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

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

### Integration by Parts

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
- Integration by Parts 0:08
- The Product Rule for Differentiation
- Integrating Both Sides Retains the Equality
- Differential Notation
- Example I: ∫ x cos x dx 5:41
- Example II: ∫ x² sin(2x)dx 12:01
- Example III: ∫ (e^x) cos x dx 18:19
- Example IV: ∫ (sin^-1) (x) dx 23:42
- Example V: ∫₁⁵ (lnx)² dx 28:25
- Summary 32:31
- Tabular Integration 35:08
- Case 1
- Example: ∫x³sinx dx
- Case 2
- Example: ∫e^(2x) sin 3x

### AP Calculus AB Online Prep Course

### Transcription: Integration by Parts

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

*Today, we are going to talk about the technique called integration by parts, let us jump right on in.*0004

*Let us recall the product rule from differential calculus.*0011

*We have the product rule for differentiation says f × g’ is equal to f’ g + fg’.*0019

*If I integrate both sides, I can retain the equality.*0049

*Integrating both sides retains the equality, as we know.*0058

*Anything we do to one side, as long as we do it to the other, everything is fine.*0067

*We have got the integral of fg’ = the integral of f’ g + fg’.*0071

*The integral of the derivative, this goes away.*0089

*That just becomes fg is equal to, and the integral operator is linear.*0093

*I can essentially distribute it out, it becomes the integral of f’ g + the integral of fg’.*0097

*Now I’m going to rearrange to get the following.*0107

*I get the integral of fg’ is equal to fg - the integral of f’ g.*0114

*This is our formula for something called integration by parts.*0129

*Again, do not worry we are going to actually explain this, much of this actually makes sense*0134

*once you actually see some examples.*0138

*In differential notation, it is going to look like this.*0141

*That is how we are actually going to be using it.*0144

*In differential notation, u is our function of x.*0147

*I’m going to arrange it this way.*0171

*u is going to be our f(x), it will make sense why I’m arranging this.*0172

*Our v is going to equal our g(x).*0178

*Our dv is equal to g’(x).*0185

*Never mind, let us stick to what we have got.*0194

*u is our f(x), v is our g(x), du is our f’(x), and dv is our g’(x).*0199

*Based on what we just wrote in the previous thing.*0213

*With that notation, what we have is the following.*0218

*We have the integral of u dv f(x) g’ fg’ is equal to uv fg - the integral of v du gf’.*0222

*This is the way we are actually going to use the integration by parts.*0244

*If we have an integrand, if we are faced with an integrand which is a product of two functions *0249

*and we can express u dv, in other words, this product, *0281

*if we can express u dv as some function 1 × some function 2, *0294

*then, we can solve the integral by using the right side of the equality.*0305

*Again, examples will make this clear.*0331

*Do not worry if you do not understand anything that I have just wrote symbolically.*0333

*Examples are going to make this immediately clear.*0337

*Let us jump right on in to our first example.*0341

*This is going to be the integral of x cos x.*0344

*It looks like we are missing a little bit of our top integral sign there.*0347

*We want to transform this to an easier integral.*0352

*We want to use integration by parts.*0356

*This is the integral of a function × another function.*0358

*One function is x, one function is cos x.*0362

*We want to use integration by parts to express as a simpler expression.*0365

*Something that we actually can integrate directly.*0390

*Here is how we are going to do it.*0407

*Let me go ahead and go to red here.*0410

*We have the integral of u dv = uv - the integral of v du.*0412

*Write the formula down and now we are going to use the formula.*0423

*Here I’m going to let u is equal to x dv.*0427

*This is our u dv, this is what we are doing.*0436

*We are looking at it as a u dv.*0438

*This is my u and this cos x, I’m going to call this my dv.*0441

*dv = cos(x).*0447

*What is du?*0453

*du is just 1, or 1 dx.*0455

*It is okay, we are using differentials.*0459

*What is v?*0460

*If dv = cos x, integrating this, what is the integral of cos x?*0461

*It is – sin x.*0468

*Here we are differentiating.*0475

*All I have is I'm looking at the integrand as some function, as some u and as some dv.*0476

