| Start | Bonnie's not here today, we're really sorry.
She's getting tested in the hospital. She's ok, right Michelle? Yeah. You promise? Yes. Ok. Did we all like the puppy picture that I posted though? That was fun. Alright. I think I did this one. Right? Let's do a different one. |
| 0:38 | Ok. x^2sin(2x)dx. Bless you.
Why don't you integrate that for a couple minutes. While I walk around. |
| 3:51 | You were a little too lit this weekend.
|
| 6:00 | I'll give you guys 2 more minutes.
|
| 7:40 | Can I wave in this camera? Will he see me?
Sup. Come support women's tennis. |
| 8:15 | Alright I gave long enough.
So, integration by parts You wanna make sure when you look at an integration by parts, and you have x to a power, and the other term is sine, cosine, or e, you're gonna have to do the integration by parts that many times. |
| 8:32 | Ok? So if you have x^3 e^x you're gonna have to do 3 rounds of integration by parts.
If you have x^5 cos(x) you'll have to do 5 rounds. You'll have to keep doing it until you reduce x down to a constant. Until you, keep taking, every time you take the derivative, the power of x drops by 1. So you have to do it that many times. We would never give you x^5, that's just way too much work. I think squared is enough for these kind of problems. Now if you had x^5 and natural log of x it's not so bad. |
| 9:02 | So we'll do that in a second.
Alright, so here what we're gonna do is, remember we wanna, we're gonna use the formula. The integral of udv is uv minus the integral of vdu. That's our formula. So, we look in here, we say we have a function and we have a derivative of a function. And we're gonna pick one part and call it the function. And the other part will be the derivative of the function. We're gonna put them together in the formula for the integration. |
| 9:31 | So we look and we say, well let's say u
is a function. dv is the derivative of a function.
Then du is 2xdx. v is -cos(2x)/2. Couple things to remember. You're differentiating this way. You're integrating this way. When you have a constant, you're gonna do the antiderivative. When you have an integral, you divide by the constant instead of multiplying by the constant. |
| 10:03 | Ok?
So if you we're doing the derivative of sin(2x) you would have to multiply by 2, since you're doing the antiderivative of sin2x you divide by 2. Ok? So now we plug it into the formula. So this is our u. This is our dv. So now we're gonna want the integral of x^2 sin(2x)dx equals u times v. So (-x^2 cos2x)/2. |
| 10:33 | Minus the integral of vdu.
So minus minus becomes plus. The integral of v, (cos(2x))/2 times du, which is (2x)dx. That simplifies to (-x^2cos(2x))/2 plus the integral of xcos(2x) |
| 11:04 | dx.
Ok? That's round 1. You've done integration by parts and you took x^2, which was a problem and turned it into x. So far so good? Well here we have a minus. And remember we're doing minus the integral, so the minus minus becomes plus. So as I've said, be very careful with constants and be very careful with minus signs. |
| 11:32 | Cause it's very easy to mess that up with this stuff.
Ok? You can always check when you're done. Take the derivative of what you've found and see if you get x^2 sin(2x). Ok. Alright, so now we're gonna do a second round. So I'm just gonna rewrite this up here. We got (-x^2 cos(2x))/2 + integral of xcos(2x)dx. |
| 12:02 | Alright, so let's do integration by parts again. Let's let u=x
dv=cos(2x)dx.
du is dx, because you take the derivative. v is the antiderivative, so v is (sin(2x))/2. Don't do that. Ok. |
| 12:31 | So now, this integral, use integration by parts.
So we've got (-x^2 cos(2x))/2 plus u times v which is (xsin(2x))/2 minus the integral of vdu. So minus the integral of this times this. Which becomes minus integral of (sin(2x))/2 dx. |
| 13:04 | This chalk is skipping.
Or to rewrite that a tiny bit, this is (-x^2 cos(2x))/2 + (xsin(2x))/2. Stray chalk mark. minus 1/2 |
| 13:30 | integral of sin(2x) dx.
Almost there. Now you just have to integrate sin(2x) which does not require integration by parts. So this is (-x^2 cos(2x))/2. This is (xsin(2x))/2 And this is -1/2 |
| 14:03 | of -cos2x
over 2
and then don't forget plus C. So, the minus cosine makes this plus.
You can combine the 1/2 and the 2 and the 4 in the denominator if you needed to. Can everybody see that down there? How'd we do on that one? Anybody get it? Yay. |
| 14:30 | Ok.
