WEBVTT mathematics/ap-calculus-ab/hovasapian
00:00:00.000 --> 00:00:04.700
Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
00:00:04.700 --> 00:00:07.400
Today, we are going to talk about trigonometric integrals.
00:00:07.400 --> 00:00:14.600
We are going to be spending a couple of lessons on this, I think two or three, because they are fairly involved.
00:00:14.600 --> 00:00:18.700
I want to make sure that there were enough examples to go around.
00:00:18.700 --> 00:00:21.700
Let us jump right on in.
00:00:21.700 --> 00:00:25.300
There are many integrals involving trig functions.
00:00:25.300 --> 00:00:29.300
Let me work in blue.
00:00:29.300 --> 00:01:06.800
There are many integrals involving trig functions that require special handling and manipulation.
00:01:06.800 --> 00:01:09.800
Examples is really the best way to present this material.
00:01:09.800 --> 00:01:13.100
We are just going to do example and the presentation.
00:01:13.100 --> 00:01:15.500
Hopefully, everything will start to make sense.
00:01:15.500 --> 00:01:34.300
Examples are the best way to present this material.
00:01:34.300 --> 00:01:44.500
Let us jump right on in, our first example is evaluate the integral of sin³ x dx.
00:01:44.500 --> 00:01:50.700
We have our nice little modified integral sign that is unique to us.
00:01:50.700 --> 00:01:53.700
How do we deal with this?
00:01:53.700 --> 00:01:54.900
Let us see what we can do.
00:01:54.900 --> 00:01:57.300
One of the things that we can do is the following.
00:01:57.300 --> 00:02:07.300
Sin³ (x) is equal to sin² x × sin x.
00:02:07.300 --> 00:02:10.500
I have pulled out a sin x.
00:02:10.500 --> 00:02:23.600
Sin² x, using one of my Pythagorean identities, I’m going to write this as 1 - cos² x × sin x.
00:02:23.600 --> 00:02:39.100
Our original integral of sin³ x dx is actually equal to the integral of 1 - cos² x sin x dx.
00:02:39.100 --> 00:02:45.900
Now we go ahead and use a u substitution.
00:02:45.900 --> 00:02:49.600
Let me just go to blue.
00:02:49.600 --> 00:03:10.100
U is equal to cos(x), du is equal to the derivative of cos is - sin(x) dx, sin(x) dx is equal to –du.
00:03:10.100 --> 00:03:22.600
Therefore, this integral becomes, we have the integral of 1 - cos²,
00:03:22.600 --> 00:03:41.700
1 - u² sin x dx × - du which is equal to the integral of u² - 1 du.
00:03:41.700 --> 00:03:46.200
I use this minus sign and I just flip this.
00:03:46.200 --> 00:03:47.400
The rest is easy.
00:03:47.400 --> 00:03:59.300
This is just, the integral is equal to u³/ 3 - u + c.
00:03:59.300 --> 00:04:02.200
Of course, we said u is cos x.
00:04:02.200 --> 00:04:10.200
This is equal to cos³ x/ 3 - cos x + c.
00:04:10.200 --> 00:04:12.300
There you go, that is it.
00:04:12.300 --> 00:04:17.100
Trigonometric integral is going to come down to manipulation.
00:04:17.100 --> 00:04:24.600
The same way that it was for the identities, when you are doing identities, proving those things in pre-calculus,
00:04:24.600 --> 00:04:28.400
moving things around, trying this identity, trying that identity.
00:04:28.400 --> 00:04:32.000
Pulling things out, recognizing when something is a derivative of the other.
00:04:32.000 --> 00:04:36.200
That is it, that is just the nature of the trigonometric integrals.
00:04:36.200 --> 00:04:39.200
Let us try another example here.
00:04:39.200 --> 00:04:43.400
This time we have, let me go back to black so I can fix this.
00:04:43.400 --> 00:04:48.800
The integral of cos⁵ x sin² dx.
00:04:48.800 --> 00:04:57.400
Here we have an odd power of cos.
00:04:57.400 --> 00:04:59.600
I’m going to pull out a cos.
00:04:59.600 --> 00:05:10.100
We have an odd power of cos.
00:05:10.100 --> 00:05:24.700
I’m going to rewrite this as the integral of cos⁴ x × sin² x cos x dx.
00:05:24.700 --> 00:05:29.200
I just pulled out one of the cos, separated it out.
00:05:29.200 --> 00:05:43.700
This is going to equal the integral of cos² x² sin² x cos x dx.
00:05:43.700 --> 00:05:48.900
Now I’m going to rewrite cos² as 1 - sin².
