:11:38.344
just got one over a single

0:11:38.344,0:11:43.080
variable. Perhaps we will have[br]to proceed, so will make a

0:11:43.080,0:11:45.976
substitution, and the[br]substitution will make is U

0:11:45.976,0:11:47.786
equals 1 - 2 X.

0:11:49.830,0:11:53.650
What will that do to the[br]integral? Well, the integral

0:11:53.650,0:11:56.706
will become the integral of[br]1 divided by.

0:11:58.020,0:12:03.870
1 - 2 X will become you much[br]simpler. We have to take care of

0:12:03.870,0:12:06.210
the DX in an appropriate way.

0:12:07.430,0:12:12.536
As before DU is[br]du DX.

0:12:14.380,0:12:15.868
Multiply by DX.

0:12:17.650,0:12:19.726
What's du DX in this case?

0:12:22.150,0:12:24.047
Well, you is 1 - 2 X.

0:12:24.630,0:12:28.998
So do you DX the derivative of[br]this quantity is just minus 2.

0:12:31.610,0:12:32.320
DX

0:12:33.820,0:12:38.560
so when we want to replace the[br]DX in the substitution process,

0:12:38.560,0:12:40.930
we can replace DX by du.

0:12:43.230,0:12:45.980
Divided by minus two, which is[br]minus 1/2 of DS.

0:12:47.560,0:12:52.735
So the DX here becomes du and[br]the factor of a minus 1/2. I can

0:12:52.735,0:12:55.840
right outside being a constant[br]factor. Now this is

0:12:55.840,0:12:59.290
straightforward to finish off[br]because the integral of one over

0:12:59.290,0:13:01.705
you do. You will be just the

0:13:01.705,0:13:03.860
logarithm. Of the[br]modulus of you.

0:13:05.820,0:13:11.507
So we've minus half natural[br]logarithm of the modulus of you.

0:13:12.700,0:13:14.190
Plus a constant of integration.

0:13:16.690,0:13:20.842
Nearly finished or we do is go[br]back to our original variables.

0:13:20.842,0:13:25.340
Now. Remember that the original[br]variable was U equals 1 - 2 X,

0:13:25.340,0:13:29.146
so this will become minus 1/2[br]natural logarithm of the modulus

0:13:29.146,0:13:30.876
of 1 - 2 X.

0:13:31.620,0:13:33.150
Plus a constant of integration.

0:13:34.610,0:13:35.678
And that's another example.

0:13:37.610,0:13:40.718
It's very easy to generalize[br]this to any cases of the form

0:13:40.718,0:13:44.085
where you have one divided by a[br]linear function of X. Let's just

0:13:44.085,0:13:47.711
see how we can do that, and then[br]we'll get a very useful general

0:13:47.711,0:13:53.820
result. Suppose we want to[br]integrate 1 divided by a X

0:13:53.820,0:13:54.790
plus B.

0:13:56.150,0:13:58.158
We want to integrate[br]with respect to X.

0:13:59.180,0:14:04.240
The substitution we make is U[br]equals a X plus B.

0:14:06.990,0:14:07.790
Do you?

0:14:09.330,0:14:13.280
Will be du DX, which in this[br]case differentiating this

0:14:13.280,0:14:18.020
function will just leave us[br]the constant a do you will be

0:14:18.020,0:14:23.550
equal to a times DX. So when[br]we want to replace the DX we

0:14:23.550,0:14:27.895
will do so by replacing it[br]with one over a du.

0:14:29.360,0:14:33.960
What will that do to our[br]integral? Well, the integral

0:14:33.960,0:14:37.640
will become the integral of[br]1 divided by.

0:14:39.050,0:14:41.700
AX plus B becomes you.

0:14:43.840,0:14:48.169
And the DX becomes one over a[br]du. I'll write the du here

0:14:48.169,0:14:51.832
and the one over a being a[br]constant factor. Uh, bring

0:14:51.832,0:14:52.498
straight outside.

