For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

### Trigonometric Integrals II

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
- Trigonometric Integrals II 0:09
- Recall: ∫tanx dx
- Let's Find ∫secx dx
- Example I: ∫ tan⁵ (x) dx 6:23
- Example II: ∫ sec⁵ (x) dx 11:41
- Summary: How to Deal with Integrals of Different Types 19:04
- Identities to Deal with Integrals of Different Types
- Example III: ∫cos(5x)sin(9x)dx 19:57

### AP Calculus AB Online Prep Course

### Transcription: Trigonometric Integrals II

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

*Today, we are going to continue our discussion of trigonometric integrals.*0004

*Let us jump right on in.*0007

*Let us recall what the integral of tan(x) looks like.*0012

*Let us work in blue, I really like blue color.*0018

*Let us recall the integral of tan x dx.*0022

*I'm going to rewrite that as the integral of sin x/ cos x dx.*0032

*And then, I'm going to let u equal cos x and du is equal to -sin x dx.*0040

*Sin x dx is equal to –du.*0052

*What we end up actually getting is, this thing is going to be the integral of -du/u.*0059

*Let us come over here = -the integral of du/u = -natlog of u + c is equal – natlog.*0071

*We set that u was cos x = x + c.*0085

*The integral of the tan is –natlog of cos x dx.*0092

*You will also see it written like this, excuse me.*0100

*This can also be written as the natlog sec x + c.*0108

*Either one of these is absolutely fine, I personally prefer that one but where this comes from is this.*0132

*-ln of cos x + c that is the same as 0 - the ln of cos x + c.*0141

*0 can be written as the natlog of 1.*0158

*Natlog of 1 - the natlog of cos x + c is equal to the natlog of this over that 1/ cos x + c,*0161

*which is equal to the natlog of sec x + c.*0180

*That is where it comes from, either one is fine.*0185

*The actual one that you go through, gives you this one.*0187

*This is just another version of it, in case you do not like that negative sign in front, totally a personal choice.*0190

*We have the integral of the sin, the integral of the cos, and now we have done the integral of the tan which we have done before.*0200

*Now let us do the integral of secant.*0207

*Now let us find, and we are doing this just so we actually have a table of the integrals of all of our fundamental trigonometric functions.*0211

*Now let us find the integral of sec x dx.*0223

*This one is just manipulation, that is all it is.*0228

*Again, the manipulation itself is not really important.*0232

*What is important is that we actually find some expression for the integral of the sec(x).*0235

*So that we have an integral of a basic trigonometric function, so that we can use it for other integrals,*0240

*if it happens to come up, which is exactly why we are doing this.*0246

*It equals the integral of sec x × sec x + tan x/ sec x + tan x dx*0249

*is equal to the integral of sec² x + tan x/ sec x + tan x dx.*0269

*We will let u equal to sec x + tan x, du = sec x tan x + sec² x.*0282

*Sec tan + sec dx/u.*0302

*This equals the integral of du/u which = the natlog of u + c which = the natlog of sec x + tan x + c.*0312

*Now we have that one.*0331

*I'm going to leave the integral of cot x dx and the integral of csc x dx to you.*0335

*Do the same way that we just did for these two.*0355

*Now we have the integrals of all six basic trig functions, that is it.*0358

*Now we can continue our discussion of trigonometric integrals.*0379

*Let us try this one, let us go to black, finish the integral, the integral of tan⁵ x.*0383

*Tan⁵ x, I know that tan² x is equal to sec² x – 1.*0392

*How about if I write this as the integral of the tan(x) × tan⁴ (x) dx?*0400

*That is going to equal the integral of the tan(x) × tan² (x)² dx.*0412

*And that is going to equal the integral of tan x × sec² x - 1 dx²,*0422

*that is equal to the integral of tan x × sec⁴ x – 2 sec² x + 1 dx.*0437

*That is going to equal the integral first one and then the second one.*0453

*It is going to equal the integral of tan x sec⁴ x dx - 2 × the integral of tan x sec² x dx + the integral of tan x dx,*0461

*that is why we did the tan and sec in first because when we simplify some other trigonometric integrals,*0483

*one of the integrals that is going to show up is the integral of tan x.*0489

*We wanted to just be able to have something to fill in there as an answer.*0492

*This is our first integral, our second integral, and our third integral.*0497

*Our first integral, the integral of tan x sec⁴ x dx, that is going to equal the integral of tan x sec x.*0506

*I pull out a sec because the sec power is even, I can pull out a sec.*0526

*That leaves me with sec³ x dx.*0531

*I’m going to let u equal the sec(x) and I’m going to let du equal sec x tan x,*0535

*that gives me the integral of u³ du is equal to u⁴/ 4 + c =,*0547

*Let me do this on the next page, I’m actually going to need that.*0571

*That actually equals sec⁴ x/ 4 + c, that is our first integral.*0575

*Our second integral is the integral of tan x sec² x dx.*0591

*We let u equal tan x, du = sec² dx.*0600

*This is actually equal to the integral of u du, that is equal to u²/ 2 + c,*0607

