okay so now that we've seen the formula we are going to do some examples we'll do five examples and one of the examples we'll do in two different ways okay so we're going to remember the formula udv is U * V minus the integral of VD the integral of U DV is U * V minus the integral of VD and remember the guiding principle that our goal is to calculate this guy so if we are unable to calculate this first guy we have to be able to calculate this second guy in order for this to help us sometimes we're going to see we have to do it more than once kind of like lal's rule remember sometimes you simplifi the limit but you had to do it more than once so this is the one written without the bounds if I had the bounds here I would have the A and B okay so let's try a first example 7x e to 8X DX okay so this is how we're going rate this um calculation so I can't find an integral immediately whose derivative who anti I can't find the anti-derivative of this thing function whose derivative is this so I have to try to simplify this to be able to calculate this guy okay so I want to choose a u and I want to choose a DV so whatever I choose for U what's left is DV because I need my integral to match this okay and I want to choose a u here's kind of a general principle whose derivative which is kind of hidden here in the DU this is the derivative of U I want to choose a u whose derivative is simpler in some way than you so given those guiding principles there are kind of two functions there's this function and there's this function so I could either choose U to be 7x or U to be e to 8X and if I choose U to be 7x its derivative is seven so that's really good so the other thing I have to check is If U is 7x then that means the DV has to be the rest right because U * DV has to match this so that means this has to be DV don't forget the DX goes with the DV so DV is e to 8X DX okay so from this u and v I have to now calculate V and du okay so du is easy that's just taking derivatives the derivative of 7x is 7 don't forget the DX in this notation I'm using du so I would either write D DX is 7 or with this notation kind of like when you do substitution du is 7 DX and I have to figure out V okay well V is the function whose derivative is this so basically V is any antiderivative of e to the 8X so this is really uh the the case but I don't usually write this because usually this is so simple that I can write it right away write what V is so the function whose derivative is e to the 8X from Cal one you know exponentials are their own derivatives except that you have to be careful about the chain rule this isn't quite right if I were taking um an anti-derivative of e to 8X I would need 1/8 here and I would have to check I need to check that the derivative of this thing that I chose is what I have here so it would be 1/8 * e 8X * 8 for the chain rule so this one 18 and E goes away so I indeed get what I have here okay if that's a little unfamiliar then maybe you need to go back and um practice some of the basic Calin integrals all right so now I'm going to apply my integration by parts formula so I'm going to have U * V so 7 x * 1/8 e 8X often we put the constants together so 7 * 1/8 I'm going to just write it as 78 x e to 8X so that's my U * V so if I wanted my bounds here like I did before I would write this I have one with bounds here so from 0 to one minus the integral so my a is zero and my B is 1 the integral of VD and I just plug in what I have here so I just take this times this and that's what goes inside this integral and I can put my constants together 7 * 1/8 is just 7/8 sometimes people even put the constant outside let's just do that since we know we can pull constants out 7 DX * 1/8 e to 8X and we usually put the DX at the end so this is what I'm left with Okay so V * du is what I have here I've put it in a slightly different order and I've pulled the constant out but it's the same thing now is when you're going to know were you successful or not am I able to integrate this or not if I am yay I can do this integral if I'm not able to to calculate this integral then either I need to do further work or I chose the wrong U and DV okay so can I can I calculate the integral of e to the 8X definitely I just did it over here actually so this is equal to 78 x e to the 8X which I didn't calculate yet but I will in a minute minus 78 the integral of e to the 8X is 1/8 e to the 8X between 0 and 1 okay so now this what I'm left with is just computation okay so let's now finish the computation I'm going to plug in one 1 * e 8 * 1 minus well I don't need to worry about zero here because 0 * anything is 0 so I'm just going to write minus 0 - 7 over 8^ 2 so 7 over 64 now you have to be a little careful about parentheses and subtraction so I'm going to put a parentheses here and I'm going to do e to the 8X I'm going to subtract this and then I'm going to do e to the 8X at one and e to the 8X at zero now the zero matters here because this is e to the e8th 8 * 1 is 8 minus E to the 0 which is 1 okay so I have to be a little careful not to in ignore this zero here because I do actually get something here here I didn't because I had x times e to 8X but here I just have e to 8X okay so this whole mess in a half is 78 e to 8 - 7 64 e to the if I want I can multiply through e to the 0 is 1 minus and minus is plus so plus 7 over 64 okay and uh that's pretty much good enough but if I was really detail oriented I could combine these first two fractions 78 - 7 over 64 that would mean I have to multiply this by 8 8 * 7 is 56 56 - 7 is 49 so I'd end up with 49 over 64 e to 8 - 7 64 and that's my answer for this one okay okay let's try a second example we're going to do five examples here's my second example let's do it in blue for fun this one a little bit harder the integral from 0 