0:14:55.870,0:14:58.710
Now this is very familiar,[br]straightforward for us to finish

0:14:58.710,0:15:00.698
this the integral of one over U.

0:15:02.500,0:15:07.336
Will be the natural logarithm of[br]the modulus of you plus a

0:15:07.336,0:15:08.545
constant of integration.

0:15:10.800,0:15:15.280
Nearly finished, we just return[br]to our original variables, one

0:15:15.280,0:15:20.656
over a natural logarithm of the[br]modulus U was X plus B.

0:15:24.480,0:15:25.870
Plus a constant of integration.

0:15:28.420,0:15:31.012
Now, this is a particularly[br]important general results, which

0:15:31.012,0:15:34.468
I'd like to get very familiar[br]with because it's going to crop

0:15:34.468,0:15:38.500
up over and over again, and you[br]want this sort of thing at your

0:15:38.500,0:15:41.956
fingertips. For example, if I[br]ask you to integrate one over X

0:15:41.956,0:15:45.700
plus one with respect to XI,[br]want you to be able to almost

0:15:45.700,0:15:49.444
just write these things down if[br]we identify the one over X plus

0:15:49.444,0:15:53.764
one with the one over X Plus B,[br]then we see that a is one.

0:15:54.570,0:15:55.570
And B is one.

0:15:56.860,0:16:02.185
Got a is one and be as one. This[br]general result will give us one

0:16:02.185,0:16:06.090
over one which is just one[br]natural logarithm of the modulus

0:16:06.090,0:16:10.350
of X Plus B, which in this case[br]is X plus one.

0:16:14.280,0:16:18.063
So our integral of one over X[br]Plus One is just the logarithm

0:16:18.063,0:16:22.672
of the denominator. Another[br]example, the integral of one

0:16:22.672,0:16:26.368
over 3X minus two with[br]respect to X.

0:16:27.850,0:16:31.282
Well, using our general result[br]we see that a is 3.

0:16:33.720,0:16:38.140
Be is minus two. Put all these[br]into the formula and we'll get

0:16:38.140,0:16:39.160
one over A.

0:16:39.800,0:16:42.212
Play being 3 means[br]we get one over 3.

0:16:43.740,0:16:48.052
Natural logarithm. Of the[br]modulus of the quantity we

0:16:48.052,0:16:50.124
started with, which was 3X[br]minus 2.

0:16:52.830,0:16:55.910
Let's see, as I say, this is[br]a particularly important

0:16:55.910,0:16:58.990
result and you should have it[br]at your fingertips. You'll

0:16:58.990,0:17:02.686
need it over and over again[br]in lots of situations as you

0:17:02.686,0:17:03.302
perform integration.

0:17:05.180,0:17:09.151
All the examples that we've[br]looked at so far have been

0:17:09.151,0:17:12.039
examples of indefinite[br]integrals. They've not been any

0:17:12.039,0:17:15.649
limits on the integral[br]whatsoever, so let's now have a

0:17:15.649,0:17:19.981
look at how we do an integration[br]by substitution when there are

0:17:19.981,0:17:24.674
limits on the integral as well,[br]and the example that I want to

0:17:24.674,0:17:29.367
look at is this one. Suppose we[br]want to integrate from X equals

0:17:29.367,0:17:32.255
1 to X equals 3, the function 9

0:17:32.255,0:17:37.228
plus X. All raised to the power[br]two and we want to integrate

0:17:37.228,0:17:38.768
that with respect to X.

0:17:41.650,0:17:46.660
Now have this just being an X to[br]the power two. There would be no

0:17:46.660,0:17:50.668
problem. Increase the power by[br]one divided by the new power and

0:17:50.668,0:17:55.344
you be finished. The problem is[br]the 9 plus X. So as before we

0:17:55.344,0:17:59.018
make a substitution, we make a[br]substitution to try to simplify

0:17:59.018,0:18:02.692
what we're doing and the[br]substitution we make is U is

0:18:02.692,0:18:04.362
equal to 9 plus X.

