1;2802;0cWEBVTT
Kind: captions
Language: en
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so derivatives and integrals
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so this is the first fundamental theorem of calculus.
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so if you have an integral
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from...
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3 to x i should do 0 to x because they like 0.
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f(t)dt
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okay and thats g of...g of x.
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right?
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then g'(x) would just equal f(x).
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right? we did this the last time.
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and g double prime of x
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will equal f'(x).
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so say i told you
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that we had something like this
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and i wanted to find g'' at uhh...
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pi/3, g''(pi/3),
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suppose i wanted to do something like that.
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g'(x)
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would be (1+cos^2x)
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times the derivative of cosine
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which is -sine.
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so if i wanted to find g'(pi/3) i now plug in pi/3 right?
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would be 1+cos^2 of pi/3
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times -sin(pi/3).
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so cosine of pi/3 is a half
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so cosine squared would be a quarter.
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so this would be 1+1/4
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times minus radical 3 over 2
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and im not interested in what that is.
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and then if i wanted to find g''(pi/3) i take the derivative of this which is product rule.
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so i have g'(x)
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would be..
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1+cos^2x
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times the derivative of -sin which is -cosine.
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plus -sin(x)
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times the derivative of 1+cos^2
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which would be 2cosx to the 1
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times -sin(x) this is very tedious.
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k? and thats..
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g''(x).
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this comes from the product rule.
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you look scared.
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got it?
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alright and then you just plug in again.
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so g''(pi/3)
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would be this.
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1+1/4(-1/2)
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plus minus radical 3/2
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2 times a half
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and you simplify it.
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so thats a typical type of thing you could ask.
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cause were testing if you can see the connection between
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the function, the integral, and the derivative.
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the derivative of the integral okay?
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so when you do the derivative of the integral the integral kind of goes away.
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especially if you just go from 0 to x or any constant to x.
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so far so good?
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so we could give you..i'll wait a minute.
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i'll wait until everybodys got that down.
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you know it'll be on the video later.
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alright so
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another thing you could do with this fundamental theorem stuff
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i did some of last time
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and i tell you this is f(x)..f(t).
00:05:06.680 --> 00:05:09.080
i give you something like that so we did this last time.
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so remember what this means.
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this means g(x) basically means the area until you get to x.
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so g(4) would be the area up to 4.
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g(6) would be the area of 6 which would be this area minus that area.
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right g(8) would be the area from here to here.
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and so on.
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so if i say
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where..
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is g increasing?
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where is g increasing?
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g is increasing..well
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remember from last semester g is increasing when g'
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is positive.
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and what is g'(x)?
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g'(x)..
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is f(x)
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so wherever f(x) is positive
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g' is increasing.
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so that would be..
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(0,4) and (6,8).
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that make sense?
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i want to know where is g increasing?
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and to know where g is increasing
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i need to know where g' is positive.
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well g' is just f.
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so f is positive between 0 and 4
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and between 6 and 8.
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so what if i said where does g concave up?
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or concave down it doesnt really matter.
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so g is concave up where g'' is greater than 0.
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so g'' is greater than 0
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well if g' is f then g'' is f'.
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and f' is just the slope.
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so that would be from here to here
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and from here to here.
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thats the 2 places where f has got a positive slope.
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so 0 to 2 and 5 to 7
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that make sense?
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i like these kinds of questions i didnt put them on the exam last semester but
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they're kind of okay.
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some people really like them some people dont like them at all.
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the AP loves them.
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theres one on every AP exam.
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okay we feel we understand that stuff?
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we're going to learn something new now i just want to make sure everyone feels okay about this.
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last chance.
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yeah? thinking about chemistry?
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no chemistry. no chem thoughts.
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we're going to learn something new.
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substitution method.
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substitution rule, same thing.
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its really a method more than it is a rule.
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what are the 4-what are my 4 favorite rules?
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product rule
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chain rule
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quotient rule
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ja rule
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but not in that order.
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so substitution so now we're going to start to learn how to do hard integrals.
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so we've really only been doing very simple integrals so far.
