INSTRUCTORS Raffi Hovasapian John Zhu

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 5 answersLast reply by: Professor HovasapianTue Jan 31, 2017 6:55 AMPost by Gautham Padmakumar on January 22, 2016Hi Raffi,I think I caught an error on your lecture. In example 5 just as you're finishing up you multiply  U^8 ( U^2 - 1) and you write it as U^10 - U^2 when you actually meant U^10 - U^8 I believe. So the final answer would be sec^11(x) / 11 - sec^9(x) / 9 + CThankss

### Trigonometric Integrals I

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Example I: ∫ sin³ (x) dx 1:36
• Example II: ∫ cos⁵(x)sin²(x)dx 4:36
• Example III: ∫ sin⁴(x)dx 9:23
• Summary for Evaluating Trigonometric Integrals of the Following Type: ∫ (sin^m) (x) (cos^p) (x) dx 15:59
• #1: Power of sin is Odd
• #2: Power of cos is Odd
• #3: Powers of Both sin and cos are Odd
• #4: Powers of Both sin and cos are Even
• Example IV: ∫ tan⁴ (x) sec⁴ (x) dx 17:34
• Example V: ∫ sec⁹(x) tan³(x) dx 20:55
• Summary for Evaluating Trigonometric Integrals of the Following Type: ∫ (sec^m) (x) (tan^p) (x) dx 23:31
• #1: Power of sec is Odd
• #2: Power of tan is Odd
• #3: Powers of sec is Odd and/or Power of tan is Even

### Transcription: Trigonometric Integrals I

Hello, welcome back to www.educator.com, and welcome back to AP Calculus.0000

Today, we are going to talk about trigonometric integrals.0005

We are going to be spending a couple of lessons on this, I think two or three, because they are fairly involved.0007

I want to make sure that there were enough examples to go around.0014

Let us jump right on in.0019

There are many integrals involving trig functions.0022

Let me work in blue.0025

There are many integrals involving trig functions that require special handling and manipulation.0029

Examples is really the best way to present this material.0067

We are just going to do example and the presentation.0070

Hopefully, everything will start to make sense.0073

Examples are the best way to present this material.0075

Let us jump right on in, our first example is evaluate the integral of sin³ x dx.0094

We have our nice little modified integral sign that is unique to us.0104

How do we deal with this?0111

Let us see what we can do.0114

One of the things that we can do is the following.0115

Sin³ (x) is equal to sin² x × sin x.0117

I have pulled out a sin x.0127

Sin² x, using one of my Pythagorean identities, I’m going to write this as 1 - cos² x × sin x.0130

Our original integral of sin³ x dx is actually equal to the integral of 1 - cos² x sin x dx.0143

Now we go ahead and use a u substitution.0159

Let me just go to blue.0166

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.0169

Therefore, this integral becomes, we have the integral of 1 - cos²,0190

1 - u² sin x dx × - du which is equal to the integral of u² - 1 du.0202

I use this minus sign and I just flip this.0222

The rest is easy.0226

This is just, the integral is equal to u³/ 3 - u + c.0227

Of course, we said u is cos x.0239

This is equal to cos³ x/ 3 - cos x + c.0242

There you go, that is it.0250

Trigonometric integral is going to come down to manipulation.0252

The same way that it was for the identities, when you are doing identities, proving those things in pre-calculus,0257

moving things around, trying this identity, trying that identity.0264

Pulling things out, recognizing when something is a derivative of the other.0268

That is it, that is just the nature of the trigonometric integrals.0272

Let us try another example here.0276

This time we have, let me go back to black so I can fix this.0279

The integral of cos⁵ x sin² dx.0283

Here we have an odd power of cos.0289

I’m going to pull out a cos.0297

We have an odd power of cos.0299

I’m going to rewrite this as the integral of cos⁴ x × sin² x cos x dx.0310

I just pulled out one of the cos, separated it out.0325

This is going to equal the integral of cos² x² sin² x cos x dx.0329

Now I’m going to rewrite cos² as 1 - sin².0344

It equals the integral of 1 - sin² x × sin² x cos x dx.0349

I suppose I could multiply it out.0367

Should I multiply first?0372

That is fine, I’m just going to go ahead and I will use a u substitution.0374

I will let u equal to sin x.0378

Let me do it over here, I want some more room.0385

I have got u was equal to sin x and I got du = cos x dx.0388

That turns into the integral of 1 - u² × u² × du which is equal to the integral of,0400