0:18:07.580,0:18:09.708
Let's see what that[br]does to the integral.

0:18:13.030,0:18:17.450
Our integral will become, well,[br]the 9 plus X becomes U so will

0:18:17.450,0:18:18.470
have you squared.

0:18:21.490,0:18:27.626
Let's deal next with the DX, as[br]we've seen already, do you is

0:18:27.626,0:18:29.514
equal to du DX?

0:18:31.670,0:18:35.311
Multiplied by DX and in this[br]particular case, this is a

0:18:35.311,0:18:38.290
nice simple case. Du DX will[br]be just one.

0:18:41.860,0:18:45.765
So whenever we see a DX, we[br]can replace it immediately

0:18:45.765,0:18:49.670
by du to USD X, so this DX[br]becomes a deal.

0:18:51.790,0:18:55.417
So far it's the same as before,[br]but now we've got to deal

0:18:55.417,0:18:57.649
specifically with these limits.[br]We have these limits.

0:18:59.260,0:19:02.874
A lower limit of one, an upper[br]limit of three, and these are

0:19:02.874,0:19:04.264
limits on the variable X.

0:19:05.170,0:19:10.149
We're integrating from X equals[br]1 to X equals 3. When we change

0:19:10.149,0:19:13.885
the variable. It's very[br]important that we changed the

0:19:13.885,0:19:18.575
limits as well, and the way we[br]do that is we use the given

0:19:18.575,0:19:22.260
substitution. We use this to get[br]new limits limits on you.

0:19:23.690,0:19:26.080
Now when X is one.

0:19:27.010,0:19:30.420
The lower limit you is going to[br]be 9 + 1.

0:19:31.540,0:19:34.900
10 so when X is one US 10.

0:19:37.560,0:19:39.808
At the upper limit, when X is 3.

0:19:41.830,0:19:45.016
Our substitution tells[br]us that you is going to

0:19:45.016,0:19:47.494
be 9 + 3, which is 12.

0:19:49.280,0:19:52.439
We're going to use the[br]substitution to change from

0:19:52.439,0:19:54.896
limits on X to limits on you.

0:19:56.640,0:19:59.335
So these limits on you are now

0:19:59.335,0:20:02.922
you Ekwe. Tools 10 to you

0:20:02.922,0:20:06.500
equals 12. And I've[br]explicitly written the

0:20:06.500,0:20:10.340
variable in here so that we[br]know we've changed from X to

0:20:10.340,0:20:10.660
you.

0:20:12.410,0:20:15.632
This is straightforward to[br]finish off because the integral

0:20:15.632,0:20:20.286
of you squared is you cubed over[br]3 plus, plus no constant of

0:20:20.286,0:20:23.508
integration because it's a[br]definite integral and we write

0:20:23.508,0:20:24.940
square brackets around here.

0:20:26.180,0:20:28.980
And we write the limits[br]on the right hand side.

0:20:31.770,0:20:36.658
We finish this off by putting[br]the upper limit in first, so you

0:20:36.658,0:20:38.914
being 12 we want 12 cubed.

0:20:39.740,0:20:40.928
Divided by three.

0:20:41.990,0:20:45.182
That's the upper limit gone[br]in. We want to put the lower

0:20:45.182,0:20:47.842
limit in you being 10, so we[br]want 10 cubed.

0:20:49.020,0:20:52.850
Divided by three and as[br]usual, we find the difference

0:20:52.850,0:20:54.382
of these two quantities.

0:20:56.630,0:20:58.058
Now 12 cubed.

0:20:58.780,0:21:00.988
Get your Calculator out[br]12 cubed.