*The u, I differentiate down.*0487

*The dv, I integrate up.*0489

*Now I use this side.*0491

*What I end up getting is the integral of x cos x is equal to this side.*0496

*They are the same thing, I can substitute uv.*0507

*Here is my u and here is my v, that is equal to -x × sin(x) - the integral of v × du.*0510

*v is -sin x × du.*0523

*du is just 1, that is easy.*0529

*This is – x sin x + the integral of sin x dx.*0534

*I get – x sin x, the integral of sin is,*0544

*I’m sorry, this is wrong.*0552

*The integral of cos is just sin.*0553

*When we integrated the cos x, the integral of cos x is sin x.*0567

*This is positive, this is positive, and this is positive.*0573

*We have used this side and we have actually created a simpler integral, based on our choice of u and dv.*0581

*The integral of sin x is just -cos x.*0593

*This becomes + cos x and +c.*0596

*With differentiation, the product rule is really simple.*0604

*It is just first × the derivative of the other + that one × the derivative of the first.*0607

*With integration, when you have a product of two functions, we do not have a complete closed form expression.*0612

*The way we do for differentiation.*0622

*But we can arrange it in such a way, such that the interval that we get on the right side*0624

*is a little bit easier than the original integral.*0630

*Something that we actually can solve, based on our choice of u and dv.*0633

*It is possible that the order is different.*0638

*Maybe this could have been the u and this could have been the dv.*0641

*Sometimes you are going to pick the wrong order and you realize that the integral you get is not any easier than the original.*0644

*You have to go back and switch the order.*0649

*Pick something else for u and pick something else for dv.*0652

*We will see some examples of that.*0655

*By integration by parts, by using integration by parts,*0660

*the original integral was equal to an expression to the right side, *0690

*where the integral was easier to handle.*0712

*That is it, that is really all we have done.*0715

*Let us try another example, this one is going to be the integral of x² sin 2x.*0721

*It seem to be missing the upper fourth of our integral signs.*0729

*In this case, let us go ahead and again write the expression.*0733

*Let me go ahead and do this in red.*0737

*We have the integral of u dv is equal to uv - the integral of v du.*0740

*Let us just go ahead and rewrite the formula over and over again, until it sticks in your mind.*0747

*We have to make a choice, this is a function x² × sin(2x).*0752

*Which one are we going to take for u, which one are we going to take for dv?*0756

*Pick one and hopefully you will get lucky.*0762

*I'm going to take u is equal to x² and I'm going to take my dv equal to sin(2x).*0764

*du is equal to 2x, and v, I’m going to integrate up.*0774

*The integral of sin 2x is equal to - ½ cos 2x.*0781

*I really hope to God that I’m getting these integrals correct.*0789

*Now we have got, let us use our formula.*0792

*The integral of u dv is equal to uv - the integral v du.*0795

*The integral of x² × sin (2x) dx is equal to uv which is x² × -1/2 cos 2x - the integral of v du.*0801

*v is -1/2 cos 2x × 2x dx.*0820

*Let us rewrite this, that = -1/2 x² cos(2x) +, the negative and negative becomes +, *0841

*the 2 and 2 cancel, + the integral of x × cos(2x) dx.*0856

*It does not necessarily look any easier, does it?*0872

*Because now all of a sudden we have this part which is nice, *0875

*but now we have another integral which is a product of a function and another function.*0878

*We are going to do another cycle of this.*0884

*Now we will just take this, we do a second round of integration by parts.*0886

*This time I’m going to set u equal to x.*0905

*I’m going to set dv equal to cos 2x.*0911

*The derivative of u = 1, and v, the integral of cos 2x is ½ sin 2x.*0918

*Therefore, this integral becomes just this integral.*0930

*It becomes uv - the integral of v du.*0937

*uv is going to be ½ x × sin(2x) - the integral of v du which is ½ sin 2x × du 1 dx.*0943

*Now let us take a look at this integral right here.*0978

*This is nice, this is just the integral of ½ sin 2x dx.*0985

*This integral, we can handle, now we can stop.*0992

*We take our first term.*0995

*This integral right here, the original integral is equal to this term which is -1/2 x² cos 2x + this which is this.*0996