to Pi of X2 sin of 2x DX okay okay again my dog is snorting next to me again I need to choose a u and a DV let's remind ourselves the integral of U DV is U * V minus the integral of VD that's my formula with bounce let's say okay so my U is going to be well let's try again what we thought worked the first time X2 because du is 2x which is a simpler function X is simpler than x^2 that means DV is sin of 2x DX all right so du is 2x don't forget the DX and V is okay what function function has as derivative sign of something well it's either cosine or minus cosine you have to kind of remind yourself so it's going to be minus cosine of 2x always the same argument inside and I have to be a little careful because if I if this were the answer and I took the derivative I'd get minus minus uh 2 sin 2X and I don't want 2 sin 2x I don't want there to be a two I want there to be nothing so I need to divide by 12 okay so this integral equals U * V again we often put the constant out front so I'm going to put this min-2 x^2 cosine 2X this is U * V that's this piece between 0 and Pi which I'll calculate later minus the integral of V du well here's a little minor bit of good news that 2 * a half is 1 and I also have minus and this product is minus something so minus and minus will give me a plus in this case so I'm going to change this minus to a plus the two and the one half disappear so I'm not going to write them I'm going to write x * cosine of 2x DX okay so I'm doing a few little calculation shorthand here if you're uncomfortable with that you can kind of keep track of everything like you could write minus and then you could have the integral of two and a minus 12 and the rest and then you could simplify it later okay okay so good news and bad news the bad news is that I actually can't calculate this immediately the good news is it's a little bit simpler than this I mean I have cosine instead of sign which is basically the same but instead of x^ S I have X so here's the thing I'm going to have to do integration by parts again on this guy okay so here's what I'm going to do I'm going to rewrite what I have so far plus and then I'll put sort of in Brackets and these curly brackets I'm going to try to squeeze it in like a crazy person what I get from doing integration by parts on this guy and I'm going to write u and v again even though they're a different u and v from before okay cuz we're not simple-minded like we used to be in Cal 1 now we're super sophisticated so I've focusing on this integral here okay so what do you think is the best choice for U here I need something either X or cosine 2X should be U and it should be something that when I take its derivative it simplifies so U is clearly X and DV is the rest cosine 2X DX okay so that means du is 1 derivative of x is 1 so 1 * DX so I'm just going to write DX that's the same as writing 1 * DX and what's a function whose d derivative is cosine 2X so notice here this is pretty sophisticated because I'm taking derivatives but here I'm taking anti-derivatives okay so I'm doing a lot of things all at once so the anti-derivative is 12 s of 2x and you can check it check that the derivative of this is indeed that okay okay so now I have inside my integral I have this formula again with this new U and V so I have U * v x * 1 12 sin 2x I'll put the 1/ 12 out front and it's evaluated between 0 and Pi minus the integral of V du which is 12 sin 2x DX and I'm going to just put the 1/2 here sorry this got too teeny okay was this a good choice or a bad choice am I able to calculate this yes or no indeed I am because I can calculate the integral of s of 2x and this one has a constant so I'll just keep it in there okay so let's do that last integration and I get I'm just going to rewrite what I have - 12x^2 cosine 2X pi plus I can take away these curly brackets now I don't have any minus sign or anything crazy I didn't really need them in the end minus and now let's actually integrate this integral of sin of 2x is - 12 cosine of 2x we actually did it already over here um -2 * -2 is + one4 so I actually get plus4 cosine 2X between 0 and pi okay and then I just have to calculate now it's just a question of calculation I've done all the sort of intellectual work now it's just the the sort of calculator work okay so I'm going to plug in this is going to be Pi pi^ 2 what's cosine of 2 pi that's one so I'm going to get min-2 pi^ 2 minus minus minus will give me plus but when I plug in zero this makes everything zero so I can ignore that let's do the same here I plug in pi s of 2 pi s of 2 pi is zero yay so that goes away at Pi minus the value at zero is zero so this is actually zero so I'm not going to write anything because I'm so sophisticated and then here I'm going to write 1 14 cosine of 2 pi when I plug in pi is 1 so I just write one minus one4 cosine of 0 cosine of 0 is also one so actually these cancel also so in the end my answer is pretty simple here- 12 pi^ 2 okay so here was an example where I had to use integration by parts twice and you know what I could have guessed it because I had x squared so one derivative brought me down to X and the next derivative brought me down to one so if I had X cubed I'd have to use integration by parts three times okay okay just as a quick aside some students really like this method I never use it but it's actually really convenient same example let's do it without the bounds this time just to practice one without the bounds same integral same integral but no bounds there's something called the tabular method which is basically a quick Shand and of keeping track of all these different choices of your U's and DVS and putting them all together so that you end up with this so I'm going to show you that shorthand and you can kind of try to think for yourself why it works okay so