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you need the substitution rule..substitution method
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when you want to work backwards from the chain rule
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we havent seen
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this is what you use when you havent a clue essentially theres 2 things in it
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and one of the things in the integrand
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is the derivative of the other thing in the integrand
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so for example
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say you have the inegral of
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i think the webassign tries to teach you how to do this.
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but i'll do my flavor of it.
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my version of it.
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so how would you do the integral of this?
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well one thing you could do is you could expand this out.
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to the tenth.
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and then distribute the x^2.
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so that would be easy if this was say squared
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because then you could just sort of foil it out.
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but the 10th is a bit messy.
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but then you could notice something.
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x^3..
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the derivative of x^3 is kind of like x^2.
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because you just, its 3x^2 right?
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so if we let u
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we like to use the letter u.
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equal 5+x^3
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and du dx the derivative of u
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would be 3x^2.
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there will be a reason for this in a second.
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and cross multiply this you get du..
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is 3x^2dx
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so 1/3du is x^2dx.
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so far so good?
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why take out the 1/3? alright
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i look at this integral and i say im going to substitute things.
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so i have 5+x^3 so if i let this be u
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this would just be the integral of u^10
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which is trivial..okay?
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but im stuck with an x^2 and a dx
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so i take the derivative and i say
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ive got 3x^2 dx's.
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but i only want to substitute for x^2dx
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so if i move the 3 to the other side
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now i can go over this integral
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and i can say this
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is u^10.
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and this..
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is 1/3du.
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pull the 1/3 out
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and thats an easy integral.
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so far so good?
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so whats the integral of u^10? u^11/11.
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so thats 33 down there right? because youve got a third and you got 11.
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you all see down there?
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and then substitute back so you would get
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1/33(5+x^3)^11+c
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okay? and the 33
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comes because this is 1/3 times 1/11.
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and double check that in your head so
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take the derivative you get 11
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thirty thirds
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(5+x^3)^10
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times 3x^2 so notice
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right? thats a third
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that cancels with the 3
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and you get the x^2 and the 5+x^3.
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so what the u substitution does
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is it undoes the chain rule.
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it basically looks at this integrand
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and says
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theres a chain rule going on in here.
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so i'm gonna work the chain rule backward.
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so lets do another example.
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suppose i had sin(x)
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alright so i would do u substitution.
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so im looking for
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something and its derivative.
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now i look at-so the derivative of sin is cosine
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and the derivative of cosine is sin
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so how would i know which to do?
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well
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this is under the radical
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so thats causing me more irritation.
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so i'm going to let that one be u.
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and let u equal 1+cos(x).
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and the derivative
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would be -sin(x)
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which is almost the same thing as sin(x) right?
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so if i cross multiply
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i would get du
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-sin(x)
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dx.
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and tonight i'll put the pages up from my book that do the u substitution.
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you can look at lots of examples.
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so now i come over here and i say well alright
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i could make this
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square root of u
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and this
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would be negative du.
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because du is -sin(x)dx so negative du is sin(x)dx.
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confused?
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who is confused?
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alright so the idea is
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you want to look at the integral
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and you want to put u into one part
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and the rest of the part has to be some form of du.
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which usually you play with the constants.
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so you can turn this now into a very simple integral.
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so i look at the beginning and i say ive got the square root of 1+cos(x).
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and i got sin(x).
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i know the derivative of 1+cos(x)
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is -sin(x).
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so this is basically -sin(x)
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just off by a minus sign.
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so if i let u be this
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and then du is -sin(x)
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then i can substitute everything in here.
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and make this
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remember this is an easy integral right? this is just u^1/2.
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we'll do like 3 more of these and then youll get good at it.
00:15:52.500 --> 00:15:54.260
that becomes u^3/2
00:15:54.860 --> 00:15:55.700
over 3/2
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plus the constant
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and then you substitute back.
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-2/3
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(1+cosx)^3/2
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plus c.
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see whats going to happen in integration
00:16:12.020 --> 00:16:17.540
is youre going to look at integrals and youre going to say i have no idea what youre doing and you start trying out all of your tricks.
00:16:17.720 --> 00:16:20.080
and see if you can come up with one that will work.
00:16:22.840 --> 00:16:24.280
what happened to the minus sign?
00:16:25.580 --> 00:16:28.820
so this is-my du is -sin(x).