When I need the knee surgery, you're the people I call. Alright, we're gonna practice this again, and this time everybody's gonna get one. Ok? How'd we do with that, alright? Got the idea? Let's do some more of these. I'll leave that there for a couple minutes for you to copy. Let's do x^5 lnx dx. |
| 15:03 | And as a clue u has to be that one.
dv has to be that. Ok? |
| 20:34 | Ok, we're ready? Do some erasing.
See now when we turn these videos into slightly better videos, we cut out all of this dead time. And then you'll just say, you know, pause and maybe we'll put the jeopardy music in, and then you'll get to do the question. I don't think we're allowed to use the jeopardy music but we can come up with some variation of that. Maybe a little Kanye. |
| 21:04 | Spice girls.
I'm sure we can think of something. We'll have the Bieb. I think Kayne's appropriate for my videos. Right, Kayne for Kahn. Right. It works. Alright. |
| 21:30 | So, I told you what u is.
When you have integration by parts This is sort of your order of priority, ok? These would be u. So if you have natural log of x in the integral that one is gonna be u. If you don't have natural log, power of x would be u. |
| 22:00 | If you don't have either natural log or x, make e^x equal to u.
And I don't think we will ever give you a problem where sine or cosine will be u. Ok, now if there's inverse trig I don't think we're covering those. Inverse trig goes here. And I'll have to ask her if she's gonna throw in one of those. Ok, so if you have like, tan inverse of x or something. Cause those are things you wanna differentiate, you don't wanna integrate inverse tan. |
| 22:30 | But you know the derivative of inverse tan because you memorized that last semester.
And when you memorize things you don't forget them, right? Ok, like the unit circle. So let's let u equal the natural log of x. dv= x^5 dx. So derivative of natural log of x is 1/x. dx. And the antiderivative of x^5 is (x^6)/6 |
| 23:05 | No plus c. Alright
we substitute in
get x^5 ln(x) dx is u times v
(x^6)/6 lnx minus integral of vdu.
Which is (x^6)/6 times 1/x |
| 23:31 | dx.
Now, this could be reduced. So this is (x^6)/6 lnx minus integral of x^5, pull out the 1/6 dx. When you're doing integration, when you have constants inside the integral take them out of the integral. Ok, clean up your integrals as much as you can. And the integral of x^5 is (x^6)/6 so this becomes |
| 24:04 | (x^6)/6 lnx - (x^6)/36 +c.
You get that one? Sort of? Close? Mess up the constants? Miss a minus sign? Yes, ok let's do another one. Now, this is the harder one that I did at the very end of class last time. |
| 24:30 | Ok, which was, I think I did e^x cosx.
Anybody have their notes from last time? It was either cosine or sine. Good. So, let's have you guys practice. Ok, so this is the type where you're gonna have to do integration by parts twice. And then you're gonna add back to the other side. |
| 31:45 | Alright, I think everybody's suffered enough.
So this is sort of the most annoying type. I'm gonna rewrite it over here so I have lots of room on the board. |
| 32:07 | So this is what we're gonna do.
We're gonna do some integration by parts, then we do it again, then we do it again, and again forever. So this is the type where you're going to do it twice and then you're gonna use a little trick to make it work. How will you recognize that? You'll have e to a power, then sine or cosine. |
| 32:34 | Those types, ok?
So if you have an integral of the form e^kx sine or cosine to the, I don't know, jx dx, that'll be this type. Ok? There's some other weird versions of that but don't worry about it for now. Like secant cubed. That's always an entertaining one. Alright. So let's let u=e^(2x) |
| 33:00 | dv=sinxdx.
I could've made it even more annoying and done something like sin(πx). That would really throw you guys. But I wouldn't do that. Ok, so du is 2e^(2x)dx. And v is -cosx. So this integral, e^(2x)sinxdx equals u times v which is -e^(2x)cosx |
| 33:40 | minus the integral of vdu, so minus minus becomes plus
2 integral e^(2x)cosxdx.
2 comes outside, minus minus becomes the plus. So far so good? So as I said on Wednesday for the small number of you who were in attendance |
| 34:02 | you look at this and you say I haven't improved my situation at all.
All I did was go from e to the something sine, to e to the something cosine. That hasn't made it any better. Right? What did we determine cosine means? What you do with a contract, right? You cosign a contract. Alright. So let's do it again. This time let's let u=e^2x dv is cosxdx. |
| 34:34 | du is, wait I made a mistake.
This is e^2x over, no it is right, never mind. 2e^(2x), good I'm glad I didn't make a mistake. And v is sinx And let's run this again. So integral of e^(2x)sinxdx equals -e^(2x)cosx + 2 times u times v |
| 35:08 | e^(2x)sinx, minus the integral of vdu
This looks like a minus sign.
