integration by parts, DI method, VERY EASY

I'm going to show you guys the easiest way to do integration by parts so check this out this is the D I method so here's my first example we have the integral X to a second power times sine of 3x we all unfortunately use substitution wouldn't work because if we let u equals two inside function 3x the derivative dies just as three we cannot cancel with the X to a second power so let's give integration by part a trial and this is going to demonstrate the first step of using the DI method the idea of the integration by parts is we are going to look at the original integral and then break down to two parts one part we are going to differentiate the other one we are going to integrate so we will put down a D and the I D stands for the part welcome to differentiate and I stand for the part we're going to integrate and we should always try to write this down first figure out which partial we integrate because it's usually harder to integrate right and then also on the side we shall put a plus minus plus minus and maybe a couple more but then let me just put on few of them to get ready make sure you have the plus minus plus minus alternate like that on the side just to get ready all right we have two functions X to a second power times sine of 3x and let me tell you we are going to integrate sine of 3x first of all we know this is doable we can totally integrate some 3x right and then I will just differentiate x to the second power and now let's just get work that's integrate sine of 3x the integral sign is negative cosine and the input is the same 3x but then we have to divided by the derivative inside 3x the during that is 3 so we have to divide by 3 so I will put down a 1/3 that and let me do it again let's integrate negative 1/3 cosine 3x and let's look at the function part first the antiderivative of cosine is positive sine right and the input stays the same let's really states the same but then it was passed this time so this negative will stay the same and will we have to divide by the street again so 1 over 3 divided by 3 we have 1 over 9 this is the small use substitution in our head ok so let's do this again just for projects antiderivative sine is negative cosine so let me put down cosine first but then negative times this negative will get positive okay and the input once again stays to say but then we have to divide by the street again 1 over 9 and then we divide by 3 we get 1 over 27 okay and it seems that we can just keep on going forever so that's positive right here and move to the D column so we are looking at x squared and we are going to differentiate that the relative of x squared is just 2x and then let's do again the realty of 2x is just a 2 and then the realty of a - yes 0 and claim we can stop right here I don't go down any further so let me erase this plus right there we just need work Rose this is the first stop when we see a 0 in the D column we stop once again when we see the 0 in a D column we stop and this is how we are going to read the answer from the DI table we have the answer already the answer is just going to be the product of the diagonals along with the sign on the side first we will have the product past the x squared times negative one-third cosine 3x and let me write it down as negative one-third x squared times cosine of 3x and this is the first part of the answer already and the second part is going to be this times that be sure you have this negative right here as will you take negative 2x times negative one over nine so you get positive 2 over nine and then we have this X and then we have the sine of 3x at the end we take this multiplied with that positive 2 times 1 over 27 so we walk first half plus 2 over 27 and then cosine of 3x and guess what we are done because you see if you take 0 multiplied with the next whatever this is gives 0 anyways right that's why we stopped when we see a 0 on the D column this is the first stop and then this is ends already at the end of course we put down a plus C that's it and check out the next one for the second stop here's the second example that's going to demonstrate the second step of using the DI method for in Turkish my parts integral X to a fourth power times Ln X so right away we are going to select something to be differentiated and then something else to be integrated this integral we have to use integration by parts board and then we also put down some plus minus plus minus be sure you alternate the signs and let me just put that down for the set up and now let's select something to be integrated first shall we integrate Ln X for example well if I put Ln X what's the answer to the integral of Ln X may be we know from the antiderivative table but to integrate our next we have to use integration by parts so I don't want to do an integration by programs inside the integration by parts right I have no choice but that's integrate X to the fourth power because we can totally do this and then let me just differentiate Ln X because to differentiate our next it's very easy so let's see I'm going to integrate this one time integral X to a fourth power is 1/5 X to the fifth power and then when I differentiate Ln X we get 1 over X that's all we'll show we keep going if you keep going right here this is not bad it's pretty easy right however if you keep differentiate one of X and fortunately this is not going to be 0 at all but it will never be 0 so first you have to remember we have to have a sense of danger we're using the DI method I'm going to stop right here because I know once I keep differentiate 1 over X this is going to be pretty crazy right and this will never end I don't know how to stop let me just stop right here in the first place how's that second of all and let's look at the product of one over X and X to the fifth power we get X to the fourth power right and can we integrate X to the fourth power of course and it's pretty easy as well right so in this case we can stop right here this is the second stop whenever we can integrate the product of a row we stop once again each row represents an integral whenever we can integrate the product of a row we stop so in this case I don't go down any further and then remember the product of the diagonal along with the sign like this it's the part of the answer already so we can write this down the first part of the answer is positive Ln X times 1 over 5x to the fifth power and we write