00:16:28.900 --> 00:16:31.620
but ive only got positive sin(x) in here.
00:16:32.020 --> 00:16:34.300
so when i substitute i use negative du.
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ok?
00:16:35.820 --> 00:16:38.120
you get rid of the minus because this is minus here.
00:16:38.540 --> 00:16:39.920
right? so -du
00:16:40.540 --> 00:16:42.300
would be positive sin(x).
00:16:45.360 --> 00:16:47.380
okay? thats why i have the minus sign here.
00:16:48.020 --> 00:16:48.520
okay?
00:16:49.080 --> 00:16:52.060
lets do some more of these after about 3 or 4 you'll start to get the hang of them.
00:16:52.360 --> 00:16:53.160
you'll see.
00:16:54.040 --> 00:16:54.920
or maybe not.
00:16:55.840 --> 00:17:00.320
so the integral is xcos(x^2)dx.
00:17:13.260 --> 00:17:15.820
alright so lets do some substituting.
00:17:15.820 --> 00:17:17.720
well i see the cosine i cant really
00:17:17.800 --> 00:17:19.720
i cant do much with the cosine.
00:17:19.720 --> 00:17:21.480
if i said i'd have u equal cosine
00:17:22.000 --> 00:17:24.260
the derivative is sin and there isnt any sin in here.
00:17:24.660 --> 00:17:26.640
but if i let u=x^2
00:17:28.940 --> 00:17:29.900
the problem is
00:17:29.900 --> 00:17:31.820
i could do the integral of cosx.
00:17:31.960 --> 00:17:32.680
thats easy.
00:17:32.760 --> 00:17:34.520
i could do the integral of x
00:17:34.740 --> 00:17:35.460
or some x^2
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but the cosine is sort of messing things up.
00:17:37.780 --> 00:17:39.660
so im going to let u=x^2
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and then du dx
00:17:42.820 --> 00:17:44.320
is going to be 2x.
00:17:51.480 --> 00:17:54.260
why, why do i want to do that? well i want to get rid of this problem.
00:17:55.160 --> 00:17:56.300
now i look at the integral
00:17:56.320 --> 00:17:59.020
and i integrand and i say i've got x dx
00:17:59.420 --> 00:18:00.860
so if i cross multiply
00:18:01.840 --> 00:18:03.120
i've got 2x
00:18:04.400 --> 00:18:04.900
dx.
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i only want to substitute for x dx so i just divide by 2.
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k now i go here and i substitute.
00:18:25.700 --> 00:18:28.640
i substitute for cos(x^2) and i get cos(u)
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and x dx
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is going to be 1/2du.
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put the half outside
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i could put the 1/2 inside but i generally put constants outside of the integral its easier that way.
00:18:41.640 --> 00:18:43.380
alright whats the integral of cosine?
00:18:43.760 --> 00:18:44.260
sin.
00:18:50.480 --> 00:18:50.980
soo..
00:18:51.480 --> 00:18:52.560
u is x^2
00:18:53.180 --> 00:18:54.380
so this is 1/2
00:18:56.080 --> 00:18:57.420
sin(x^2)
00:18:58.320 --> 00:18:58.820
plus c.
00:18:58.820 --> 00:19:01.600
and if you took the derivative of sin(x^2) what would happen?
00:19:01.600 --> 00:19:03.120
you'd have cosine of x^2
00:19:03.460 --> 00:19:04.620
times 2x.
00:19:05.540 --> 00:19:07.860
so thats where that x appears from.
00:19:09.260 --> 00:19:11.740
so u substitution is working this backwards
00:19:11.740 --> 00:19:12.920
over and over again.
00:19:12.920 --> 00:19:14.880
so lets give you guys one or two to practice.
00:19:17.220 --> 00:19:18.260
nice easy ones.
00:19:25.100 --> 00:19:27.580
i know its new for almost all of you so
00:19:28.000 --> 00:19:30.080
i wont give you killer ones yet.
00:19:31.200 --> 00:19:33.920
thats for the second half of the lecture.
00:19:38.640 --> 00:19:41.520
believe me you guys will get the hang of this very soon its really not that bad.
00:20:20.060 --> 00:20:20.560
okay?