0:21:03.800,0:21:11.504
Is 1728 so we have 1728[br]/ 3 subtract 10 cubed which

0:21:11.504,0:21:17.464
is 1000. Divided by three,[br]so our final answer is

0:21:17.464,0:21:19.356
going to be 728.

0:21:21.440,0:21:22.520
Divided by three.

0:21:24.390,0:21:28.090
And that's our first example of[br]a definite integral using

0:21:28.090,0:21:31.420
integration by substitution.[br]Always remember that you use the

0:21:31.420,0:21:35.490
given substitution to change the[br]limits on the integral as well.

0:21:35.490,0:21:37.340
Don't forget to do that.

0:21:42.640,0:21:46.312
OK, all the examples that we've[br]looked at so far have been

0:21:46.312,0:21:48.454
fairly straightforward, and[br]they've required a substitution

0:21:48.454,0:21:52.126
of the form you equals X plus be[br]a linear substitution. We're

0:21:52.126,0:21:54.574
going to now have a look at some

0:21:54.574,0:21:57.902
more complicated ones. All[br]the examples I'm going to

0:21:57.902,0:21:59.444
look at or of this form.

0:22:00.830,0:22:06.526
So all going to take the form of[br]the integral of F of G of X.

0:22:08.130,0:22:09.558
G dash devex.

0:22:11.370,0:22:14.672
DX now this looks horrendous. So[br]what will do is will try and

0:22:14.672,0:22:17.720
take it apart a little bit to[br]see what's going on here.

0:22:19.270,0:22:24.158
Suppose we have a function[br]G of X which is equal to 1

0:22:24.158,0:22:25.286
plus X squared.

0:22:28.780,0:22:31.730
That's going to be this[br]function in here, and you'll

0:22:31.730,0:22:35.270
notice from this expression[br]that the G of X is used as

0:22:35.270,0:22:36.745
input to another function F.

0:22:37.840,0:22:39.628
Suppose the function F is the

0:22:39.628,0:22:42.605
square root function. The[br]function which takes the

0:22:42.605,0:22:45.938
square root of the input.[br]So I'm going to write that

0:22:45.938,0:22:49.574
like this. Suppose F of you[br]is the square root of you.

0:22:51.430,0:22:56.170
Then what does this mean? F of G[br]of X? Well, this is called a

0:22:56.170,0:22:59.014
composite function or the[br]composition of the functions F&G

0:22:59.014,0:23:04.070
the Function G is used as the[br]input to F, so F of G of X.

0:23:05.710,0:23:08.834
Well, the F function square[br]roots the input, so we want

0:23:08.834,0:23:12.526
to square root the input[br]which is G of X, which is one

0:23:12.526,0:23:13.378
plus X squared.

0:23:15.950,0:23:18.560
So in this particular case,[br]we're looking at an example

0:23:18.560,0:23:21.692
that's going to be something[br]like this, but the integral of F

0:23:21.692,0:23:22.736
of G of X.

0:23:23.940,0:23:27.765
Is the square root[br]of 1 plus X squared.

0:23:29.520,0:23:34.070
What about the G dash of XG dash[br]trivex is the derivative of G

0:23:34.070,0:23:35.370
with respect to X.

0:23:36.230,0:23:40.689
Well, his GG of X is one plus X[br]squared. What's its derivative?

0:23:41.820,0:23:43.750
The derivative G Dash Devex.

0:23:46.230,0:23:47.898
Is just 2X.

0:23:49.680,0:23:53.554
So if I substitute for G, Dash,[br]Devex as 2X, I'll be dealing

0:23:53.554,0:23:55.044
with an integral like that.

0:23:56.920,0:24:01.041
So as a first example, we're[br]going to look at an integral of

0:24:01.041,0:24:05.162
this particular form, and I hope[br]that you can now see it's of

0:24:05.162,0:24:09.283
this family of integrals. We've[br]got a function G of X in here,

0:24:09.283,0:24:12.770
one plus X squared, that's that,[br]bit its derivative G dash,

0:24:12.770,0:24:16.891
devex. The two X appears out[br]here and the G itself is input

0:24:16.891,0:24:20.061
to another function, which is[br]the square routing function F.

