AP Calculus AB Solving Integrals by Substitution

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

Today, we are going to introduce a technique called solving integrals by substitution.0004

This is a very powerful technique.0009

For all practical purposes, for the rest of you, the work that you do as an engineer or physicist or mathematician,0012

this solving by substitution is probably the one technique that you are going to use more than any other,0018

if you are going to do anything by hand.0023

Most of the time, you are going to be using software to do it.0027

But this technique is really what we use, most of the time.0028

Let us jump right on in.0033

I’m going to do this lesson not by doing theory because the theory will make sense by doing the examples.0036

It is the examples that I think make it clear, what it is that is going on.0045

If we just sort of launch into theory, this is one of those things0050

where I think the symbolism gets in the way of what is actually happening which is very simple.0053

I do not want to off you skate that.0057

Let us take a look at what it is that we do know, as far as antiderivatives and solving integrals.0062

This gives us a list of the antiderivatives and integrals that we know how to find.0066

Pretty straightforward.0072

We know how to do polynomials, things with powers.0074

We know how to do basic trigonometric functions, things like this.0077

However, we are going to be dealing with functions that are a lot more complicated than this.0080

We want to be able to integrate those functions.0085

In order to integrate those functions, what we are going to be doing is using the substitution rule,0088

in order to transform these complex integrals into something that looks like one of these.0093

This is our home base, we are going to transform them into something that looks like one of these, by using the substitution.0098

We are going to integrate it because we know how to do these.0105

And then, we are going to substitute back and get back our final answer.0108

Having said that, let us jump right on in with some examples and I think it will make sense.0115

Evaluate the following indefinite integral, the integral of tan(x) dx.0123

We know that we do not have a basic integral for tan x dx, it is not in that list.0129

Let me go ahead and write that out.0138

I think I will work in red here.0139

We do not have a basic integral or antiderivative whichever you prefer, the derivative for the tan(x).0147

The question is how can we transform this and turn it into something we recognize, that we can integrate.0170

How can we transform this into something we recognize, something on our basic integral list?0174

A lot of this substitution technique is going to be based on just start trial and error.0198

You remember back in pre-calculus, when you are doing those identities,0203

when you have to prove that two identities are equivalent.0205

Sometimes you might start on a path and you end up hitting a wall, not a problem.0208

There is no way of knowing beforehand until you gain some experience,0213

what to choose for your substitution how to go about this.0218

You are just going to try things and follow the mathematical logic.0220

If you end up at some place which is good, you are good.0225

If you end up hitting a wall, just go back to the beginning, like everything else in life.0227

If you lose your way, just go back to the beginning, that is it, nice and simple.0231

Do not think that you have to just look at it and know exactly what to do.0235

You are going to try different things.0238

Calculus is sophisticated mathematics.0240

We do not just look at it and see, we know what to do.0242

Let us go ahead and change this around.0247

We know what tan x is, in terms of the basic functions sin x and cos x.0249

I’m going to write this as the integral of sin x.0256

Let us make this a little clearer.0260

The integral of sin x/ cos x dx.0262

I’m going to do a substitution, I'm going to call cos x and I’m going to call it u.0270

Let me do this in red.0280

I'm going to call cos x u.0284

I'm going to take du dx, that equal cos x, I'm sorry.0288

That is not equal cos x, the derivative of cos x du dx is -sin x.0294

I’m going to write du is equal to -sin x dx.0301

All I have done is I have moved this over here.0306

This is a ratio, I can just move that over.0308

Now I have du = - sin x dx.0311

I’m going to move the negative sign over and I’m going to write sin x dx is equal to –du.0315

I’m going to plug those back in.0323

Sin x dx, that is right there, that is sin x dx.0327

This is the integral of, I’m going to put this in for that – du.0333

The denominator, I have cos x.0341

Cos x is just u, that is that.0343

I will pull the negative sign out and I will get du/u.0349

I have transformed this into something I recognize.0353

I recognize that in the list in the previous page, I know that dx/x, the integral of that is just ln x.0356

The variable does not matter.0364

du/u, dx/x, dz/z, dp/p, it is the same thing.0367

It is just a symbol.0373

I have transformed it by making an appropriate substitution.0374

I have called something, something, I have taken the derivative of it.0379

By doing that, I ended up having something which looks like,0382

Now I put that back in, I have transformed it into something I recognize.0387

I just go ahead and integrate this.0391

-the integral of du/u = the natlog of the absolute of u + c.0395

I substitute back in, u is equal to cos(x).0404

ln of the absolute of cos x + c, then I have my negative sign, and I’m done.0412

That is all I have done, taken something, I have transformed it and then I have made a u substitution.0421

Transformed it into something I do recognize, for which I do have an antiderivative, an elementary antiderivative.0431

I have taken the antiderivative and I have substituted it back to turn it back into a function of x.0436