00:05:48.900 --> 00:06:07.100
It equals the integral of 1 - sin² x × sin² x cos x dx.
00:06:07.100 --> 00:06:12.200
I suppose I could multiply it out.
00:06:12.200 --> 00:06:13.800
Should I multiply first?
00:06:13.800 --> 00:06:18.000
That is fine, I’m just going to go ahead and I will use a u substitution.
00:06:18.000 --> 00:06:25.000
I will let u equal to sin x.
00:06:25.000 --> 00:06:28.600
Let me do it over here, I want some more room.
00:06:28.600 --> 00:06:40.200
I have got u was equal to sin x and I got du = cos x dx.
00:06:40.200 --> 00:06:55.400
That turns into the integral of 1 - u² × u² × du which is equal to the integral of,
00:06:55.400 --> 00:07:04.300
I’m sorry this is² and this is², sorry about that.
00:07:04.300 --> 00:07:16.800
We have got 1 - 2u² + u⁴ × u² du.
00:07:16.800 --> 00:07:27.300
That is going to equal the integral of u² - 2u⁴ + u⁶ du.
00:07:27.300 --> 00:07:56.700
And that is equal to u³/ 3 – 2u⁵/ 5 + u⁷/ 7 + c.
00:07:56.700 --> 00:08:00.600
Let us go ahead and we said that u is equal to sin x.
00:08:00.600 --> 00:08:19.400
Our final answer is sin³ x/ 3 – 2 sin⁵ x/ 5 + sin⁷ x/ 7 + c.
00:08:19.400 --> 00:08:26.100
There you go.
00:08:26.100 --> 00:08:32.400
This was with an odd power of cos.
00:08:32.400 --> 00:08:40.700
An odd power of sin works the same way.
00:08:40.700 --> 00:08:46.100
In other words, pull out one of the sins, express.
00:08:46.100 --> 00:08:55.300
Now the even power of the cos, in terms of the sin, sin, and the cos.
00:08:55.300 --> 00:09:02.200
What we did here is, odd power of cos, we pulled out a cos, we express the cos as a sin.
00:09:02.200 --> 00:09:05.600
And then, we simplify it by using u substitution, by u = sin x.
00:09:05.600 --> 00:09:07.500
An odd power of sin works the same way.
00:09:07.500 --> 00:09:14.900
You are to pull out a sin, you are going to express the sin as the cos, and then, you are going to let u equal cos.
00:09:14.900 --> 00:09:18.300
But I’m going to be actually be summarizing all this, after I do a few examples.
00:09:18.300 --> 00:09:23.100
It is not a problem.
00:09:23.100 --> 00:09:30.300
Let us see, this time we have sin⁴ x.
00:09:30.300 --> 00:09:47.100
Now we have an even power of a trig function.
00:09:47.100 --> 00:09:49.700
How do we deal what this one?
00:09:49.700 --> 00:09:59.200
With even powers of trig functions, we are going to employ the following identities.
00:09:59.200 --> 00:10:08.500
We employ the following identities.
00:10:08.500 --> 00:10:12.700
Let us go ahead and put them in blue.
00:10:12.700 --> 00:10:36.900
Sin² x = ½ of 1 - cos 2x and cos² x = ½ of 1 + cos 2x.
00:10:36.900 --> 00:10:41.000
In this case, we have the integral of sin⁴.
00:10:41.000 --> 00:10:53.400
The integral of sin⁴ x dx is equal to the integral of sin² x² dx,
00:10:53.400 --> 00:10:57.300
which is equal to the integral, sin², we said is ½.
00:10:57.300 --> 00:11:10.700
This is going to be ½ of 1 - cos 2x² dx.
00:11:10.700 --> 00:11:39.100
That is going to equal the integral of ¼ × 1 - 2 cos 2x + cos² 2x dx = ¼ the integral of dx -1/4 × -2.
00:11:39.100 --> 00:12:02.300
This is going to be -1/2 × the integral of cos 2x dx + ¼ × the integral of cos² x dx.
00:12:02.300 --> 00:12:04.600
This is easy to solve, this is easy to solve.
00:12:04.600 --> 00:12:08.700
Here we are going to have to do another round of using this.
00:12:08.700 --> 00:12:11.400
What we get is the following.
00:12:11.400 --> 00:12:13.200
Let me rewrite what it is that I had.
00:12:13.200 --> 00:12:30.600
I had ¼ the integral of dx - ½ the integral of cos of 2x dx + the integral of cos² 2x dx.