00:20:20.660 --> 00:20:22.020
theres your first 2.
00:20:22.160 --> 00:20:24.240
give you guys a couple minutes.
00:20:26.820 --> 00:20:29.740
so i look at this i see sin^4xcosx.
00:20:30.620 --> 00:20:32.940
so im going to need u substitution.
00:20:33.520 --> 00:20:34.720
and i say to myself well
00:20:35.420 --> 00:20:37.500
if i let u=cos
00:20:38.200 --> 00:20:39.160
the problem is
00:20:39.560 --> 00:20:40.960
du would be sin
00:20:41.380 --> 00:20:41.880
sort of
00:20:41.880 --> 00:20:45.060
and that would be du^4 which doesnt make any sense.
00:20:45.300 --> 00:20:48.420
right? im only used to one dx at a time or one du.
00:20:48.660 --> 00:20:50.740
so du^1/4 would be sort of painful.
00:20:51.440 --> 00:20:53.660
and if i let u=sin^4
00:20:54.140 --> 00:20:57.500
the derivative of that is 4sin^3cos
00:20:57.500 --> 00:20:58.760
thats not what this is.
00:20:59.640 --> 00:21:01.500
but if i let u just equal sin
00:21:02.540 --> 00:21:04.660
then du would be cosine.
00:21:07.320 --> 00:21:11.340
instead of writing the dx down here im just gonna cross multiply right away.
00:21:12.460 --> 00:21:12.960
ok?
00:21:12.960 --> 00:21:15.320
you can skip that step if you feel the need to.
00:21:18.300 --> 00:21:21.780
and now its easy to substitute i say well this is u^4.
00:21:23.340 --> 00:21:23.840
this
00:21:24.620 --> 00:21:25.480
is du.
00:21:29.280 --> 00:21:29.780
easy?
00:21:30.500 --> 00:21:32.660
so thats now u^5/5.
00:21:36.800 --> 00:21:42.380
and i substitute back and i get sin^5x/5
00:21:43.340 --> 00:21:43.840
plus c.
00:21:45.620 --> 00:21:47.220
how did we do on that one?
00:21:47.680 --> 00:21:49.820
of course its easy if i tell you to use sin.
00:21:49.820 --> 00:21:53.440
so in the beginning the mistake people make is theyre not sure which one
00:21:53.440 --> 00:21:56.520
or theyre trying to substitute for the whole expression
00:21:56.520 --> 00:21:58.640
which you dont need to do you dont need the power.
00:21:58.640 --> 00:22:00.900
because you can always do u to a power
00:22:00.900 --> 00:22:03.360
so dont worry about substituting for the power.
00:22:03.860 --> 00:22:04.360
ok?
00:22:05.900 --> 00:22:07.820
but you understand the idea?
00:22:09.700 --> 00:22:11.140
so some general rules
00:22:11.140 --> 00:22:12.940
if you have something to a power
00:22:12.940 --> 00:22:15.560
and you have something else thats not to any power
00:22:15.560 --> 00:22:16.600
just the power of 1
00:22:16.740 --> 00:22:17.620
still a power
00:22:17.800 --> 00:22:18.300
okay?
00:22:18.300 --> 00:22:21.300
then the thing with a higher power is almost always going to be u.
00:22:21.480 --> 00:22:24.440
but usually you can ignore what the power is.
00:22:25.380 --> 00:22:25.880
ok?
00:22:26.620 --> 00:22:29.820
the second thing is you have 2 things in the integral and theyre 1 power apart
00:22:30.020 --> 00:22:31.500
x^10 and x^9
00:22:31.500 --> 00:22:34.840
then the higher power will be u and the lower power will be du.
00:22:34.840 --> 00:22:35.960
well do an example of that in a minute.
00:22:37.260 --> 00:22:38.940
alright what about here?
00:22:39.500 --> 00:22:42.260
i see cos(e^x) and i see e^x.
00:22:43.400 --> 00:22:45.700
so if i let u=cos
00:22:46.700 --> 00:22:48.140
theres no sin in here.
00:22:48.580 --> 00:22:50.200
but if i let u=e^x
00:22:54.200 --> 00:22:54.700
du
00:22:55.580 --> 00:22:58.480
is e^xdx.