*I will take this term + ½ x sin 2x.*1015

*I will go ahead and I will actually integrate and that.*1026

*The integral of ½ sin 2x is going to be + 1/4 cos(2x) + c.*1030

*That is my final answer from my original integral.*1051

*My original integral was a product.*1055

*I used u dv, I used the integration by parts formula to find this first expression right here.*1057

*This one was another integral where you had a product function × a function.*1065

*I did a second round of integration by parts and I have got this expression.*1071

*This time around, the integral in this expression, for the right side of the integration by parts formula,*1075

*happen to be a simple integral that we can solve.*1081

*We took this term and this term, and then we find the integral and we ended up with our final answer. *1083

*There you go, integration by parts.*1090

*Sometimes, integration by parts is going to require multiple rounds.*1092

*Let us do e ⁺x cos x dx.*1100

*Let us see what we have got.*1106

*Once again, let us write our integral formula.*1108

*The integral of u dv is equal to uv - the integral of v du.*1114

*I'm going to take my u equal to e ⁺x.*1123

*I’ m going to take my dv equal to cos x.*1130

*My du is going to equal e ⁺x.*1137

*My v is going to equal sin x.*1139

*Therefore, the integral of u dv is equal to uv – v du.*1145

*You know what, I will just write out the whole thing.*1155

*The integral of e ⁺x cos x dx = uv e ⁺x × sin x - the integral of v du e ⁺x sin x dx.*1156

*Again, we have a function of two.*1179

*We are going to do a second round here.*1182

*Once again, we are going to do a second round.*1184

*Let me do this second round in blue.*1190

*This time u is equal to e ⁺x dv is equal to sin x du is equal to e ⁺x v.*1196

*The integral of sin x is -cos x.*1207

*Therefore, this integral is going to yield, it is going to be uv - the integral v du.*1211

*uv – e ⁺x cos x - the integral of v du – v du e ⁺x cos x.*1219

*- and -, I will turn this into a +.*1237

*It is going to be e ⁺x cos x dx.*1239

*That is interesting.*1245

*What have I got?*1250

*Now I have got this thing which is a solution of this thing.*1257

*Let me bring back my first term.*1263

*I have the integral, the original integral of e ⁺x cos x dx is equal to this first term e ⁺x sin x - this integral *1265

*which is this thing – e ⁺x cos x + the integral of e ⁺x cos x dx.*1287

*Now I have this right here.*1304

*Negative and negative becomes +.*1310

*This negative distributes over here that becomes -.*1312

*We end up with the integral of e ⁺x cos x dx which is our original.*1315

*I get e ⁺x sin x + e ⁺x cos x - the integral of e ⁺x cos x dx.*1324

*Notice that and that are the same.*1339

*I’m going to move this over to that side because this is an equality.*1351

*I can move it over to that side.*1355

*I’m going to get 2 × the integral of e ⁺x cos x dx is equal to e ⁺x sin x + e ⁺x cos x, then I divide by 2.*1357

*The integral of e ⁺x cos x dx, I’m going to get a closed form expression.*1372

*e ⁺x sin x + e ⁺x cos x/ 2 + c.*1380

*There you go, I hope that made sense.*1389

*This time, what we did in the second round ended up giving me something that actually,*1393

*the integral was a repeat of the original integral.*1399

*Because it is an equality, I can just bring whatever term is here over to the other side *1403

*and divide by whatever constant is in front to leave me an integral, and then on the right side, a closed form expression.*1407

*I hope that made sense.*1415

*Let us see what we have got here.*1420

*Once again, let us go ahead and finish our integral.*1425

*The integral of inv sin x dx.*1428

*In this case, you notice that it is not a product, it is just one function, inv sin x dx,*1432

*but we can use integration by parts.*1437

*The integral, because there is no elementary integral for that.*1442

*Let us see what we have got.*1445

*The integral of u dv is equal to u × v - the integral of v du.*1446

*We write our formula.*1455

*There is a 1 here, let us try u is equal to 1.*1458

*dv = integral sin x, I did not really do anything because I can integrate this because that is what I’m trying to integrate.*1466

*That order is not going to work.*1478

*Let us try u is equal to inv sin(x) and dv is equal to 1.*1497

*This we can do, the du, we know that the derivative of the inv sin that is just 1/√1 - x².*1506