So that becomes minus integral of 2 e^(2x)sinxdx |
| 35:30 | So you clean this up a little bit.
You get integral of e^(2x)sinxdx. = -e^(2x)cosx + 2e^(2x)sinx -4 integral e^(2x)sinx dx. And now you look at this and you say I'm back where I started. You say how can I be back where I started? That doesn't do me any good. |
| 36:03 | But notice, you're subtracting 4 of that integral
and you have 1 of that integral over here.
So if we take this and we add it to the other side we'll have 5 integral e^(2x)sinx dx = -e^(2x)cosx + 2e^(2x)sinx. |
| 36:32 | So far so good?
Then you just divide by 5. You get integral of e^(2x)sinx dx is -e^(2x)cosx + 2e^(2x)sinx over 5 plus a constant. I hope you people in the corner can see that. Ok? So you just take this and you divide it through by 5. |
| 37:00 | And you throw in the plus c at the end.
So as I said, so it's sort of a trick. Because you say to yourself, well I never really integrated it. But you did, ok. You did a couple iterations of integration by parts. And then you'll know you did it correctly when you have the original integral but it's subtracted on the other side. So if you have a plus sign here you messed up your minus sign somewhere in the middle. Ok, you have to have a minus here so you can add it back to the other side. |
| 37:33 | So far so good?
Yes. Sure. Remember so, I figured the first round of integration by parts I had a 2e^(2x)cosx. So I do a second round and where I had the e^(2x)cosx I now have e^(2x)sinx-2 integral of e^(2x)sinx. So I distribute the 2 and I get -e^(2x)cosx |
| 38:02 | +2e^(2x)sinx
2 times -2 is -4 integral of e^(2x)sinx.
Then I add 4 of these to both sides. So, if this is, you know 4 I for integral, and I have to add +4I to both sides. Which would give you 5 of the integral. Ok? A lot of people have trouble with this type. |
| 38:31 | And when you put it on the exam that means you have to grade it.
So I was never excited about putting them on the exam. But we do them. Alright. One last one. Which we'll do as a team. That's enough for the moment. |
| 39:09 | Integral of inverse tan also known as arctan.
dx So now I have to let u=inverse tan |
| 39:30 | du is 1/(1+x^2) because we remember that.
Because we remember everything we study. Me neither. dv is just gonna be dx. Cause remember it's really 1dx. And v is gonna be the integral of dx which is x. Because the derivative of x is 1dx. So far so good? Ok, now let's do integration by parts. |
| 40:00 | I have the integral of tan inverse of x.
This chalk is skipping. Equals u times v, xtan^-1x minus the integral of vdu which is x/(1+x^2)dx. Now how do we integrate that? |
| 40:34 | Any ideas?
Don't do integration by parts, how do we integrate that? Substitution. Let u=x, I'm sorry let u=1+x^2 du is 2xdx. |
| 41:01 | Don't confuse these u's with those u's. You can use a different letter if you want.
Maybe a "w" or something. So 1/2du x dx So this then becomes integral of x tan^-1 x minus 1/2 integral of du/u. That's a u. |
| 41:37 | You all see that?
And the integral of du/u is just natural log. So this becomes x tan^-1 x -1/2 ln(u) +c and we plug back in. And since 1+x^2 is always positive we don't really need the absolute value bars. x tan^-1(x) -1/2 ln(1+x^2) + a constant. |
| 42:08 | Ok?
We're lost on that one? Really? Do it again. I let u= tan^-1 du is 1/(1+x^2) dv is dx so v is x So now I have u times v which is this. Minus the integral of vdu which is x/(1+x^2). |
| 42:33 | Ok?
Now if I wanna integrate that I use u substitution. Not integration by parts. How do I know? Because the derivative of x^2 is x. So I look at this and I say I have a function and the derivative of that function is in the integral. So u is 1+x^2, du is just 2xdx. The derivative, then I put the 2 on the other side I get 1/2du is xdx. And I substitute. |
| 43:00 | So this becomes xdx becomes 1/2du, and 1+x^2 becomes u.
And then the integral of du/u is just natural log. Don't forget u substitution. Ok? So keep practicing. Next type of integration technique. Or types. There's lots of types of integration, I mean you could just spend all day doing this stuff. |
| 43:31 | What would be a fun one?
How about We're gonna take a detour for a second. And remind you of more fun stuff that you've learned in the past. |
| 44:01 | Ok.