00:23:00.000 --> 00:23:02.400
so this now just becomes integral of
00:23:02.880 --> 00:23:04.320
thats cosine of u
00:23:05.700 --> 00:23:07.440
and e^xdx
00:23:08.560 --> 00:23:09.340
is du.
00:23:11.620 --> 00:23:12.820
you like that one?
00:23:13.200 --> 00:23:14.640
so thats just sin of u
00:23:18.120 --> 00:23:19.240
plus a constant.
00:23:19.640 --> 00:23:20.880
so thats sin...
00:23:21.360 --> 00:23:22.200
e^x
00:23:23.580 --> 00:23:24.340
plus c.
00:23:25.420 --> 00:23:28.560
by the way some professors are very fussy about the whole plus c thing.
00:23:28.640 --> 00:23:31.840
they take off points if you dont write the plus c.
00:23:32.060 --> 00:23:33.500
so try to remember it.
00:23:33.500 --> 00:23:35.780
when youre done, when youre done with your exam
00:23:35.780 --> 00:23:37.120
look at all your integrals
00:23:37.120 --> 00:23:40.120
and dont forget to write, just write plus c after all of them.
00:23:40.120 --> 00:23:43.220
but not for the ones where you take the derivative of the integral.
00:23:43.660 --> 00:23:45.260
okay? dont mess that up.
00:23:45.560 --> 00:23:47.320
lets do some more of these.
00:24:10.520 --> 00:24:13.320
so remember what i told you, you look for something wrong
00:24:13.320 --> 00:24:15.300
and if theyre 1 power apart
00:24:15.800 --> 00:24:18.680
the thing thats to the higher power will be u
00:24:18.680 --> 00:24:21.680
and the thing thats to the lower power will be du.
00:24:23.820 --> 00:24:25.660
okay? higher power will be u
00:24:25.840 --> 00:24:27.360
lower power will be du.
00:24:30.320 --> 00:24:33.040
lets let u=6+x^10.
00:24:37.660 --> 00:24:39.020
why do i choose that?
00:24:39.020 --> 00:24:42.180
because the derivative of x^10 is something x^9.
00:24:42.180 --> 00:24:44.560
and the derivative of 6 is just going to go away.
00:24:46.020 --> 00:24:46.520
so du
00:24:47.640 --> 00:24:51.120
would be 10x^9dx.
00:24:51.120 --> 00:24:53.840
you all see why i can put the dx over here right?
00:24:53.960 --> 00:24:56.680
its du dx and then i just multiply across.
00:24:59.880 --> 00:25:01.480
so now i look at the front
00:25:01.520 --> 00:25:02.720
and i say well this
00:25:03.140 --> 00:25:04.640
wont become u^4
00:25:04.680 --> 00:25:07.000
so i just have to play with this a bit.
00:25:09.140 --> 00:25:10.680
so i have 10x^9
00:25:11.020 --> 00:25:12.540
so i have to divide by 10
00:25:14.200 --> 00:25:15.260
and get x^9.
00:25:19.840 --> 00:25:21.280
and now i substitute.
00:25:22.340 --> 00:25:23.820
this is u^4
00:25:24.600 --> 00:25:28.920
x^9dx is 1/10du.
00:25:32.220 --> 00:25:33.520
so thats 1/10...
00:25:35.500 --> 00:25:37.500
u^5/5
00:25:39.640 --> 00:25:40.920
plus the constant.
00:25:44.200 --> 00:25:46.920
so u is 6+x^10 so this is now
00:25:47.380 --> 00:25:50.240
(6+x)^10
00:25:51.080 --> 00:25:51.860
over 50
00:25:53.220 --> 00:25:58.080
plus c and yes you could have written 1/10 and the 1/5 separated thats fine.
00:25:58.520 --> 00:25:59.160
okay?
00:25:59.280 --> 00:26:00.720
howd we do on this one?
00:26:01.160 --> 00:26:02.540
starting to get the hang of it?
00:26:02.940 --> 00:26:06.020
good should i kill you with some hard ones now? sure.
00:26:06.880 --> 00:26:07.600
thank you.