this down first 1 over 5x to the fifth power and then the own X like that and then remember the product of a row with the sign is still an integral just like the original one to see X to the fourth power times Ln X and then you was a positive integral to begin with isn't it the product this right here he is still an integral and first of all it's a minus integral minus integral let me put this down right here for you guys 1 over x times that 1 over 5x to the fifth power and because this is the integral we put on a DX and what's this look at this right here is just 1 over 5 and then 1 over X and X to the fifth power is X to the first power and earlier I was just looking at the function part okay once again whenever we can integrate the product of a row we stop this right here is the answer the first part of the answer 1 over 5x to the fifth power Ln X and then we just have to worked out this integral it's going to be minus the integral 1 over 5x to the fourth power we add 1 to the 4 which is 5 and divided by 5 so we get 1 over 25 and then we have X to the fifth power and we are done so put plus C this is it the second stop and let's move on to the third stop now this example is going to demonstrate the third stop of using the di method for integration by parts we have the integral equal to x times sine X so let's get it right away we have D and I we are going to pick something to be differentiated and something else to be integrated and also that's put on a sign on the side plus minus plus minus I don't know how many I need let me just put on view to set up okay the functions are equal yet and then we also have sine X which one should we put to be integrated can we easily integrate e to the X of course right can we also easily integrate sine X yes in fact in this situation it doesn't matter which one we'll pick to be integrated doesn't matter let me integrate sine X and seriously if you want to integrate e to the X right here at the end you and I will get the same answer okay but I chose sine X to be integrated so of course I will differentiate e to the X and let's get to work integrating sine X we get negative cosine X and what's the integral negative cos x we are we know integrating Kozak's with their positive sides but then this was negative originally so we maintain the same negative sign once again all together integrating negative cosine X we will end up negative sine X and let's do one more to practice integrating sine X is negative cosine but then we have to account for this negative as well negative negative together we get positive cosine so be sure when you do this integrating sine cosine things like that do it carefully maybe you want to do it backwards double-check the relative of cosine is negative sine right derivative negative sine X is negative cosine derivative of negative cosine will end up parsley side so we could and you see seriously this one never end and it's just you know a lot of alternating these things so let's move to the D column let's differentiate e to the X well it's just e to the X e to the X e to the X and so on so on so on so right now here is the third staff stir staff look at the original integral we have e to the x times an X and pay attention to the function part in this row we have e to the X sine X just a function part okay I don't care about the positive sign X design when the function part repeats we stop you see e to the X sine X it's the same as the original so we stopped right here on this row I don't need to have this right here anymore so let me just cross them out okay and now we can construct the answer part and also the integral part that's C this is going to be make sure you do the product of the diagonal along with this side we will first have positive e to the x times negative cosine X so we will have negative e to the X cosine X and this is the first part of the answer already and the second part of the answer we will take negative e to the x times negative sine X so we get positive ETA X sine X the product the diagonals is the part of the ends already lastly I'm going to multiply this third row and then provided when we have the third row it is still an integral so look at this right here this right here it is still an integral the green dysentry stand out by its okay positive times negative we have a negative integral e to the X sine X and because this is the integral of course we put down the X at the end so now you see this is the original and then we also have this right here on the other side right this is the function part for the answer already and this is the stead of doing such a problem whenever we see the function power repeat in a part of the row we are going to add the integral e to the X sine X DX on both sides so the between right here as well plus integral e to the X sine X DX this way this and that or cancel each other out right so let's see on the left hand side originally app one of them but then I just add another one so altogether there are two of this integral to integral e to the X sine X DX and this is the same as this is the function part for the answer negative e to the X cosine X plus e to the X sine X now we're done now because we have this extra tube we don't want originally we don't have the tune in front right but it's okay because this means we have two times this integral I just need to multiply this by 1/2 I just need to multiply this by 1/2 I just need to by multiply this by 1/2 I multiply everything by 1/2 so that listen that will cancel and then I get the integral e to the X sine X DX which we know it will be negative 1/2 e to the X cosine X plus 1/2 e to the X sine X and then we are done plus C and this is it and this is the third situation this is the third stop we are doing integration by parts when we are using the TI method okay and hopefully you guys like this video if you liked it make sure you subscribe don't make me really happy thank you that's it

Channel: blackpenredpen

Views: 657,672

Keywords: Another way to set up the integration by parts, integral of x^4*ln(x), integration by parts, DI methods, easiest integration by parts, blackpenredpen, Tabular Integration, integral of e^x*sinx, easy integration by parts, blackpenredpn, bprp integrals, di method for integration by parts, calculus 2 integration by parts, integration by parts examples, integration by parts tricks

Id: 2I-_SV8cwsw

Length: 16min 58sec (1018 seconds)

Published: Sat Feb 20 2016