0:24:20.910,0:24:23.560
So it looks a bit[br]complicated, but I hope we

0:24:23.560,0:24:26.210
can see what all the all the[br]different ingredients are.

0:24:27.220,0:24:32.722
Now, the way we tackle a problem[br]like this is to always make the

0:24:32.722,0:24:37.438
substitution you equals G of X[br]whatever the G of X was.

0:24:38.440,0:24:42.710
Well, in this particular case, G[br]of X was one plus X squared, so

0:24:42.710,0:24:45.760
I'm going to make this[br]substitution. You equals 1 plus

0:24:45.760,0:24:49.396
X squared. Let's see what that[br]will do to this integral.

0:24:52.940,0:24:55.822
Letting you be one plus X[br]squared will just have the

0:24:55.822,0:24:57.394
square root of you in here.

0:25:01.710,0:25:06.416
Now we've got it handled all[br]this. The two X DX. Well in

0:25:06.416,0:25:10.760
terms of Differentials, we know[br]already that du is du DX DX.

0:25:10.760,0:25:13.656
Well do you DX in this case is

0:25:13.656,0:25:17.870
just 2X. So do you. DX DX[br]is 2X DX.

0:25:19.230,0:25:22.180
Now this is very fortunate[br]because we see that the

0:25:22.180,0:25:23.655
whole of two X DX.

0:25:24.900,0:25:26.208
Can be replaced by.

0:25:27.380,0:25:31.488
Do you so the whole of two X[br]DX becomes the du and

0:25:31.488,0:25:34.648
integrals of this form? This[br]will always happen when we

0:25:34.648,0:25:37.492
make a substitution like[br]this. Now this is very

0:25:37.492,0:25:40.020
straightforward to finish,[br]provided you know a little

0:25:40.020,0:25:43.812
bit of algebra. The square[br]root of U is due to the

0:25:43.812,0:25:46.656
power half, so we're[br]integrating you to the half

0:25:46.656,0:25:47.920
with respect to you.

0:25:49.190,0:25:50.320
How do we do this?

0:25:51.660,0:25:53.400
We increase the power by one.

0:25:54.660,0:25:57.740
That will give us you to[br]the half plus one that's

0:25:57.740,0:25:59.420
1 1/2 or three over 2.

0:26:01.270,0:26:05.288
And we divide by the gnu power[br]and the new power. In this case

0:26:05.288,0:26:09.593
is 3 over 2, so we're dividing[br]by three over 2, and we add a

0:26:09.593,0:26:10.454
constant of integration.

0:26:12.310,0:26:16.434
Nearly finished. Dividing by[br]three over 2 is like multiplying

0:26:16.434,0:26:18.625
by 2/3, so I'll write that as

0:26:18.625,0:26:21.485
2/3. And in terms of our

0:26:21.485,0:26:25.220
original variable. You was one[br]plus X squared.

0:26:28.180,0:26:30.988
And all that needs to be raised[br]to the power three over 2.

0:26:32.730,0:26:34.319
And we need a constant of[br]integration.

0:26:37.340,0:26:40.370
So that's the first example of[br]a much more complicated

0:26:40.370,0:26:44.006
integral, and I want to go[br]back and just point a few

0:26:44.006,0:26:46.733
things out. Remember, a[br]crucial step was to recognize

0:26:46.733,0:26:50.066
that the 2X in here was the[br]derivative of this quantity,

0:26:50.066,0:26:51.884
one plus X squared in there.

0:26:54.019,0:26:58.051
The G dash of X in the general[br]case is the derivative of the

0:26:58.051,0:26:59.779
input to the F function here.

