WEBVTT
00:00:01.760 --> 00:00:09.440
In this lesson, weβll learn how to use integration by substitution to evaluate indefinite integrals.
00:00:09.440 --> 00:00:18.880
At this stage, you should feel comfortable finding the antiderivative of a variety of functions, including polynomials, trigonometric, and logarithmic functions.
00:00:19.920 --> 00:00:27.120
In this lesson, weβll look at how to apply these rules to find the antiderivative, or the integral, for more complicated functions.
00:00:28.200 --> 00:00:34.000
Because of the fundamental theorem of calculus, itβs important to be able to find the antiderivative.
00:00:34.000 --> 00:00:46.640
But our formulae donβt tell us how to evaluate integrals such as the integral of π₯ to the fifth power times π₯ to the sixth power plus nine to the seventh power with respect to π₯.
00:00:47.280 --> 00:00:54.000
To find this integral, we use a special strategy of introducing something extra, a new variable.
00:00:54.800 --> 00:00:58.000
This is called integration by substitution.
00:00:58.000 --> 00:01:01.240
And itβs sometimes referred to as the reverse chain rule.
00:01:02.040 --> 00:01:05.000
The first step is often to get the integral into this form.
00:01:05.640 --> 00:01:10.000
Note we have a function π of π₯ and its derivative π prime of π₯.
00:01:10.720 --> 00:01:15.240
And as itβs often the case, itβs sensible just to have a look at an example of how this works.
00:01:16.200 --> 00:01:24.480
Determine the integral of π₯ to the fifth power multiplied by π₯ to the sixth power plus nine all to the seventh power with respect to π₯
00:01:25.160 --> 00:01:31.200
This is not a polynomial thatβs nice to integrate using our standard rules for finding the antiderivative.
00:01:31.760 --> 00:01:37.040
And we certainly do not want to distribute our parentheses and find the antiderivative for each term.
00:01:37.760 --> 00:01:40.600
Instead, we spot that the integral is set up in this form.
00:01:41.280 --> 00:01:44.920
We have some function π of π₯ and its derivative π prime of π₯.
00:01:46.600 --> 00:01:55.240
If we look carefully, we see that π₯ to the fifth power is a scalar multiple of the derivative of π₯ to the power of six plus nine.
00:01:55.240 --> 00:02:00.400
And this means we can use integration by substitution to evaluate our indefinite integral.
00:02:01.200 --> 00:02:21.040
The substitution rule says that if π’ is equal to π of π₯ is a differentiable function whose range is some interval π and π is continuous on this interval, then the integral of π of π of π₯ multiplied by π prime of π₯ with respect to π₯ is equal to the integral of π of π’ with respect to π’.
00:02:21.680 --> 00:02:26.480
Weβre going to let π’ be equal to the function that we originally defined π of π₯.
00:02:26.840 --> 00:02:30.000
So, π’ is equal to π₯ to the sixth power plus nine.
00:02:30.640 --> 00:02:40.040
Now, this is great, as when we differentiate this function π’ with respect to π₯, we see that dπ’ by dπ₯ is equal to six π₯ to the fifth power.
00:02:40.920 --> 00:02:45.160
In integration by substitution, we think of the dπ’ and dπ₯ as differentials.
00:02:45.160 --> 00:02:51.277
And we can alternatively write this as dπ’ equals six π₯ to the fifth power dπ₯.
00:02:51.360 --> 00:02:57.560
Notice that whilst dπ’ by dπ₯ is definitely not a fraction, we do treat it a little like one in this process.
00:02:58.120 --> 00:03:04.120
We divide through by six, and we see that a sixth dπ’ is equal to π₯ to the fifth power dπ₯.
00:03:04.680 --> 00:03:07.200
And now, letβs look back to our original integral.
00:03:07.920 --> 00:03:13.280
We see that we can replace π₯ to the fifth power dπ₯ with a sixth dπ’.
00:03:13.880 --> 00:03:18.200
And we replace π₯ to the sixth power plus nine with π’.
00:03:18.920 --> 00:03:26.320
And we now see that the integral weβre evaluating is the integral of a sixth of π’ to the seventh power dπ’.
00:03:27.000 --> 00:03:37.015
If we so choose, we can take this factor of a sixth outside of the integral sign, and then weβre evaluating a sixth of the integral of π’ to the seventh power with respect to π’.
00:03:37.445 --> 00:03:46.960
The integral of π’ to the seventh power is π’ to the eighth power divided by eight plus, since it is an indefinite integral, π the constant of integration.