*v is equal to the integral of 1 is just x.*1515

*Therefore, the integral of inv sin(x) dx is equal to uv.*1521

*u × v is equal to x × the inv sin(x) - the integral of v du x/ 1 - x², under the radical dx.*1530

*This one, we can actually probably solve with just the u substitution.*1555

*Let me go to red.*1560

*I’m going to let u equal to 1 - x², du = -2x dx.*1561

*I have got x dx equal to du/-2.*1574

*Therefore, this integral is actually equal to, the integral of -du/2 × u⁻¹/2*1582

*= -1/2 the integral of u⁻¹/2 du, which is equal to -1/2 u ^½ / ½.*1602

*That cancels, I'm left with just u ^-½ which is equal to -√1 - x².*1625

*This integral is equal to that.*1640

*Therefore, our answer is the integral of inv sin(x) dx is equal to *1645

*x × inv sin(x) - this integral which is -√1 - x².*1660

*It is equal to x × inv sin(x) + √1 - x² + c.*1672

*Now we have an integral for the inverse sin using integration by parts.*1685

*Again, just because you have one function here, you can still use the technique of integration by parts *1691

*by choosing one of them to equal 1, usually that dv to equal 1 and seeing where you go.*1698

*Let us see what is next here.*1707

*The integral from 1 to 5, this is a definite integral of ln x².*1710

*Let me see.*1717

*Let us go back to black here, let us rewrite our formula.*1722

*The integral of u dv is equal to u × v - the integral of v du.*1725

*Again, I'm going to take this time on u is equal to ln(x)².*1735

*I’m going to take dv equal to 1.*1743

*du is equal to 2 ln x/ x.*1749

*v is equal to x.*1758

*It is kind of convenient.*1763

*Therefore, the integral from 1 to 5 of ln x² dx is equal to this, uv, u × v.*1765

*It is equal to x × ln(x)² - the integral of v du.*1780

*v du x × 2 ln x/ x dx.*1797

*That worked out really beautifully, = x × ln x² - 2 × the integral of ln x dx.*1806

*Let us see, might have to do a second round here.*1824

*Let us go to blue. We have got ln x, what is the integral of ln x?*1834

*Let us take u equal to ln x.*1838

*Let us take dv equal to 1, du = 1/x.*1843

*v is equal to x.*1851

*Therefore this is going to give us uv, u × v.*1854

*This is going to be x ln x - the integral of v du.*1859

*x × 1/x dx, this is equal to x ln x.*1867

*x/x cancels, the integral of 1 is just x.*1875

*Therefore, our final answer is equal to that x ln x² from 1 to 5,*1880

*this is a definite integral, -2 × this integral which is this thing x ln x - x from 1 to 5.*1906

*That is our answer.*1923

*Of course, you have to evaluate this upper and lower thing.*1926

*I will leave that to you.*1932

*But that is it, first application, second application, final answer.*1933

*And then, go ahead and work it out.*1938

*Let us see what is next.*1944

*The idea is always the same.*1951

*The idea is always the same.*1957

*Mainly, given the integral of f × g, choose an appropriate u and dv.*1965

*Either f is going to be u and g is going to be dv or g is going to be u and f is going to be dv,*1988

*whichever gives you the simplest integral on the right, *1996

*that you can either deal with directly or do another round, dv, and see where things go.*1999

*If you hit a wall, in other words, if you end up with an integral on the right which is intractable by any technique, *2028

*just switch original choice.*2040

*Just switch your choice of u and dv, and see where that one goes.*2052

*Again, the integral that you get on the right is either going to be directly integrable*2060

*or you would have to go through a second round.*2065

*Or you might have to use another technique, whatever it is, *2069

*u substitution or trigonometric substitution, whatever it is that you need.*2072

*That is pretty much what we are doing.*2078

*Now I want to introduce something called tabular integration.*2081

*From what it looks like, this is definitely not something that is actually taught anymore.*2086

*I always taught it when I learn calculus and I think it is a really nice powerful technique *2092

*to make your life a lot easier and a lot faster.*2096

*I will leave it up to you and I will leave it up to your teacher to decide*2099

*whether this is something that they will accept on an exam or something like that.*2102

*Let us talk about tabular integration.*2106

*There is a technique, this technique, greatly reduces the work involved in working with integration by parts.*2121