0:27:01.009,0:27:03.409
What will do is will have a[br]look at another example and

0:27:03.409,0:27:05.809
then we'll try and start to do[br]some of this purely by

0:27:05.809,0:27:08.409
inspection because at the end[br]of the day we want you to be

0:27:08.409,0:27:10.609
able to get to the stage where[br]you're happy, almost just

0:27:10.609,0:27:11.609
recognizing some of these[br]integrals.

0:27:17.789,0:27:20.469
OK, let's have another[br]look at another example

0:27:20.469,0:27:23.149
of one of these[br]complicated things. F. Of

0:27:23.149,0:27:24.154
G. Of X.

0:27:26.099,0:27:27.407
G dash devex.

0:27:30.919,0:27:36.679
And in this case I'm going to[br]take as my G of X function. The

0:27:36.679,0:27:41.671
following G of X is going to be[br]2 X squared plus one.

0:27:44.939,0:27:48.929
This is going to be used as[br]input to the F function, and in

0:27:48.929,0:27:52.919
this case I'm going to choose F[br]as this function F of you is

0:27:52.919,0:27:56.339
going to be the function which[br]takes an input, finds it square

0:27:56.339,0:27:59.759
roots, and then finds its[br]reciprocal. So F of you is one

0:27:59.759,0:28:01.469
over the square root of you.

0:28:03.859,0:28:07.079
What would that look like in[br]terms of this integral?

0:28:07.079,0:28:10.299
Well, this integral will be[br]F of G of X.

0:28:11.899,0:28:16.043
All this means that the G of X[br]function is used as input to

0:28:16.043,0:28:20.187
F. So if we use two X squared[br]as one as input to this

0:28:20.187,0:28:24.035
function, F will get one over[br]the square root of 2 X squared

0:28:24.035,0:28:26.403
plus one. So I'll be[br]integrating something like

0:28:26.403,0:28:26.699
this.

0:28:34.449,0:28:36.689
We also need to look[br]at the G Dash Devex.

0:28:38.189,0:28:42.869
Well, because G of X is[br]2 X squared plus one, if

0:28:42.869,0:28:45.989
we differentiate GG[br]dash, devex will be just

0:28:45.989,0:28:47.939
simply to choose of 4X.

0:28:49.929,0:28:53.369
So for G Dash Devex, I'm going[br]to write 4X.

0:28:54.519,0:28:55.487
And with the DX.

0:28:57.049,0:29:00.052
Look in this example. It's an[br]integral like this now this

0:29:00.052,0:29:03.055
looks very complicated. It might[br]have actually been given to you

0:29:03.055,0:29:06.331
in a form that looked more like[br]this because we might have

0:29:06.331,0:29:10.153
written the 1 * 4 X all as the[br]single term. We might have

0:29:10.153,0:29:13.702
actually written it down like 4X[br]divided by the square root of 2

0:29:13.702,0:29:16.432
X squared at one, all integrated[br]with respect to X.

0:29:18.809,0:29:21.955
Just look a bit complicated, but[br]we'll see by making a

0:29:21.955,0:29:24.529
substitution that we can make[br]some progress and the

0:29:24.529,0:29:27.961
substitution that will make as[br]before is we let you be the

0:29:27.961,0:29:29.105
function G of X.

0:29:30.219,0:29:32.307
We let you be this[br]function in here.

0:29:33.499,0:29:34.549
So if you.

0:29:35.979,0:29:39.464
Is G of X or G of X? In this[br]case was two X squared plus one.

0:29:42.199,0:29:43.427
What's going to happen?

0:29:44.169,0:29:46.297
Well, let's put all this[br]into this integral.

0:29:47.559,0:29:51.387
Don't worry about the 4X for[br]a minute, but the two X

0:29:51.387,0:29:54.896
squared plus one here in the[br]denominator becomes a U, so

0:29:54.896,0:29:58.086
we'll have a square root of[br]you in the denominator.