00:26:08.320 --> 00:26:09.120
no problem.
00:26:13.440 --> 00:26:15.360
oh yeah im sorry i wrote that wrong.
00:26:17.540 --> 00:26:18.260
thank you.
00:26:20.040 --> 00:26:21.240
thats what i meant.
00:26:54.720 --> 00:26:55.520
theres one.
00:26:56.840 --> 00:26:58.600
what should we let u equal?
00:27:00.140 --> 00:27:01.020
good choice.
00:27:01.380 --> 00:27:01.880
ln(x)?
00:27:03.200 --> 00:27:04.400
also known as log.
00:27:06.100 --> 00:27:09.340
why is it so hard to look at the letters of ln and say log?
00:27:10.480 --> 00:27:10.980
i mean
00:27:10.980 --> 00:27:14.120
lots of things in english arent not pronounced the way they are spelled.
00:27:16.940 --> 00:27:19.560
none of you can pronounce february correctly so.
00:27:20.900 --> 00:27:22.620
some of you can pronounce february correctly.
00:27:22.620 --> 00:27:26.380
most of you say febUary and thats not the way its spelled so there you go.
00:27:26.900 --> 00:27:27.540
alright.
00:27:27.640 --> 00:27:29.080
so the derivative of u
00:27:30.180 --> 00:27:33.140
is 1/x(dx) and look
00:27:33.360 --> 00:27:35.820
i have 1/x(dx) thats this.
00:27:36.860 --> 00:27:38.740
right? that is 1/x(dx).
00:27:39.560 --> 00:27:41.080
so this is just sin of u.
00:27:48.060 --> 00:27:49.180
integral of that
00:27:50.180 --> 00:27:52.120
is -cos(u)
00:27:53.220 --> 00:27:54.580
plus a constant.
00:27:54.980 --> 00:27:56.660
and then substitute back
00:27:57.380 --> 00:28:00.020
-cos(lnx)
00:28:01.760 --> 00:28:02.880
plus a constant.
00:28:04.460 --> 00:28:05.980
how did we do on that one?
00:28:07.560 --> 00:28:09.220
yes no lovin life?
00:28:10.040 --> 00:28:11.080
not lovin life?
00:28:17.340 --> 00:28:19.820
time to give you one slightly harder.
00:28:40.460 --> 00:28:40.960
okay.
00:28:41.780 --> 00:28:42.900
u substitution.
00:28:50.740 --> 00:28:52.340
now try u substitution.
00:28:59.180 --> 00:29:01.320
lets let u=cosx now
00:29:01.320 --> 00:29:03.200
how do i know to let u equal the denominator?
00:29:03.460 --> 00:29:05.380
well if i let u=sinx
00:29:05.380 --> 00:29:07.440
i'll have du in the denominator
00:29:07.440 --> 00:29:08.680
and that doesnt do me any good.
00:29:08.680 --> 00:29:10.660
i dont know what to do with the du down there.
00:29:10.660 --> 00:29:12.000
its got to be up in the numerator.
00:29:12.480 --> 00:29:14.600
but if i let u=cosx
00:29:16.300 --> 00:29:21.580
du is -sinxdx.
00:29:23.240 --> 00:29:24.840
so i can now rewrite this
00:29:26.560 --> 00:29:28.800
as -du/u.
00:29:31.860 --> 00:29:33.800
and that is ln-yes?
00:29:42.020 --> 00:29:43.540
thats a very good rule.
00:29:43.540 --> 00:29:46.240
so yes if you have a function over a function
00:29:46.660 --> 00:29:49.540
u will be in the denominator it cant be in the numerator.
00:29:50.200 --> 00:29:50.700
k?
00:29:50.920 --> 00:29:52.820
because if u is in the numerator
00:29:53.620 --> 00:29:54.360
du above
00:29:54.360 --> 00:29:55.300
almost always.
00:29:55.720 --> 00:29:58.040
okay? because here
00:29:58.800 --> 00:29:59.300
right?
00:30:00.180 --> 00:30:02.720
sort of was in the denominator but not really.
00:30:02.900 --> 00:30:03.400
ok?