*We will do it with cases.*2153

*Case 1, when one of the functions that you choose, when one of the functions differentiates down to 0, *2155

*one of the functions under the integrand differentiates down to 0, and the other integrates up infinitely.*2175

*This is the first case, an example.*2197

*Now if you are faced with the integral, the integral of x³ sin(x) dx.*2204

*Let us go to red here.*2211

*This x³ will eventually differentiate down to 0.*2213

*x³, 3x², 6x, 6, 0.*2216

*Sin x, it integrates up infinitely.*2222

*Sin x - cos x -, it goes on and on and on.*2226

*Here is how we actually do this.*2231

*We let u equal to, this is really great.*2234

*Here is our u and here is our dv.*2242

*u is equal to x³, dv is equal to sin x, differentiate this straight all the way down to 0.*2246

*The first one gives you 3x².*2254

*The second gives you 6x.*2258

*The third gives you 6, the last gives you 0, integrate up.*2261

*1, 2, 3, 4, 1 differentiation, 2 differentiation, 3 differentiation, 4 differentiation, so integrate.*2268

*The integral of sin x is -cos x, the integral of -cos x is – sin x.*2275

*The integral of –sin x is cos x and the integral of cos x is sin x.*2284

*Now you put them together like this.*2292

*With a + sign, with a – sign, alternating.*2298

*With a + sign, with a – sign.*2304

*This, when you do integration by parts, you are going to end up doing 1, 2, 3 rounds, 3, 4 rounds.*2309

*If you use tabular integration, your final answer is going to be this.*2317

*It is going to equal x³ – cos.*2322

*You are going to put these two together.*2325

*It is going to be - x³ cos x.*2326

*Let me do blue, you are going to put this and this together with a + sign.*2332

*It is going to be, you are going to put this and this together with the – sign.*2338

*- and - becomes + 3x² sin x.*2342

*This and this, +6x cos x.*2348

*And then, this and this, -6 sin x + c.*2355

*That is it, look at how much time I have saved.*2365

*I probably saved about 10 minutes of work.*2367

*Tabular integration, when you have one function, when one of the function is under the integrand *2369

*actually differentiates down to 0, the other one integrates infinitely, go ahead and do that.*2374

*Differentiate it down to 0, that many times, integrate that many times up infinitely.*2380

*And then, connect the original with the next one down, the original with the next one down, *2385

*the original with the next one down, the original until you reach the end of this one.*2390

*And then alternate signs, + -, +-, that gives you what you would have gotten,*2395

*if you had gone through four rounds of integration by parts.*2399

*This is tabular integration and it works really beautifully.*2405

*The first case is when one of the functions differentiates down to 0 and the other integrates infinitely.*2408

*Case 2, I actually could have presented this as one case, which I probably should have, now that I'm looking at it.*2416

*But that is okay, I will do it like this.*2422

*Case 2, this is when both functions,*2428

*I better write it out explicitly.*2442

*When one function differentiates infinitely and the other integrates infinitely.*2447

*Here is an example, example that we are going to be looking at is the integral of e ⁺2x sin 3x.*2474

*Here is how we deal with this one.*2491

*Let us go ahead and go to red.*2493

*I'm going to let u, I’m going to have dv.*2495

*u, I’m going to call e ⁺2x u, it is my choice for u.*2500

*sin(3x) is going to be my dv.*2505

*You are going to differentiate, they differentiated and integrate infinitely.*2511

*The question is where do you stop?*2514

*Before, you can stop at 0, you cannot go any further.*2515

*Here is how you are going to do it.*2518

*Differentiate, this is going to be e ⁺2x.*2520

*Integrate, this is going to be -1/3 cos 3x.*2524

*Differentiate, it is time to go back and fourth.*2535

*This is 2e ⁺2x, my apologies. *2538

*When we differentiate again, this is going to be 4e ⁺2x.*2542

*When we integrate again, we are going to get -1/9 sin(3x).*2546

*Here is where you stop.*2559

*Stop when you have integrated dv to a point where you have a version of the original dv.*2563

*Our original dv was sin 3x.*2599

*I integrate, I integrate, I have sin 3x again.*2603

*It is -1/9 sin 3x but it is still sin 3x.*2606

*It is a version of the original, that is when you stop.*2609

*What do you put together, here is what you put together.*2613

*You do the same thing, except one little extra of this + alternating this - the last one is this + with an integral sign.*2616