0:29:59.529,0:30:04.482
We've got to. Look after the[br]four X DX as well, but in terms

0:30:04.482,0:30:09.071
of differentials do you remember[br]is du DX, which is 4X times DX

0:30:09.071,0:30:13.307
and you'll see because of the[br]nature of this sort of problem

0:30:13.307,0:30:17.896
that the four X DX quantity is[br]exactly what we've got over here

0:30:17.896,0:30:23.544
for X DX. So the whole of four X[br]DX becomes du the whole of that.

0:30:25.029,0:30:25.977
Becomes a du.

0:30:27.549,0:30:29.988
That's the nice thing about[br]problems like this. That

0:30:29.988,0:30:32.698
substitution will always make[br]them drop out in this way.

0:30:34.539,0:30:38.007
Again, provided that you know a[br]bit of algebra, you can finish

0:30:38.007,0:30:42.053
this off. Let's just do that.[br]The square root of U is due to

0:30:42.053,0:30:46.619
the power half. One over the[br]square root will give us you to

0:30:46.619,0:30:49.864
the minus 1/2. So the problem[br]that we're integrating here is

0:30:49.864,0:30:51.339
you to the minus 1/2.

0:30:52.579,0:30:53.699
With respect to you.

0:30:56.409,0:31:00.069
You to the minus 1/2 integrated[br]we increase the power by one.

0:31:00.779,0:31:05.015
So we increase minus 1/2 by one[br]will give us plus 1/2.

0:31:07.229,0:31:09.287
And we divide by the new power.

0:31:10.979,0:31:14.367
And we need a constant of[br]integration and for all intents

0:31:14.367,0:31:17.447
and purposes that's finished,[br]but we can revert to our

0:31:17.447,0:31:20.527
original variable X through the[br]substitution. What did we have?

0:31:21.419,0:31:22.448
We had you.

0:31:23.669,0:31:29.073
What's 2 X squared plus one? So[br]this is going to become two X

0:31:29.073,0:31:33.705
squared plus one all raised to[br]the power half plus C and

0:31:33.705,0:31:37.951
division by 1/2 is like[br]multiplication by two, so I have

0:31:37.951,0:31:42.583
a two there at the beginning and[br]that's the solution of this

0:31:42.583,0:31:45.285
fairly complicated integral done[br]through a substitution.

0:31:52.749,0:31:56.397
So let me try and extract from[br]all that are fairly general

0:31:56.397,0:32:00.045
results and the general result[br]is this that if we have an

0:32:00.045,0:32:02.173
integral of F of G of X.

0:32:03.319,0:32:06.089
He dashed of X DX.

0:32:08.379,0:32:10.147
Then the[br]substitution U

0:32:10.147,0:32:11.915
equals G of X.

0:32:13.419,0:32:18.326
Do you is G dashed[br]of XDX?

0:32:19.349,0:32:21.665
That substitution will[br]always transform this

0:32:21.665,0:32:26.683
integral into simply F of G[br]of X, which is F of U.

0:32:29.059,0:32:31.299
And the whole of G dash DX.

0:32:32.379,0:32:33.969
Will become simply do youth.

0:32:36.049,0:32:39.352
And this hopefully will[br]be an integral which is

0:32:39.352,0:32:40.820
much simpler to evaluate.

0:32:42.109,0:32:45.124
Now really what we want to be[br]able to do is get you into the

0:32:45.124,0:32:47.737
habit of spotting some of these[br]and being able to write down the

0:32:47.737,0:32:50.350
answer straight away. So let me[br]just give you one or two more

0:32:50.350,0:32:52.360
examples where hopefully we can[br]actually spot what's going on.

0:32:52.929,0:32:56.878
Supposing we looking at[br]the integral of E to the X

0:32:56.878,0:33:00.468
squared multiplied by two[br]X and we want to integrate

0:33:00.468,0:33:02.263
that with respect to X.