00:03:47.600 --> 00:03:49.240
We distribute our parentheses.
00:03:49.240 --> 00:03:56.640
And we see that the integral is equal to one over 48 times π’ to the eighth power plus πΆ.
00:03:56.640 --> 00:04:02.320
And notice Iβve chosen this to be a capital πΆ because our original constant of integration has been multiplied by one-sixth.
00:04:03.080 --> 00:04:07.320
But, remember, we were originally looking to evaluate our integral in terms of π₯.
00:04:07.960 --> 00:04:10.089
So, we look to our original definition of π’.
00:04:10.275 --> 00:04:13.360
And we said that π’ was equal to π₯ to the sixth power plus nine.
00:04:13.720 --> 00:04:22.200
And we then see that our integral is equal to one over 48 times π₯ to the sixth power plus nine to the eighth power plus πΆ.
00:04:23.760 --> 00:04:33.000
In this example, we saw that we should try and choose π’ to be some factor of the integrand whose differential also occurs, albeit some scalar multiple of it.
00:04:33.720 --> 00:04:39.040
If thatβs not possible though, we try choosing π’ to be some more complicated part of the integrand.
00:04:39.240 --> 00:04:42.720
This might be the inner function in a composite function or similar.
00:04:43.400 --> 00:04:45.400
Letβs have a look at an example of this form.
00:04:46.440 --> 00:04:54.400
Determine the integral of eight π₯ times eight π₯ plus nine squared with respect to π₯ by using the substitution method.
00:04:55.320 --> 00:05:01.040
In this example, weβve been very explicitly told to use the substitution method to evaluate this integral.
00:05:01.760 --> 00:05:10.920
Usually, we will look to choose our substitution π’ to be some factor of the integrand whose differential also occurs, albeit some scalar multiple of it.
00:05:10.920 --> 00:05:13.680
Here though, itβs not instantly obvious what that might be.
00:05:14.480 --> 00:05:21.160
Instead, then, we try choosing π’ to be some more complicated part of the function, perhaps the inner function in a composite function.
00:05:21.880 --> 00:05:24.360
Letβs try π’ equals eight π₯ plus nine.
00:05:25.080 --> 00:05:28.638
This means that dπ’ by dπ₯ is equal to eight.
00:05:29.160 --> 00:05:31.800
And we can treat dπ’ and dπ₯ as differentials.
00:05:31.800 --> 00:05:38.200
Remember, dπ’ by dπ₯ is not a fraction but we certainly treat it like one when performing integration by substitution.
00:05:38.400 --> 00:05:41.320
We can say that dπ’ is equal to eight dπ₯.
00:05:41.680 --> 00:05:45.680
Or, equivalently, an eighth dπ’ is equal to dπ₯.
00:05:46.200 --> 00:05:48.280
Now, this isnβt instantly helpful.
00:05:49.040 --> 00:05:59.924
As if we replace dπ₯ with an eighth dπ’ and eight π₯ plus nine with π’, weβll still have part of our function, thatβs the eight π₯, which is in terms of π₯.
00:06:00.413 --> 00:06:04.720
But if we look back to our substitution, we see that we can rearrange this.
00:06:04.720 --> 00:06:10.160
We subtract nine from both sides, and we see that eight π₯ is equal to π’ minus nine.
00:06:10.960 --> 00:06:17.600
Then, our integral becomes π’ minus nine times π’ squared multiplied by an eighth dπ’.
00:06:18.240 --> 00:06:26.640
Letβs take this eighth outside of the integral and then distribute the parentheses, and we see we have a simple polynomial that we can integrate.
00:06:27.040 --> 00:06:30.720
The integral of π’ cubed is π’ to the fourth power divided by four.
00:06:31.320 --> 00:06:35.800
The integral of negative nine π’ squared is negative nine π’ cubed divided by three.
00:06:36.280 --> 00:06:39.080
And we mustnβt forget πΆ, our constant of integration.
00:06:39.840 --> 00:06:44.880
We can simplify nine π’ cubed divided by three to three π’ cubed.
00:06:44.880 --> 00:06:50.320
But we mustnβt forget to replace π’ with eight π₯ plus nine in our final step.
00:06:50.960 --> 00:07:01.160
When we do, we see that our integral is equal to an eighth times eight π₯ plus nine to the fourth power divided by four minus three times eight π₯ plus nine cubed plus πΆ.
00:07:01.720 --> 00:07:04.840
When we distribute our parentheses, we see we have our solution.
00:07:04.840 --> 00:07:13.640
Itβs one over 32 times eight π₯ plus nine to the fourth power minus three-eighths times eight π₯ plus nine cubed plus πΆ.