Do not worry, we will do plenty of these.0445

Evaluate the following indefinite integral, let us go to blue.0451

Let us do just a little bit of trial and error here.0457

I need a substitution, I need to set u this variable, substitution that I’m making to equal something in here.0460

Let us do a little trial and error.0471

I’m going to let u = cos x.0480

My choices are, I can let u = cos x, I can let u = sin x, or I can let u = sin³ x.0486

Just to see if I get something that works.0492

If I let u = cos x, du is equal to -sin x dx.0495

-sin x dx = du, is there a -sin x dx in here?0506

No, there is a sin³ x.0510

This does not really work too well because we need sin³ x to be some differential du.0512

Let us try something else.0540

Let us try u = sin x.0542

du = du dx = cos x which means that du = cos x dx.0553

Do I have a cos x dx?0569

I do, there is a cos x dx.0571

I’m going to let that equal u.0578

Now I have the integral, cos x, that is this part.0582

I’m going to call that du, I’m going to put that in here.0585

u is equal to sin x.0592

This is sin³ x, that is just u³.0595

I’m going to rewrite it in the order that I’m accustomed to.0600

The function and the differential, u³ du.0603

I recognize that, I can integrate that.0606

The integral of that is u⁴/ 4 + c.0608

U is equal to sin x, it becomes 4x/4 + c.0617

That is it, nice and simple.0628

I choose my appropriate u, I see if it gives me something that I can convert into, that I recognize.0631

I chose u = sin x, I got down to du = cos x dx.0637

Yes, cos x dx is in the integrand, I will call that du.0642

For this part, I call it u³.0646

This is now an integral that I do know how to solve.0649

It is an elementary integral.0652

I solve it then I substitute back.0653

Evaluate the following indefinite integral.0658

We have a couple of choices, we can let u = x, we can let u = ln x, we can let u = ln³ x of ln(x)³.0662

Again, experience will dictate what choices you are going to make for your u.0671

In this case, I’m just going to go ahead and go directly.0675

u = natlog(x).0677

du dx is equal to 1/x, which implies that du is equal to dx/x.0682

dx/x, here is my dx/x, this is equal to, the integral of,0698

this is my dx/x is my du, I will put the du there.0706

I have ln x³, ln x is u, it is just u³.0711

I ended up with the same thing, u³ du.0716

I will write it in a form that I’m used to, which is the function first and the differential afterwards.0720

This is just u⁴/ 4 + c, u is ln x, it becomes ln x⁴/ 4 + c.0725

I’m done, this is the integral of that via this process of substitution.0740

Turning it into an elementary integral that I do recognize.0746

Evaluate the following indefinite integral.0756

This time I'm going to let u equal to x ^½, that is just √x du dx.0758

I will just go ahead and do it straight.0770

I will put the dx over there.0773

du dx = ½ x⁻¹/2 which means that du = ½ × x⁻¹/2 dx, which is equal to 1/ 2 × √x dx.0779

I’m going to bring this 2, multiply it out over here.0807

This is going to be 2 du = dx/ √x.0810

Here is my dx/ √x, that is equal to 2 du.0819

I’m going to write 2 du.0823

Cos(√x), √x is equal to u, this is cos(u).0829

I’m going to pull the 2 out, 2 × the integral of cos u du, that is an elementary integral, I know how to do that.0837

That is equal to 2 × sin(u) + c.0845

I know what u is, u is equal to √x.0852

Therefore, my final answer is 2 × sin √x + c.0855

The integral of cos x/ 1 + sin² x.0872

u could be cos x, u can be sin x, u could be 1 + sin² x.0875

Let us see what works, I’m going to go ahead and take u is equal to sin x.0881

Again, there is no way for you to know beforehand, what u was supposed to be.0886

Try things, if you try it and it works out well, then you can convert it an integral of the elementary type.0891

You are good, if not, then start again, pick another u.0897

It is that simple, that is all.0900

You are not supposed to just look at it and know.0901

Experience will dictate, looking at it, and knowing eventually.0904

Which is why I just immediately chose sin x.0908

du = cos x d(x), cos x d(x) = du.0912

Here is my cos x d(x), I’m going to rewrite that as the integral of du/ 1 + sin² x.0922

Sin is u, this is 1 + u².0932

This is an elementary integral, du/ 1 + u², that is equal to the inv tan(u) + c.0939

u is sin x, I substitute back in, the inv tan of sin(x) + c, I’m done.0947

Evaluate the following, we do the same thing, except now we actually evaluate the integral of the n.0961