*that is going to be u is tan, it is going to be tan² x/ 2 + c.*0617

*That is our second integral.*0623

*Our third integral, that was the integral of tan x dx.*0628

*We already solved that equals - the natlog of cos x + c.*0635

*We put them all together.*0643

*The integral of tan⁵ x dx is going to end up equaling sec⁴ x/ 4.*0647

*The c’s, I can combine it to 1c, that is not a problem.*0668

*I have this one, and then + tan² x/ 2.*0672

*I have this one, - the natlog of the cos(x) + c.*0685

*There you go, always manipulation, always manipulation.*0696

*It seems like that is all we do in mathematics, is not it?*0701

*That is not all we do, that is a big part of it.*0703

*Now evaluate sec⁵.*0713

*How do we deal with sec⁵?*0717

*In the previous lesson, we dealt with even powers of sec, it also has odd power of sec.*0720

*Let us see what we can do.*0728

*I'm going to rewrite this as sec⁵ x, sec² x dx.*0730

*This is how we deal with an odd power of sec, you try something.*0748

*This is what somebody tried in, it actually ended up working.*0754

*Let us use integration by parts.*0758

*Let me work in blue.*0769

*We will call this u and we will call this dv.*0772

*Let us start all over again.*0793

*The best advice I have ever heard, when you lose your way, go back to the beginning.*0798

*Sec⁵ x, I'm going to write it as the integral of sec³ x sec² x dx.*0802

*There you go, now we are going to use integration by parts.*0812

*Now I go back to blue.*0819

*I’m going to call sec³ x my u, my sec² x is going to be my dv.*0821

*u is equal to sec³ x, dv is equal to sec² x.*0827

*du is equal to 3 sec² x sec x tan x.*0836

*v is just equal to tan x, very nice.*0848

*We know what the integration by parts formula is.*0853

*It is the integral of u dv is equal to uv - the integral of v du.*0856

*Here is my u and here is my v.*0864

*Now it is going to equal, this thing is equal u, u × v.*0866

*It is going to be sec³ x tan(x) - the integral of v du.*0871

*It is going to be the integral of tan x × 3 sec² x sec x tan x dx.*0888

*It is going to be sec³ x tan x - 3 × the integral of sec⁴ x tan² x dx, very nice.*0907

*We have this one already, we just need to deal with this integral right here.*0938

*Let us go ahead and do that.*0945

*I have got 3 × the integral of, we had sec⁴ x tan² x dx, that is equal to 3 × the integral of,*0948

*I have got an even power of sec, I’m going to pull out a sec².*0969

*I’m going to write that as sec² x, sec² x tan² x dx.*0972

*Sec² is equal to 1 + tan², this is actually equal to 3 × the integral of, let me rewrite this.*0985

*The sec², I’m going to write as 1 + tan² x × tan² x × sec² x dx.*1000

*u substitution, I’m going to let u equal tan x.*1016

*I’m going to let du equal sec² x dx.*1020

*Therefore, that gives me 3 × the integral of 1 + u², 1 + tan², u is tan.*1028

*1 + u² × u² × du = 3 × the integral of u² + u⁴ du = 3 × u³/ 3 + u⁵/ 5 + c*1035

*which = 3 ×, u is tan(x), tan³ (x)/ 3 + tan⁵ x/ 5 + c.*1063

*Our final answer, we pull the values that we had.*1093

*The integral of sec⁵ x dx is equal to, we had a sec³ x tan x, that was - this thing.*1096

*-3 tan³ x/ 3 - 3 tan⁵ x/ 5 + c.*1115

*There you go, that is it, just manipulation, that is all we are doing.*1134

*Let us discuss our last set of problems that we might see under the integral sign.*1143

*How to deal with integrals of this type?*1148

*When you have the integral of sin(ax) cos(bx).*1151

*The integral of sin ax sin bx, or the integral of cos ax cos bx.*1156

*You can use the following identities.*1165

*Sin(ax) cos bx, sin and cos where a and b are different.*1169

*Is the same as ½ sin of a - bx + sin a + bx.*1174

*Sin(ax) sin(bx) is ½ cos a - bx - cos a + bx.*1181

*Cos and cos is ½ cos, a - + cos a + b.*1188

*That is it, just using these identities.*1194

*Let us finish off with an example.*1197

*Evaluate the integral cos 5x sin 9x.*1199

*Let us rewrite it to make it look like it was, with sin first, it does not really matter.*1208

*It is sin(9x) × the cos(5x) dx.*1216

*A is equal to 9 and b is equal to 5.*1231

*The identity we are going to be using is sin(ax) cos(bx) = ½ × sin of a - b × x + sin of a + bx.*1237

*This equals the integral, this thing = this thing.*1263

*½ the sin of a – b, 9 – 5, 4x + sin of 9 + 5 is 14x dx = ½ the integral of sin 4x dx + ½ × the integral of sin 14x dx.*1271

*This equals 1/8 - 1/8 the cos of 4x – 1/28.*1303

*½, 1/14, 14 come out.*1317

*Cos(14x) + c, that is it, just using the identities.*1321

*Thank you so much for joining us here at www.educator.com.*1329

*We will see you next time, bye.*1331

1 answer

Last reply by: Professor Hovasapian

Fri Jun 3, 2016 7:35 PM

Post by Acme Wang on May 27, 2016

Hello Professor,

1. When you wrote the process of finding integral secxdx, I caught a little error. When you wrote secx* (secx+tanx/secx+tanx), the result should be (secx^2+tanxsecx)/(secx+tanx) right?

2. Also, in Example II, I found a error. The integral(tanx*3(sec^2)x*secx*tanx*dx) should equal to 3integral(sec^3)x(tan^2)x*dx, right? But you wrongly wrote into (sec^4)x(tan^2)xdx.

Thank you very much!

Acme