00:12:30.600 --> 00:12:35.400
I think it was ¼, if I’m not mistaken, yes.
00:12:35.400 --> 00:12:46.700
That is going to equal ¼ x, that takes care of that one, -1/4 sin of 2x,
00:12:46.700 --> 00:12:56.800
that takes care of that one, + ¼ × the integral of cos² 2x dx.
00:12:56.800 --> 00:13:01.300
This is the integral that we have to deal with, right here.
00:13:01.300 --> 00:13:13.800
Now we are going to use cos² x is equal to ½ of 1 + cos 2x.
00:13:13.800 --> 00:13:31.800
This integral comes down here and we end up with ¼ the integral of ½ of 1 + cos 2x dx,
00:13:31.800 --> 00:13:45.200
which is equal to ¼ × ½ the integral of dx, that is that one.
00:13:45.200 --> 00:14:02.300
+ ½ the integral of cos² x = 1 + cos 2x.
00:14:02.300 --> 00:14:05.700
We have cos² 2x.
00:14:05.700 --> 00:14:14.600
This is cos 4x, the integral of cos 4x dx, my apologies.
00:14:14.600 --> 00:14:31.300
This is equal to 1/8 x + ¼ × ½ is 1/8.
00:14:31.300 --> 00:14:39.900
The integral ¼ of 4 comes out as ¼.
00:14:39.900 --> 00:14:50.500
It is going to be 8 × 4 is 32 + 1/32 × sin(4x).
00:14:50.500 --> 00:15:05.700
Putting it all together, this was that.
00:15:05.700 --> 00:15:08.100
We cannot forget this and this.
00:15:08.100 --> 00:15:32.000
Our final integral is equal to, I will write integral was equal to ¼ x which is this one,
00:15:32.000 --> 00:15:51.000
-1/4 sin 2x + 1/8 x + 1/32 sin 4x + c.
00:15:51.000 --> 00:16:00.000
There you go, that is it.
00:16:00.000 --> 00:16:05.000
Our summary for evaluating trigonometric integrals of the following type.
00:16:05.000 --> 00:16:14.900
When we have integrals of the type sin to some power m, let me work in red, to some power m × cos to some power p.
00:16:14.900 --> 00:16:22.000
Here are our choices, if the power of the sin is odd, you pull one of the sin x.
00:16:22.000 --> 00:16:26.200
You use sin² x = 1 - cos² x.
00:16:26.200 --> 00:16:29.800
You express everything in terms of cos x.
00:16:29.800 --> 00:16:34.600
You pull a sin x, expressive everything in cos x.
00:16:34.600 --> 00:16:41.100
Then set u equal to cos x, simplify and integrate using u substitution.
00:16:41.100 --> 00:16:46.200
If the power of cos is odd, you pull a cos x.
00:16:46.200 --> 00:17:03.200
You use cos² = 1 – sin², express everything in terms of sin x and set u = sin x, and then, u substitution and simplify.
00:17:03.200 --> 00:17:07.500
If the powers of both sin and cos are odd, either of the procedures above will work.
00:17:07.500 --> 00:17:11.100
It does not really matter which one you choose.
00:17:11.100 --> 00:17:17.900
If the powers of both sin and cos are even, we use the following identities.
00:17:17.900 --> 00:17:22.400
Sin² is equal to this, cos² is equal to this.
00:17:22.400 --> 00:17:29.300
When sin is odd, when cos is odd, when both are odd, either one is fine, or when both are even.
00:17:29.300 --> 00:17:35.500
There are four possibilities.
00:17:35.500 --> 00:17:41.200
Let us deal with our next set of examples and then we can summarize those.
00:17:41.200 --> 00:17:54.900
We have the integral of tan⁴ sec⁴ x dx.
00:17:54.900 --> 00:18:19.000
Let us do this, because the derivative of ddx of tan x = sec² x and the derivative of sec x = sec x tan x,
00:18:19.000 --> 00:18:27.700
we might try something similar to what we did before, pull out something.
00:18:27.700 --> 00:18:42.200
You might try pulling out certain powers of one or the other.
00:18:42.200 --> 00:18:48.800
I’m going to try this, I’m going to rewrite this.
00:18:48.800 --> 00:18:52.700
Let me work in blue.
00:18:52.700 --> 00:19:01.000
The integral of tan⁴ x sec⁴ x dx.
00:19:01.000 --> 00:19:03.500
I’m going to pull out, I’m going to break up the sec⁴.
00:19:03.500 --> 00:19:08.900
I’m going to pull out a sec² because I have the relationship between tan is sec².