00:07:16.320 --> 00:07:24.240
Determine the integral of 48 minus six π₯ over the fifth root of 16 minus two π₯ dπ₯.
00:07:24.920 --> 00:07:28.400
Itβs not instantly obvious how we need to evaluate this integral.
00:07:29.080 --> 00:07:42.120
However, if we look carefully, we see that the numerator is a scalar multiple of the inner function on the denominator such that 48 minus six π₯ is three times 16 minus two π₯.
00:07:42.920 --> 00:07:48.920
This is a hint to us that weβre going to need to use integration by substitution to evaluate this indefinite integral.
00:07:49.840 --> 00:07:52.760
Weβre going to let π’ be equal to 16 minus two π₯.
00:07:53.440 --> 00:08:00.040
Weβve chosen this part for our substitution as 16 minus two π₯ is the inner function in a composite function.
00:08:00.600 --> 00:08:05.760
We differentiate π’ with respect to π₯, and we see that dπ’ by dπ₯ is equal to negative two.
00:08:06.360 --> 00:08:13.440
Now, remember, dπ’ by dπ₯ is not a fraction, but we treat it a little like one when performing integration by substitution.
00:08:13.440 --> 00:08:19.040
And we can see that this is equivalent to saying negative one-half dπ’ is equal to dπ₯.
00:08:19.600 --> 00:08:22.600
Letβs substitute what we now have into our original integral.
00:08:23.160 --> 00:08:28.680
We saw that 48 minus six π₯ is equal to three times 16 minus two π₯.
00:08:28.680 --> 00:08:31.400
So, the numerator becomes three π’.
00:08:31.500 --> 00:08:34.000
The denominator becomes the fifth root of π’.
00:08:34.000 --> 00:08:36.880
And we replace dπ₯ with negative a half dπ’.
00:08:37.400 --> 00:08:40.280
Letβs take out a factor of negative three over two.
00:08:40.960 --> 00:08:43.560
And weβll write our denominator as π’ to the fifth power.
00:08:44.120 --> 00:08:47.720
Weβre dividing π’ to the power of one by π’ to the power of one-fifth.
00:08:47.920 --> 00:08:53.160
So, we subtract one-fifth from one, and weβre left with π’ to the four-fifths.
00:08:54.120 --> 00:08:59.960
The antiderivative of π’ to the four-fifths is π’ to the nine-fifths divided by nine-fifths.
00:08:59.960 --> 00:09:03.280
Thatβs the same as five-ninths times π’ to the nine-fifths.
00:09:03.920 --> 00:09:07.920
And, remember, this is an indefinite integral, so we add that constant of integration π.
00:09:08.480 --> 00:09:15.240
When we distribute the parentheses, we have negative five-sixths times π’ to the nine-fifths plus capital πΆ.
00:09:15.680 --> 00:09:26.160
Since our original constant has been multiplied by negative three over two, and since we were evaluating an integral in terms of π₯, we must replace π’ with 16 minus two π₯.
00:09:26.160 --> 00:09:32.520
And we see that our integral is negative five-sixths times 16 minus two π₯ to the nine-fifths plus πΆ.
00:09:34.200 --> 00:09:41.800
In our previous two examples, we saw that we can perform integration by substitution, even if itβs not instantly obvious what that process might look like.
00:09:42.960 --> 00:09:47.560
Weβll now see how we can use the process to integrate a more complicated trigonometric function.
00:09:49.080 --> 00:10:02.640
Determine the integral of negative 24π₯ cubed plus 30 sin six π₯ times negative six π₯ to the fourth power minus five cos of six π₯ to the fifth power with respect to π₯.
00:10:03.400 --> 00:10:20.040
To evaluate this integral, we need to spot that negative 24π₯ cubed plus 30 sin six π₯ is the derivative of the inner part of this composite function negative six π₯ to the fourth power minus five cos six π₯.
00:10:20.800 --> 00:10:25.000
This tells us we can use integration by substitution to evaluate this integral.
00:10:25.320 --> 00:10:32.480
Weβll let π’ be the inner function in our composite function, and then we use the general result for the derivative of cos ππ₯.
00:10:32.480 --> 00:10:42.880
And we see that dπ’ by dπ₯, the derivative of π’ with respect to π₯, is negative 24π₯ cubed plus 30 sin six π₯.
00:10:43.360 --> 00:10:49.960
Remember, dπ’ by dπ₯ is not a fraction, but we do treat it a little like one when performing integration by substitution.