here's the method you put U your first choice of U and all its derivatives in the fir in this second column okay U and it's word is hard to read U and all its derivatives and here I'm going to put your first choice of DV and all its anti-derivatives I'm going to explain to you what I mean and then you're going to have this altern signs this alternation of signs plus minus plus minus Etc as long as you need it okay so in my first choice I do what I would do to begin so this is U and this is the function that's associated with my DV usually we don't write the DX on this side okay so U is X2 and DV is s of 2x and now what I do is in this column I take derivatives so derivative of x^2 is 2x derivative of 2x is 2 derivative of 2 is zero this tabular method works when you eventually come down to a zero and here I'm going to take anti-derivatives so anti-derivative of s of 2x you can see this is mimicking my first step my anti-derivative of sin of 2x is this my derivative of x^2 was 2x so this first step is getting Mimi mimicked here so this is min-2 cosine of 2x and I do it again I take the anti derivative of this which is basically what I did here except in my case I kind of ignored the 1/2 and put it outside okay so here I'm including the one2 min-2 so I'm going to get min-2 integral of any derivative of cosine 2X is 1 12 sin 2X and there's another minus one2 here so it's - one4 sin 2X okay and then I'll do it one more time which is what I did at the very last step when I integrated s of 2x so that's what I'm doing here basically I kind of got rid of some of these constants in my previous calculation but I take the anti-derivative of s of 2x which is - 12 cosine of 2x and - 12 * -4 qu is + 1/8 so this ends up being +8 cosine of 2x okay so here's my table I'm going to stop at the row where I had the zero so this would be really convenient if I had X cubed okay I could just do this really quickly and then what you do is well what did I do over here I multiplied V and du I took the original choice that I had and du so what I'm going to do is so this was my original DV so this was my v and this was my um this was my U so I have all these U's this is my original du my my first du and then I kept taking derivatives so what you're going to do is you're going to multiply across and tabular Method always ends when this guy is zero because if you had anything more down here it would not it would give you zero what's happening is that this integral I'm just going to write it exactly using this multiplication and taking into account these signs this alternating between minus and plus has to do with in the integration by parts formula you have a minus here and if you're at the second part of the integ ation by parts and you use it again you have another minus here and minus and minus is plus so you have to alternate minus and plus I'm just giving you a quick feel for why this tabular method works but I'm not really writing it down in all detail so you might have to convince yourself if you want to understand it fully but here's what I'm going to write I'm just going to multiply these and I'll put the constants out front so I'm going to have minus a half x^2 cosine 2X that's this multiplication minus 2x * - A4 minus and minus becomes a plus 2 * a quter is a half so this ends up being uh well let me write it this way to really see - 2x * -4 s 2X and then I have plus 2 * this and this is an indefinite integral the way I've written it here so if I had this example I would write plus C okay and if I simplified that I would have - 12x^2 cosine of 2x - 2 4 is plus 12 x sin 2x 2 * 1/8 is 1/4 cosine 2X plus C okay and this is my this would be be my formula for the indefinite integral if I were doing the definite integral I would then have to evaluate all these between 0 and Pi is what I had in the P question okay so this thing here is the same as these guys here you can see this one4 cosine 2X is here the 12 x sin 2x it's here the minus a half X cosine 2X here okay so that's kind of a fun method you can use it if you like it I almost never use it just haveit I guess but it's a fun method and it works so it works when you eventually get down to zero okay okay so let's try um we have three more examples to go through so let's do one more and then we'll take a little pause let's do this one Negal of Ln let's say my variable wasn't X but it was s okay this doesn't really look like an integration by parts formula but it is oh I really don't have much choice for you because I only have one function so I'm going to put the one function I have to be U I'm calling things s instead of X so that means DV is DS okay so du what's the derivative of natural log of s it's 1 / s and don't forget the DS since I'm writing things in terms of s now there are no X's they're S's and if DV is DS that's like if I had a one times a DS so what's a function whose derivative is one it's if I had X I would write X but I'm calling it s so it's s okay and you can check D DS of s is one and that's what I want this one here okay so this is what the integration by parts formula gives me I'm doing an indefinite integral now it tells me let me just rewrite it again U * V minus the integral of v it tells me U * V is Ln of s * s for some reason we Al often write the s first I'll just write it that way but I could have also written Ln of s * s minus the integral of V du so that's s * 1 / s DS well this is excellent for me because what's s * 1 / s that's just one right and what's the integral of one it's just the variable so it's s don't forget not derivative integral and then I have an indefinite here so I'm going to write plus C so this is we learned in Cal one mysteriously the integral we often wrote it in terms of X the integral of Ln X DX is X Ln x - x okay so that's um the way to prove that integral