You guys remember the double angle formula? You're supposed to. Ok, double angle formula, the cos2x was cos^2x-sin^2x It's also 2cos^2x -1 1-2sin^2x. Those are all cos2x. |
| 44:31 | You can take that formula, rearrange it a little bit, and isolate the squared terms. We don't care about this one.
Ok. If you rearrange that you'll get that cos^2(x) is 1+ cos2x all over 2. Or you could break it up. And the sin^2(x) is 1-cos2x, divide everything by 2. |
| 45:11 | Ok?
Those formulas are very useful. Because now we're gonna do some integrals. |
| 45:30 | Suppose I want to do the integral of sin^2(x)dx.
You go I don't know how to do sin^2, I know how to do sin. I want to do sin^2, I would use that formula. I'd say well that's the integral of 1/2 - (cos2x)/2 dx. It's a trig substitution. No calculus on that step. |
| 46:04 | Now I could break that up into two integrals, right?
The integral of 1/2 is just 1/2x. And the integral of cos2x times 1/2, right? So this is (1/2)x - (1/2)(sin2x)/2 + a constant. |
| 46:31 | Also known as (1/2)x - (sin2x)/4.
Ok? So again, you can take sin^2 and rewrite it as 1/2 - (1/2)cos2x. Break it into 2 integrals. The 1/2 is just 1/2x. And the (1/2)cos2x is (1/2)(sin2x)/2. That's not so bad. Of course I could make this harder. |
| 47:09 | How could I make it harder?
Or I could make it easier. Why make it easy though? Then you guys would just be happy. What if I gave you sin^3(x)? Why would you ever wanna integrate sin^3? You know why you'd wanna integrate sin^3. Cause we'd put it on the exam. |
| 47:32 | That's why you'd wanna integrate sin^3.
So how would you do that? You'd say well I could make that integral of sinx times sin^2(x). How is that doing me any favors? Well I don't really wanna use that. I could, but that would make a mess. But what I could do, is take this and make that 1-cos^2. |
| 48:11 | Why would I wanna do that? Now I could just use u substitution.
You let u=cosx du= -sinxdx So this becomes minus the integral of 1-u^2 du. |
| 48:40 | No no, don't confuse this substitution with that substitution.
This is sin^2 + cos^2 is 1 when you see sin^2 you can replace it with 1-cos^2. Isn't this confusing? So, when you just see plain old sin^2 you need to use that substitution. But if you see sin^2 or cos^2 mixed in with sinx, then you can use the ones that you learned from the Pythagorean Theorem. |
| 49:07 | Which is
sin^2 + cos^2 =1.
So sin^2(x) is 1-cos^2. cos^2(x) is 1-sin^2. |
| 49:31 | Ok?
So how will you know? Look for the odd power rather than the even power. Ok? Once you have this as 1-u^2 this is u-(u^3)/3. Plus a constant. So that's -cosx +cos^3(x) |
| 50:00 | over 3 plus a constant.
Yes. Sinx, well remember it's du. We're doing u substitution. Not integration by parts. U substitution. I know. It's a lot. It is, it's a lot to keep track of. You have my sympathy. I've done, I did my count at one point. I think I've done about 10,000 integrals in my lifetime. Some number like that. So of course it's easy for me, ok? |
| 50:30 | You guys are up to about 7, so it takes you a while to get the hang of these.
Ok, you've probably never done sin^3 before. So, it'll take you a little while. Alright. Let's do some more. What's another fun one to do? |
| 51:05 | What if I had sec^4(x)?
Remember when you were in high school and you said when am I gonna use this stuff? Now. You're using it now. You might go we're never gonna use this stuff in life. Well now. |
| 51:32 | How about in real life? Oh, never.
If you're an engineer you'll use it. If you're a physicist you'll use it. I don't think economists will use this. Biology, this stuff doesn't really show up in biology. A little bit. There's some, there's some sine, cosine stuff, right? E^x stuff shows up all the time. Ok? So when you become research biologists you use it. Chemistry it shows up. But not too much sine and cosine, because you can turn sine and cosine into e. |
| 52:01 | There's something you probably don't realize.
So, you can have a lot of fun with this. But this one, well why don't I rewrite this as sec^2(x)sec^2(x) dx? And then you have another trig identity. 1+tan^2(x) is sec^2(x). Or, well no. Fine. So I can now rewrite this |
| 52:32 | as (1+tan^2(x))sec^2(x)
dx.
Whew, a lot of these. I don't know. People starting working on these a long time ago and just kept coming up with all sorts of crazy answers. There's some really clever integration tricks out there. |
| 53:00 | There are.