00:10:49.960 --> 00:10:57.640
And we see that this is equivalent to saying that dπ’ is equal to negative 24π₯ cubed plus 30 sin six π₯ dπ₯.
00:10:58.200 --> 00:11:05.074
We, therefore, replace negative 24π₯ cubed plus 30 sin six π₯dπ₯ with dπ’.
00:11:05.487 --> 00:11:10.880
And we replace negative six π₯ to the fourth power minus five cos six π₯ with π’.
00:11:11.400 --> 00:11:13.960
And we see that our integral becomes really nice.
00:11:13.960 --> 00:11:16.440
Itβs the integral of π’ to the fifth power dπ’.
00:11:17.080 --> 00:11:21.440
Well, the antiderivative of π’ to the fifth power is π’ to the sixth power over six.
00:11:21.960 --> 00:11:28.640
So, the integral of π’ to the fifth power dπ’ is π’ to the sixth power over six plus the constant of integration π.
00:11:29.160 --> 00:11:37.840
Remember though, our integralβs in terms of π₯, so we replace π’ with negative six π₯ to the fourth power minus five cos of six π₯.
00:11:38.480 --> 00:11:40.400
And weβve evaluated our integral.
00:11:40.400 --> 00:11:47.720
Itβs a sixth of negative six π₯ to the fourth power minus five cos of six π₯ to the sixth power plus π.
00:11:49.400 --> 00:11:55.080
In our final example, we will look at how we can use integration by substitution to integrate a logarithmic function.
00:11:56.640 --> 00:12:03.476
Determine the integral of negative 11 over six π₯ times the cube root of the natural log of π₯ dπ₯.
00:12:04.054 --> 00:12:15.800
In order to evaluate this integral, we need to spot that the derivative of the natural log of π₯ is one over π₯ and that a part of our function is a scalar multiple of one over π₯.
00:12:16.400 --> 00:12:21.040
This tells us we can use integration by substitution to evaluate our integral.
00:12:21.600 --> 00:12:24.080
We let π’ be equal to the natural log of π₯.
00:12:24.640 --> 00:12:28.320
And weβve seen that dπ’ by dπ₯ is, therefore, one over π₯.
00:12:29.000 --> 00:12:35.960
Remember, dπ’ by dπ₯ is not a fraction but we treat it a little like one when performing integration by substitution.
00:12:35.960 --> 00:12:40.320
And we see that this is equivalent to saying dπ’ is equal to one over π₯dπ₯.
00:12:40.760 --> 00:12:43.160
Letβs substitute what we now have into our integral.
00:12:43.880 --> 00:12:51.000
If we take out a factor of negative eleven-sixths, we see that we can replace one over π₯ dπ₯ with dπ’.
00:12:51.280 --> 00:12:54.289
And we can replace the natural log of π₯ with π’.
00:12:54.781 --> 00:12:59.800
To make this easy to integrate, we recall that the cube root of π’ is the same as π’ to the power of one-third.
00:13:00.400 --> 00:13:09.520
And we know that the antiderivative of π’ to the power of one-third is π’ to the power of four-thirds divided by four-thirds, or three-quarters π’ to the power of four-thirds.
00:13:09.960 --> 00:13:18.560
We distribute our parentheses, and we see that our integral is equal to negative eleven-eighths times π’ to the power of four-thirds plus capital πΆ.
00:13:18.560 --> 00:13:26.520
And Iβve changed it from a lowercase π to a capital πΆ, as weβve multiplied our original constant of integration by negative eleven-sixths, thereby changing the number.
00:13:27.160 --> 00:13:32.080
It is, of course, important to remember that our original integral was in terms of π₯.
00:13:32.080 --> 00:13:34.680
So, we replace π’ with the natural log of π₯.
00:13:35.160 --> 00:13:41.280
And we see that our answer is negative eleven-eighths times the natural log of π₯ to the power of four-thirds plus πΆ.
00:13:43.240 --> 00:13:49.280
In this video, weβve seen that we can introduce a substitution to find the integral of more complicated functions.
00:13:49.920 --> 00:13:58.600
Weβve learned that we usually try to choose π’ to be some factor of the integrand whose differential also occurs, albeit some scalar multiple of it.
00:13:59.240 --> 00:14:03.800
If thatβs not possible though, we try choosing π’ to be some complicated part of the integrand.
00:14:03.800 --> 00:14:07.040
This might be the inner function in a composite function or similar.
00:14:08.160 --> 00:14:15.280
We also saw that this method can be used to integrate functions involving roots, trigonometric functions, and logarithmic functions.