00:30:03.400 --> 00:30:05.340
theyre really multipled this is really
00:30:05.840 --> 00:30:11.140
sin of lnx times 1/xdx.
00:30:12.580 --> 00:30:13.540
but generally
00:30:13.540 --> 00:30:15.540
you cant have du in the denominator
00:30:15.540 --> 00:30:17.880
because theres nothing you can do with that its nonsense.
00:30:17.880 --> 00:30:19.700
so you have to have it up in the numerator.
00:30:20.640 --> 00:30:21.680
alright so this
00:30:22.060 --> 00:30:22.560
is..
00:30:23.480 --> 00:30:25.360
-ln(u)
00:30:25.720 --> 00:30:27.880
plus a constant substitute back.
00:30:31.320 --> 00:30:32.840
i suggest you memorize
00:30:32.840 --> 00:30:34.420
what the integral of tangent is.
00:30:35.100 --> 00:30:35.600
k?
00:30:36.960 --> 00:30:38.920
so one to add to your pile
00:30:48.160 --> 00:30:51.320
and you should be able to figure out cotx once you know tanx.
00:31:12.040 --> 00:31:12.540
okay.
00:31:13.020 --> 00:31:13.520
yes?
00:31:17.140 --> 00:31:18.660
how did i get from here to here?
00:31:19.060 --> 00:31:21.220
well whats the derivative of ln?
00:31:22.280 --> 00:31:23.920
1/x so this is 1/u.
00:31:25.020 --> 00:31:25.660
right? so
00:31:26.140 --> 00:31:32.500
the integral of -du/u right? is the same as -1/udu
00:31:32.740 --> 00:31:34.660
so thats why its natural log.
00:31:37.680 --> 00:31:39.440
heres one thats annoying.
00:31:46.140 --> 00:31:48.060
so we'll do this one as a team.
00:31:48.060 --> 00:31:50.260
cause this one is slightly annoying.
00:31:53.440 --> 00:31:56.500
what if i let u=tanx?
00:31:56.920 --> 00:32:00.880
well du would be sec^2 which doesnt do me much good because ive got sec^10.
00:32:01.940 --> 00:32:04.260
what if i let u=sec?
00:32:04.260 --> 00:32:06.400
whats the derivative of secant?
00:32:07.180 --> 00:32:09.420
sectan, so what if i rewrote this?
00:32:11.040 --> 00:32:21.940
as sec^9x(secxtanx)dx i just pulled out one of the secants.
00:32:24.720 --> 00:32:27.120
we can be quite nasty if we need to be.
00:32:27.420 --> 00:32:29.520
we've got even more evil ones up our sleeves.
00:32:31.580 --> 00:32:32.080
now
00:32:32.480 --> 00:32:34.780
let u=secx.
00:32:36.440 --> 00:32:43.000
du is secxtanxdx.
00:32:47.460 --> 00:32:48.420
so far so good?
00:32:50.540 --> 00:32:51.740
once i showed you.
00:32:52.760 --> 00:32:54.240
well you'll get there on your own.
00:32:54.780 --> 00:32:58.680
so this now becomes u^9du.
00:33:01.260 --> 00:33:02.600
which is u^10/10.
00:33:06.180 --> 00:33:07.060
plus c.
00:33:09.800 --> 00:33:10.600
and thats..
00:33:12.620 --> 00:33:16.560
sec^10x/10+c.
00:33:16.920 --> 00:33:19.320
i have one more type to show you guys.
00:33:41.740 --> 00:33:44.220
suppose you have something that looks like that.
00:33:53.080 --> 00:33:53.580
now
00:33:53.940 --> 00:33:55.680
if i let u=x
00:33:56.300 --> 00:33:57.820
du is just dx.
00:33:58.420 --> 00:33:59.880
that doesnt do me much good.
00:33:59.880 --> 00:34:01.740
what if i put the x back under the radical?
00:34:01.740 --> 00:34:04.400
then i'd have x^2 and x^3 so that doesnt do me any good.
00:34:05.780 --> 00:34:08.140
if i let u=1+x
00:34:08.480 --> 00:34:10.320
what if i let u=1+x?
00:34:12.180 --> 00:34:12.900
then du
00:34:13.920 --> 00:34:15.800
is just dx right?
