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INSTRUCTORS Raffi Hovasapian John Zhu
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Lecture Comments (6)

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? :)

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

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 sense0134

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 side0624

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⁻¹/21582

= -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 integrable2060

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 decide2099

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