I’m sorry this is² and this is², sorry about that.0415

We have got 1 - 2u² + u⁴ × u² du.0424

That is going to equal the integral of u² - 2u⁴ + u⁶ du.0437

And that is equal to u³/ 3 – 2u⁵/ 5 + u⁷/ 7 + c.0447

Let us go ahead and we said that u is equal to sin x.0477

Our final answer is sin³ x/ 3 – 2 sin⁵ x/ 5 + sin⁷ x/ 7 + c.0480

There you go.0499

This was with an odd power of cos.0506

An odd power of sin works the same way.0512

In other words, pull out one of the sins, express.0521

Now the even power of the cos, in terms of the sin, sin, and the cos.0526

What we did here is, odd power of cos, we pulled out a cos, we express the cos as a sin.0535

And then, we simplify it by using u substitution, by u = sin x.0542

An odd power of sin works the same way.0545

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.0547

But I’m going to be actually be summarizing all this, after I do a few examples.0555

It is not a problem.0558

Let us see, this time we have sin⁴ x.0563

Now we have an even power of a trig function.0570

How do we deal what this one?0587

With even powers of trig functions, we are going to employ the following identities.0590

We employ the following identities.0599

Let us go ahead and put them in blue.0608

Sin² x = ½ of 1 - cos 2x and cos² x = ½ of 1 + cos 2x.0613

In this case, we have the integral of sin⁴.0637

The integral of sin⁴ x dx is equal to the integral of sin² x² dx,0641

which is equal to the integral, sin², we said is ½.0653

This is going to be ½ of 1 - cos 2x² dx.0657

That is going to equal the integral of ¼ × 1 - 2 cos 2x + cos² 2x dx = ¼ the integral of dx -1/4 × -2.0671

This is going to be -1/2 × the integral of cos 2x dx + ¼ × the integral of cos² x dx.0699

This is easy to solve, this is easy to solve.0722

Here we are going to have to do another round of using this.0724

What we get is the following.0729

Let me rewrite what it is that I had.0731

I had ¼ the integral of dx - ½ the integral of cos of 2x dx + the integral of cos² 2x dx.0733

I think it was ¼, if I’m not mistaken, yes.0750

That is going to equal ¼ x, that takes care of that one, -1/4 sin of 2x,0755

that takes care of that one, + ¼ × the integral of cos² 2x dx.0767

This is the integral that we have to deal with, right here.0777

Now we are going to use cos² x is equal to ½ of 1 + cos 2x.0781

This integral comes down here and we end up with ¼ the integral of ½ of 1 + cos 2x dx,0794

which is equal to ¼ × ½ the integral of dx, that is that one.0812

+ ½ the integral of cos² x = 1 + cos 2x.0825

We have cos² 2x.0842

This is cos 4x, the integral of cos 4x dx, my apologies.0846

This is equal to 1/8 x + ¼ × ½ is 1/8.0854

The integral ¼ of 4 comes out as ¼.0871

It is going to be 8 × 4 is 32 + 1/32 × sin(4x).0880

Putting it all together, this was that.0890

We cannot forget this and this.0906

Our final integral is equal to, I will write integral was equal to ¼ x which is this one,0908

-1/4 sin 2x + 1/8 x + 1/32 sin 4x + c.0932

There you go, that is it.0951

Our summary for evaluating trigonometric integrals of the following type.0960

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.0965

Here are our choices, if the power of the sin is odd, you pull one of the sin x.0975