You guys, this is just the tip. The tip of the iceberg. Ok. So, distribute that and you have sec^2(x)dx + second integral of tan^2(x)sec^2(x) dx. So this is just tangent because the derivative of tangent is sec^2. You guys know that right? Supposed to know your derivatives. |
| 53:33 | And tan^2 sec^2, you do u substitution.
You let u=tanx. du is sec^2(x)dx. So this becomes u^2. So again, you took sec^4, you make it sec^2 sec^2. |
| 54:03 | Then you replace one of the sec^2's with 1+tan^2.
Then distribute. So now you have sec^2 times 1and sec^2 times tan^2. Now for this one you do u substitution, because if u is tan, du is sec^2. So the integral of u^2 is u^3/3. And that's, u is tangent of x. So it's tan(x) +(tan^3(x))/3 + a constant. |
| 54:50 | These are annoying but you get the hang of them.
It's just lots of playing with trig. |
| 55:09 | What other ones did I like to do?
How about Could I come up with some more annoying ones? Oh sure. |
| 55:38 | How about
cos^4(x)dx?
Now, you don't see an odd power. You say what if I broke that up into cos^3 and cosine? I can't really do anything with that. But I can make it cos^2 times cos^2. |
| 56:03 | Ok, and that, I can now use that substitution.
So this would now become 1/2 + cos2x/2 times 1/2 +(cos2x)/2 dx. I can do this all day. |
| 56:31 | And this, now FOIL it out.
Equals 1/4 + (1/2)cos2x + (1/4)cos^2(2x) dx. Loving this? You see what I did? All I did was I took the cos^2's and made them each using this substitution |
| 57:05 | Ok?
So 1/2 + (cos2x)/2, 1/2 + (cos2x)/2 And then I FOILed it. 1/2 times 1/2 is 1/4. (1/4)cos2x + (1/4)cos2x is (1/2)cos2x. And (cos2x)/2, (cos2x)/2 is cos^2(2x) over 4. And this, I'm gonna have to do one more round of substituting. |
| 57:32 | I know.
How do you think I felt the first time I did this? Same as you guys. Ok, so cos^2(2x) is 1/2 + (1/2)cos4x dx before you distribute. You get 1/4 +1/2 cos, hey I have to write this on the board. |
| 58:01 | Plus 1/8 + (1/8)cos4x dx.
Why is it 4? Because it's double whatever that angle is. So 2x becomes 4x. If it was 10x it would become 20x. If it was πx it would be 2πx. Ok. Now those are easy to integrate. The integral of 1/4 is 1/4x. |
| 58:32 | Integral of cos2x
is (sin2x)/2.
Integral of 1/8 is 1/8 of x. And cos4x is (sin4x)/4. Plus c. You guys all have this I hate my life face. Yes. |
| 59:15 | Here.
Sec^2 is 1+tan^2. So sec^2 times 1 sec^2 times tan^2 Ok? This is where you rewind the video. |
| 59:31 | And you watch it, and watch it again.
Maybe somebody clever can get something going there. How we doing on these? Yes. Ah, so what's the trick? How do I know I have to do for both? It's an even power. So with even powers of sine and cosine, you're gonna have to use these. |
| 60:03 | Ok, with odd powers you don't. Let's give you guys a much easier one.
Suppose I had sin^3(x) cos^3(x) dx. |
| 60:30 | So the key to the odd powers is you take all of the even powers out except for 1.
So for example we leave sin^3 alone. And we say, I wanna have one cosine left over, because the derivative of sine is cosine. So I can make this into a u substitution. I just have to change, I just have to get rid of all the other cosine's. So, I make this cos^2(x) cosxdx, because then this will be my du. |
| 61:05 | And I know that cos^2 is 1-sin^2.
Ok? So this becomes sin^3(x) (1-sin^2(x)) cosx dx. |
| 61:34 | Ok, now do u substitution. Let u=sinx.
du=cosx dx. This becomes the integral of u^3 times 1-u^2 du. That equals u^3 - u^5 du. |
| 62:05 | Which is (u^4)/4
minus (u^6)/6 + a constant.
u is (sin^4(x))/4 |
| 62:30 | minus (sin^6(x))/6 + a constant.
I do some of these in the pages in my book, I put them up, advanced trig integrals. That's in the documents. I'll see if I can find some more practice ones of these. One of the problems with the Stewart book is he does a tiny number of these. So there's not really a lot to practice. So I'll try to find you guys some more. Once you do the 5 or 6 basic types which I kind of did today That's it, that's all there really is. Except the really sadistic ones. |
| 63:03 | Ok. I'll see everybody on Wednesday.
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