*Here is what it looks like.*2636

*It is going to be -1/3 e ⁺2x cos 3x + 2/9, e ⁺2x sin 3x.*2638

*Put these two together under an integral sign.*2669

*It is going to be, I will just write this as + the integral - 4/9 e ⁺2x sin 3x.*2673

*Notice what you have got, you have e ⁺2x sin 3x 4/9.*2691

*The original was e ⁺2x sin 3x.*2697

*You can do that where you move it over to the other side and you divide.*2700

*What you get here is the following.*2704

*We have the original e ⁺2x sin 3x and we set that is equal to -1/3 e ⁺2x cos(3x) + 2/9 e ⁺x sin 3x - 4/9.*2710

*The integral of e ⁺x sin 3x dx.*2739

*This and this, because this is an original version of that, you can move it over here.*2746

*What you are going to end up with is 4/9 + 9/9, you are going to end up with 13/9 ×*2754

*the integral of e ⁺2x sin 3x dx is equal to -1/3 e ⁺2x cos 3x + 2/9 e ⁺x sin 3x.*2773

*Of course, you divide by 13/9, and then you get your final answer.*2793

*Your final answer is going to be the integral of e ⁺2x sin 3x dx is going to equal,*2798

*After you divide by that, you are going to end up with -3/13 e ⁺2x cos 3x + 2/13 e ⁺x sin 3x + c.*2810

*There you go.*2828

*Now case 1 is just a special case of case 2, here is why.*2831

*Remember with case 1, what we had was this.*2850

*We had u and we had dv, and we had x³, and we had sin(x).*2854

*This goes to 3x², this goes to 6x, this goes to 6, and this goes to 0.*2861

*The sin x goes to -cos x, -cos x goes to -sin x.*2871

*-sin x integrates to cos x, and cos x integrates to sin x.*2879

*We said, what you are going to take is the following.*2886

*You are going to take this and that, with a + sign.*2889

*This and that, with a – sign.*2894

*This and that, with a + sign.*2896

*This and that, with a – sign.*2899

*The final one, it is going to be this and that, with a + sign under the integral sign.*2901

*Because case 2, it was always the last two.*2909

*As long as you have reached some version of the original dv,*2914

*that last thing that you get is multiplied by this and it falls under the integral sign.*2919

*If I wrote this out, I would get the following.*2927

*I would get x³ - x³ cos x, that takes care of that and that.*2929

*I would get + 3x² sin x, that takes care of that and that.*2938

*I would get + 6x cos x, it takes care of those two.*2946

*I would get -6 sin x, that takes care of those two.*2952

*I would get + the integral of 0 × sin x dx.*2957

*The integral of 0 does not really matter.*2963

*It is 0, it is a constant, when you integrate that.*2971

*There is your constant.*2975

*This case 1 is really just a special case of case 2.*2977

*Differentiate down to 0, integrate up.*2984

*Put them together, alternating signs.*2987

*Differentiate, integrate up until you end up getting a version of the original dv.*2992

*The last one, you put together under the integral sign.*2998

*This is tabular integration, I use it all the time.*3002

*Case 1 is the one you actually use most often.*3007

*Case 2, I do not actually use case 2 all that often.*3009

*I tend to use it more and more now.*3015

*This whole last part, I do not know, some people are a little weary of that but it actually does work.*3018

*I would not worry about it all that much.*3024

*In any case, I hope that helps.*3026

*Take good care and thank you for joining us here at www.educator.com.*3028

*We will see you next time, bye.*3031

1 answer

Last reply by: Mohammed Jaweed

Fri Mar 18, 2016 12:08 AM

Post by Mohammed Jaweed on March 18 at 12:07:14 AM

The integral of cos(x) is not -sin(x). It is sin(x).

3 answers

Last reply by: Professor Hovasapian

Thu Jan 7, 2016 11:01 PM

Post by Gautham Padmakumar on January 2 at 09:23:42 PM

at 2:24 for the differential notation, why do we write it as du = f'(x)

why not du = f'(x) dx

thanks! helpful videos as always. also curious as to when the Calculus BC course by you will be uploaded? :)