You use sin² x = 1 - cos² x.0982

You express everything in terms of cos x.0986

You pull a sin x, expressive everything in cos x.0990

Then set u equal to cos x, simplify and integrate using u substitution.0994

If the power of cos is odd, you pull a cos x.1001

You use cos² = 1 – sin², express everything in terms of sin x and set u = sin x, and then, u substitution and simplify.1006

If the powers of both sin and cos are odd, either of the procedures above will work.1023

It does not really matter which one you choose.1027

If the powers of both sin and cos are even, we use the following identities.1031

Sin² is equal to this, cos² is equal to this.1038

When sin is odd, when cos is odd, when both are odd, either one is fine, or when both are even.1042

There are four possibilities.1049

Let us deal with our next set of examples and then we can summarize those.1055

We have the integral of tan⁴ sec⁴ x dx.1061

Let us do this, because the derivative of ddx of tan x = sec² x and the derivative of sec x = sec x tan x,1075

we might try something similar to what we did before, pull out something.1099

You might try pulling out certain powers of one or the other.1108

I’m going to try this, I’m going to rewrite this.1122

Let me work in blue.1129

The integral of tan⁴ x sec⁴ x dx.1133

I’m going to pull out, I’m going to break up the sec⁴.1141

I’m going to pull out a sec² because I have the relationship between tan is sec².1143

I’m going to write this as the integral of tan⁴ x × sec² x dx.1149

Now I'm going to express this, in terms of tan.1161

I have = the integral of tan⁴ x × 1 + tan² x which is sec² x × sec² x dx.1166

I’m going to let u = tan x and my du = sec² x dx.1184

That takes care of that right there.1193

I end up with the integral of u⁴ × 1 + u² du, nice and easy.1196

= the integral of u⁴ + u⁶ du.1211

That is equal to u⁵/ 5 + u⁷/ 7 + c.1220

I get u is equal to tan(x).1228

My final integral is equal to tan⁵ x/ 5 + tan⁷ (x)/ 7 + c.1233

There you go, nice.1246

The same thing, we just pulled something out.1247

Use the u substitution and express it.1251

This time, let us see, what do we got?1258

We have the integral of sec⁹ tan³.1264

Let us pull out sec x tan x.1268

This integral is going to equal the integral of sec⁸ x tan² x × sec x tan x, that is what I pulled out, dx.1286

We know an identity, we have sec² x = 1 + tan² x.1318

Therefore, tan² x is equal to sec² x – 1.1325

Wherever I see a tan, I will put that in.1332

That is going to equal the integral of sec⁸ x × sec² x – 1 × sec x tan x dx.1336

I will let u = sec x, my u substitution.1350

My du = sec x tan x dx.1355

I’m left with the integral of u⁸ × u² – 1 × du = the integral of u ⁺10 – u² du.1361

It = u ⁺11/ 11 – u³/ 3 + c.1384

I’m left with my integral, my final answer is going to be u is sec(x).1391

It is going to be sec ⁺11 (x)/ 11 – sec³/ 3 + c.1397

There you go, nice and simple, pretty straightforward.1407

Let us summarize this.1411

When we have an integral of the form sec to something power m, tan to some power p, my choices are as follows.1415

If a power of secant is even, pull out a sec².1424

Use sec² = tan² + 1, express everything in terms of tan x.1431

Pull out a sec², express everything in terms of tan x.1436

Set u = tan x and use u substitution to solve.1440

That is if the power of secant is even.1445

If the power of tangent is odd, pull a sec x tan x.1447

Use tan² = sec² – 1, express everything in terms of sec.1452

Let u equal secant.1456

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.1459

That one is a toss up to see what you can do.1471

Welcome to the wonderful world of trigonometric integrals.1475

That takes care of that.1479

With that, thank you so much for joining us here at www.educator.com.1482

We will see you next time for a continuation of trigonometric integrals.1487

Take care, bye.1490