00:19:08.900 --> 00:19:21.400
I’m going to write this as the integral of tan⁴ x × sec² x dx.
00:19:21.400 --> 00:19:25.800
Now I'm going to express this, in terms of tan.
00:19:25.800 --> 00:19:44.300
I have = the integral of tan⁴ x × 1 + tan² x which is sec² x × sec² x dx.
00:19:44.300 --> 00:19:53.600
I’m going to let u = tan x and my du = sec² x dx.
00:19:53.600 --> 00:19:56.200
That takes care of that right there.
00:19:56.200 --> 00:20:11.500
I end up with the integral of u⁴ × 1 + u² du, nice and easy.
00:20:11.500 --> 00:20:19.900
= the integral of u⁴ + u⁶ du.
00:20:19.900 --> 00:20:28.200
That is equal to u⁵/ 5 + u⁷/ 7 + c.
00:20:28.200 --> 00:20:33.300
I get u is equal to tan(x).
00:20:33.300 --> 00:20:46.400
My final integral is equal to tan⁵ x/ 5 + tan⁷ (x)/ 7 + c.
00:20:46.400 --> 00:20:47.600
There you go, nice.
00:20:47.600 --> 00:20:51.300
The same thing, we just pulled something out.
00:20:51.300 --> 00:20:58.300
Use the u substitution and express it.
00:20:58.300 --> 00:21:04.000
This time, let us see, what do we got?
00:21:04.000 --> 00:21:08.400
We have the integral of sec⁹ tan³.
00:21:08.400 --> 00:21:26.200
Let us pull out sec x tan x.
00:21:26.200 --> 00:21:57.900
This integral is going to equal the integral of sec⁸ x tan² x × sec x tan x, that is what I pulled out, dx.
00:21:57.900 --> 00:22:05.300
We know an identity, we have sec² x = 1 + tan² x.
00:22:05.300 --> 00:22:12.200
Therefore, tan² x is equal to sec² x – 1.
00:22:12.200 --> 00:22:15.800
Wherever I see a tan, I will put that in.
00:22:15.800 --> 00:22:30.400
That is going to equal the integral of sec⁸ x × sec² x – 1 × sec x tan x dx.
00:22:30.400 --> 00:22:34.800
I will let u = sec x, my u substitution.
00:22:34.800 --> 00:22:41.400
My du = sec x tan x dx.
00:22:41.400 --> 00:23:04.000
I’m left with the integral of u⁸ × u² – 1 × du = the integral of u ⁺10 – u² du.
00:23:04.000 --> 00:23:10.900
It = u ⁺11/ 11 – u³/ 3 + c.
00:23:10.900 --> 00:23:16.800
I’m left with my integral, my final answer is going to be u is sec(x).
00:23:16.800 --> 00:23:27.300
It is going to be sec ⁺11 (x)/ 11 – sec³/ 3 + c.
00:23:27.300 --> 00:23:30.900
There you go, nice and simple, pretty straightforward.
00:23:30.900 --> 00:23:34.700
Let us summarize this.
00:23:34.700 --> 00:23:44.300
When we have an integral of the form sec to something power m, tan to some power p, my choices are as follows.
00:23:44.300 --> 00:23:50.800
If a power of secant is even, pull out a sec².
00:23:50.800 --> 00:23:56.500
Use sec² = tan² + 1, express everything in terms of tan x.
00:23:56.500 --> 00:24:00.100
Pull out a sec², express everything in terms of tan x.
00:24:00.100 --> 00:24:04.800
Set u = tan x and use u substitution to solve.
00:24:04.800 --> 00:24:06.900
That is if the power of secant is even.
00:24:06.900 --> 00:24:11.900
If the power of tangent is odd, pull a sec x tan x.
00:24:11.900 --> 00:24:16.100
Use tan² = sec² – 1, express everything in terms of sec.
00:24:16.100 --> 00:24:19.300
Let u equal secant.
00:24:19.300 --> 00:24:31.300
If the power of secant is odd and or the power of tan is even, you are going to utilize whatever resources you have at your disposal.
00:24:31.300 --> 00:24:35.100
That one is a toss up to see what you can do.
00:24:35.100 --> 00:24:39.300
Welcome to the wonderful world of trigonometric integrals.
00:24:39.300 --> 00:24:42.600
That takes care of that.
00:24:42.600 --> 00:24:46.800
With that, thank you so much for joining us here at www.educator.com.
00:24:46.800 --> 00:24:49.800
We will see you next time for a continuation of trigonometric integrals.
00:24:49.800 --> 00:24:50.000
Take care, bye.