0:33:03.639,0:33:07.359
Let's try and compare what we've[br]got here with what's up here.

0:33:09.259,0:33:13.666
If you think of the X squared as[br]being the G of X.

0:33:15.019,0:33:17.819
So this bit is the G of X.

0:33:21.089,0:33:24.953
If we differentiate, X squared[br]would get 2X and you see that

0:33:24.953,0:33:29.139
appears out here, so this, but[br]in here is the G Dash, Devex.

0:33:31.109,0:33:33.908
And there's another function[br]involved in here as well.

0:33:33.908,0:33:37.018
It's the F function, and in[br]this particular example, the

0:33:37.018,0:33:40.439
F function is the exponential[br]function, and we see that G

0:33:40.439,0:33:43.860
is input to F because X[br]squared is input to the

0:33:43.860,0:33:44.482
exponential function.

0:33:45.759,0:33:49.269
So if we want to tackle this[br]particular integral, the

0:33:49.269,0:33:53.481
substitution U equals whatever[br]the G of X was, which in this

0:33:53.481,0:33:55.587
case is X squared will simplify

0:33:55.587,0:34:01.519
this integral. You being X[br]squared do you will be 2X DX.

0:34:02.289,0:34:05.835
And immediately this integral[br]will become the integral of.

0:34:07.329,0:34:11.164
E to the power X squared, which[br]was E to the power you.

0:34:12.479,0:34:18.108
And automatically the two X DX[br]is taken care of in the du.

0:34:20.809,0:34:24.150
Now this is just the integral of[br]the exponential function E to EU

0:34:24.150,0:34:27.234
with respect to you and we can[br]write the answer straight down

0:34:27.234,0:34:30.318
as the same thing eater, the you[br]and a constant of integration.

0:34:31.639,0:34:35.755
If we return to our original[br]variables, you was X squared, so

0:34:35.755,0:34:40.557
will hav E to the X squared plus[br]a constant. So this quantity E

0:34:40.557,0:34:45.702
to the X squared plus a constant[br]is the integral of two XE to the

0:34:45.702,0:34:46.731
X squared DX.

0:34:47.789,0:34:53.081
And as I say, we want to get[br]into the habit of being able

0:34:53.081,0:34:57.239
to almost spotless, and you[br]should get into the habit of

0:34:57.239,0:35:01.019
spotting that the quantity[br]here the two X is the

0:35:01.019,0:35:04.043
derivative of this function[br]in this other composite

0:35:04.043,0:35:07.823
function. Here, let's see if[br]we can do one straightaway.

0:35:07.823,0:35:11.225
Supposing, supposing that I[br]ask you to integrate the

0:35:11.225,0:35:16.139
cosine of three X to the[br]power 4 * 12 X cubed. Suppose

0:35:16.139,0:35:19.163
we want to integrate this[br]horrible looking thing.

0:35:20.229,0:35:24.103
Now the thing I want you to spot[br]is that if you differentiate

0:35:24.103,0:35:27.977
this function in here 3X to the[br]power four, you'd actually get 4

0:35:27.977,0:35:31.553
threes or 12X cubed. You get the[br]concert if it's out here.

0:35:32.229,0:35:34.357
So this is like the G of X.

0:35:35.779,0:35:39.310
And this is like the G Dash,[br]Devex, and making the

0:35:39.310,0:35:42.199
substitution you equals 3X plus[br]four would immediately reduce

0:35:42.199,0:35:46.372
you to a much simpler integral,[br]just in terms of you. I'm hoping

0:35:46.372,0:35:50.545
that you by now you can spot[br]that if we integrate, this will

0:35:50.545,0:35:52.150
actually just get the sign.

0:35:53.859,0:35:57.579
Of three X to the four plus a[br]constant of integration. Suggest

0:35:57.579,0:36:01.919
you go back and look at that[br]again if you not too happy about

0:36:01.919,0:36:05.029
it. And that's integration[br]by substitution.