00:34:16.360 --> 00:34:17.880
what do i do with this x?
00:34:19.520 --> 00:34:21.380
well if u is 1+x
00:34:21.580 --> 00:34:24.460
then let u-1=x.
00:34:26.120 --> 00:34:27.800
and now this would become
00:34:28.660 --> 00:34:33.820
(u-1) times the square root of u du
00:34:44.520 --> 00:34:46.440
i look at this i say if i let u=x
00:34:47.120 --> 00:34:48.240
du is dx
00:34:48.240 --> 00:34:49.920
which is just 1 that doesnt do me any good.
00:34:50.640 --> 00:34:52.760
but if i let u=1+x
00:34:53.600 --> 00:34:55.640
so this will now become the square root of u.
00:34:55.640 --> 00:34:59.900
this is just u-1 now i could just distribute that square root of u.
00:35:01.580 --> 00:35:03.640
this will become u^3/2
00:35:04.140 --> 00:35:05.520
minus u^1/2
00:35:07.300 --> 00:35:07.800
du.
00:35:09.320 --> 00:35:11.000
so how do i know to do that?
00:35:11.020 --> 00:35:12.540
theyre the same power.
00:35:13.160 --> 00:35:13.660
i look
00:35:14.560 --> 00:35:18.960
and i say if you basically have an x and an a+/- x
00:35:18.960 --> 00:35:20.740
thats the trick that you use.
00:35:21.360 --> 00:35:21.860
okay?
00:35:22.460 --> 00:35:24.640
if you look at the integral and you have
00:35:25.340 --> 00:35:27.500
one of them is some constant +/- x
00:35:27.500 --> 00:35:28.360
or the other way around
00:35:28.360 --> 00:35:29.600
x+/- a constant
00:35:29.700 --> 00:35:30.880
and the other is just plain x
00:35:30.880 --> 00:35:32.460
this is the trick that youre going to use.
00:35:33.460 --> 00:35:36.260
the whole next month is going to consist of
00:35:36.260 --> 00:35:37.780
techniques of integration.
00:35:37.780 --> 00:35:39.640
its like you have little tools in your tool box
00:35:39.640 --> 00:35:40.940
figure out which one to use.
00:35:41.160 --> 00:35:43.640
so right now you've got a second tool.
00:35:44.060 --> 00:35:45.600
we're going to do integration by parts
00:35:45.600 --> 00:35:48.000
parts by fractions we're going to have about 5 tools when we're done.
00:35:48.680 --> 00:35:50.840
then the trick is just which one do you use.
00:35:50.840 --> 00:35:54.020
my advice is you try one after the other until you get one that works.
00:35:56.240 --> 00:35:58.240
okay so this
00:35:58.240 --> 00:35:59.860
you can integrate see here
00:35:59.860 --> 00:36:01.440
if you distributed the x at the beginning
00:36:01.640 --> 00:36:03.160
you would have had this
00:36:04.060 --> 00:36:06.440
and that you cant actually do anything with.
00:36:07.220 --> 00:36:07.720
okay?
00:36:08.300 --> 00:36:09.420
but here you can.
00:36:10.120 --> 00:36:10.840
cause this
00:36:13.060 --> 00:36:13.860
this is just
00:36:15.360 --> 00:36:18.440
u^5/2 over 5/2
00:36:20.340 --> 00:36:20.900
minus..
00:36:21.980 --> 00:36:25.120
u^3/2 over 3/2
00:36:26.040 --> 00:36:26.760
plus c
00:36:26.760 --> 00:36:28.360
and then substitute back.
00:36:29.220 --> 00:36:43.060
so you get 2/5 (1+x)^5/2-2/3 (1+x)^3/2+c.
00:36:50.380 --> 00:36:51.980
teachers like that one.
00:36:52.600 --> 00:36:55.240
sort of an advanced u substitution one.
00:36:56.640 --> 00:36:59.500
i know i had it when i did integration back in the day.
00:37:01.020 --> 00:37:02.740
taught it right here at stony brook.
00:37:06.060 --> 00:37:08.300
you guys have a chem exam tonight?
00:37:08.300 --> 00:37:10.300
alright so we'll stop a few minutes early.