Because now it is a definite integral from 0 to 2.0968

Do the exact same thing that you did.0970

Now instead of c, you are just going to evaluate the integral, f(b) – f(a).0972

u is equal to x² + 2, du is equal to 2x dx.0980

I’m just going to do the whole thing.0991

du/dx is equal to 2x which means that du is equal to 2x dx.0996

I have an x and dx, I’m going to move the 2 over.1007

du/2 is equal to x dx.1010

My x dx is right here, this is going to be the integral from 0 to,1014

I’m not going to put those in yet.1021

I’m just going to do the integral, first part.1023

x dx that is my du/2 × u which is x² + 2.1025

x² + 2 ⁺15, it is u ⁺15.1033

I'm going to pull the ½ out and I’m going to write this as u ⁺15 du, switch the order,1037

in order to turn it into the order that I’m accustomed to which is integrand times the differential element.1044

It will always be that way, there are books that actually put the du first, especially in science.1051

Do not worry about the order, the order itself does not matter.1057

This is just multiplication.1059

Now I solve it, nice and straightforward.1062

This is just equal to ½ × u ⁺16/ 16.1065

I put the u back in, I get ½ of x² + 2 ⁺16/ 16.1085

I evaluate that from 0,2.1102

That is it, I just put my values in.1107

I’m going to leave the ½ out for a minute.1109

I put 2 into here, it is going to be 2 × 2² is 4, 4 + 2 is 6.1112

It is going to be 6 ⁺16/ 16 -, then I put a 0 in.1120

It is just going to be 6 ⁺16/ 32, nice and simple.1129

Evaluate the following, what do we pick for u?1144

This is the integral from e ⁺e³ of 1/ x × 3√ln x, this is not x³.1148

This is x × 3√ln x.1156

I’m going to let u equal to ln x which means that du dx is 1/x, which implies that du is equal to dx/x.1160

There we go, I have got my dx/x, right here.1176

This equals the integral of, I will write my du.1183

This 3√ln x, the 3√ln(x) is just ln(x)¹/3.1192

1/ ln(x)¹/3 is nothing more than ln x⁻¹/3.1205

Therefore, 1/ ln x¹/3 is ln x⁻¹/3.1218

Ln x is equal to u, we are dealing with u, this is just u⁻¹/3.1225

I will write in a form that I’m accustomed to, u⁻¹/3.1233

du is going to equal u²/3 / 2/3 which is going to be 3/2 u²/3, which is equal to 3/2.1241

u is equal to ln x, it is going to be 3/2 ln x²/3.1260

I'm going to evaluate it from e ⁺e³, that is going to equal 3/2 ln(e³) is just 3.1270

It is 3²/3 – ln (e) is just 1 - 1²/3.1287

It is just equal to 3/2 × 3²/3 – 1.1298

Evaluate the following, nice and simple.1311

u can equal inv sin(x), u can equal 1 - x², or u can equal √1 - x², which one do we choose?1315

This is going to be trial and error.1323

That will save a little time, in terms of the trial and error, I'm just going to go ahead and say u is equal to inv sin(x).1326

du dx = 1/ 1 - x² under radical which means that du is equal to 1/ 1 – x² under the radical × dx.1337

There we go, I have my dx × √1 – x².1357

This is equal to the integral of my du, this is du.1363

Inv sin(x) is u.1370

This one is easy, u du this is equal to u²/ 2.1373

u is inv sin(x)²/ 2.1383

I’m going to evaluate that from 0 to 1.1394

This is going to be inv sin of 1²/ 2 - inv sin of,1401

I will slow down and write everything out properly.1426

Inv sin(0)²/ 2.1429

The inv sin(1) is π/2.1438

It is going to be π/2²/ 2.1444

The inv sin(0) is 0.1452

My answer is just π²/4/ 2.1456

This is going to be π²/ 8.1463

That is it, nice and straightforward.1471

Nothing strange happening here.1473

Evaluate the following, sin x × cos of cos x dx.1477

In this case, I'm going to let u equal cos x.1484

I'm going to let du dx = -sin x, which means that du = -sin x dx,1490

which means -du = sin x dx.1504

Sin x dx, here is my sin x dx.1509

I’m going to write the integral, this is my –du.1514

I have u is equal to cos x, it is cos(u).1520

I rewrite that by pulling the negative out, putting the cos u first, du.1527

I know how to do that, that is just equal to –the integral of cos is sin u.1533

This is going to be -sin of cos x because u is cos x.1542

I get –sin(cos x) evaluated from 0 to π/2.1551

I will go ahead and leave it like this which is equal to -,1564

cos(π/2), I put this in for here.1572

cos(π/2) is 0, the sin(0) is 0, that is 0.1576

cos(0) is 1, sin(1) is 0.841.1582

I hope that make sense.1597

The whole idea behind u substitution is to find some appropriate u.1599

Take the derivative of it, move the dx over and see if you can find something that actually matches this,1605

so we can start substituting back in and turning the variable into everything in terms of u,1617

some elementary function of u.1623

Once we have an elementary function of u, we just integrate it.1626

Once we get our final answer, we substitute back and turn it back into a function of x.1630

At that point, we can either have a c, stop, we are indefinite integral, or evaluate the integral itself.1636

Thank you so much for joining us here at www.educator.com, we will see you next time, bye.1643