 So first we will talk about integration in general, how important is your integration chapter and how many concepts will directly be linked to your understanding of integration. Okay, so integration as a chapter itself how important is it integration. Yeah. So, if you just talk about the overview of the integration chapter, we will be talking about two types of integration. One is indefinite. Okay, indefinite integrals. Okay, in this particular chapter you are more mostly going to learn the process of integration. So it's a process driven chapter. Okay. So what is the raw process? What is the methods that we use to integrate different types of functions? Okay. And we are going to basically talk about how when a function is integrated, what does it give rise to? Okay. So what is the integral? What is the area function? And how do we find out by different tricks and tactics and procedures? So mostly that will be covered under indefinite integral. Okay, so today we'll be starting with this chapter itself. So by this we'll be learning how to find area function, or you can say primitive, okay, or antiderivative, whatever. Many books call it as antiderivative, some books will call it primitive, some books will call it as the area function. The second chapter which is directly linked to integration is definite integrals. Definite integrals or definite integration. This chapter is property driven. Okay. Because the process will be already known to you. Right, so the process you would have already learned all the processes in your indefinite integral. So why are you learning new indefinite integral? Nothing, but you are going to learn certain properties by which you are going to actually find area. Okay, or area related, you know, you can say, you know, expressions. Okay, so this is linked to area, I would not say exactly area, it is linked to area. Okay, those who have done definite integrals in your school would know that this is mostly driven by some certain set of properties. Next chapter which is directly linked is your application of integrals, application of integrals. And in this we are going to basically talk about area under curves. So this chapter is actually the one where you are calculating the area between two curves. And in the previous chapter which is definite integrals, you're going to evaluate area related integrals, so integrals which are basically related to area. Okay, both are different things to a certain extent. Okay, so we'll talk about that in our application of integrals. And lastly, we have differential equations. Okay. In this chapter, we are going to talk about the combination of your understanding of differentiation. And of course, integration. So basically it's the time to involve now solving equations which are based on differential coefficients. Okay, so dy by dx e to y by dx square or the derivative of any dependent variable with respect to independent variable when you form a equation out of it. How do we basically form the equation first of all and how do we basically solve those equations that will be covered under differential equation are very important aspects of you can say engineering and physics. In fact, in chemistry also differential equation is used and of course you are going to talk about it in our maths as well how to solve this differential equations is going to play a very, very heavy role in your undergrad topics also. So I remember in electrical engineering, we had a lot of application of differential equations and mostly like solving of circuits etc we use differential equations a lot of course we convert it to a simpler form by applying transform calculus. But yes, very important concept. In fact, you would have already realized in your chapters like simple harmonic motion, maybe in chapters like in a solving RLC circuit or transient states in case of you know AC currents. So all those concepts will involve the understanding of differential equations. Okay. So our main agenda today is starting with indefinite integral. So basically I take around three classes to four classes to finish off indefinite. We'll see how fast you are because you've already done this in school. So initially I'll be going slightly faster because you have done things in school indefinite integration indefinite integration indefinite integration basically is a process right. And from this process, we get something which we call as the primitive offer function primitive offer function primitive offer function so this is the outcome from the process. Okay, this is the outcome from the process. Now what is primitive offer function see whenever you, you have a function. Okay. So you initiate the function. Okay, so let's say you apply the derivative on that function with respect to x, let's say you end up getting something called the derivative of the function. So this is called the derivative of the function. Okay, derivative of f of x. And this f of x is actually called the primitive primitive of f dash x. Primitive the word itself literally means something which existed before primitive man. Okay, the man, the men who existed not much before maybe I'm talking about Ice Age and all those things. Okay, so the process of obtaining the primitive from the derivative. Okay, is through the process of integration. So we apply the process of integration on the derivative to end up getting the primitive. Okay, just to give you a simple real life example, you can say coffee beans. Okay, coffee beans are basically primitive of coffee, the coffee that you have early in the morning. Okay, that is actually a derivative of so derivative of this is your cup of coffee. It's derived out of it. It is derived out of it isn't it. Let some teams come out. Okay, and then if you want to obtain the coffee beans back from the coffee, you will have to do the integration. Okay, not literally by the way, this is just a funny example to just talk about it. Okay, so this is the primitive or let me write it in a correct English format also primitive. This is the primitive. Okay, primitive and the derivative they are all the same thing. Okay. So first of all I would like to discuss with you why it is called indefinite. Okay, indefinite let me give you an example for the same. Let us consider a simple scenario let's say I talk about 3x squared. This is derivative of some function. Let me name it as f of x. Now please understand this itself is called a first degree differential first order first degree differential equation. So this is actually a, I don't know how many of you have done differential equation in school. This is actually an example of a first order differential equation differential equation. Why it is called first order because if you see the highest differential coefficient order that is occurring over there is one. That means you differentiate it f of x only once with respect to x that is the highest times you have differentiated it. That is why it is called first order. Don't worry about order too much we'll be talking about order in our differential equation chapter in much more detail. To solve this differential equation you have to use the process of integration in calculus. Okay, so from this process of integration you will end up getting the solution to this differential equation, which is nothing but the primitive, which is nothing but the function. Okay, so which is nothing but the primitive of this guy. So how do you solve it, when you solve it. Anybody who can tell me the answer that you might get when you solve three x square. Whose derivative is three x square. Can you guess that right many answers will come from there some of you will say excuse some of you say excuse plus one some of you can say excuse minus one that there could be so many answers. That means there is not a single function whose derivative is three x square there are many such functions. In fact you will say there's a family of curves. You can say there's an indefinite. There's nothing definite here so there are an indefinite number of curves whose derivative is going to be three x square. So this gives you a family of you can say curves here. And because of that, the answer that you will get from here is excuse plus a C, where C is called the constant of integration also called the arbitrary constant. Okay, these all things are known to you already. I'm just, you know, just revising it so that everybody's on the same page. So this C is called the constant of integration. Am I breaking up. I mean, is my voice breaking up or is it fine. I was breaking up. Okay. Okay, let me switch off my camera for the time being maybe a little later on I will switch it on. So you get so many answers so there is no definite there is no definite answer. There is no definite curve. Okay, and hence the concept of indefinite. So this is why we call it as indefinite integrals. That means there could be so many curves like x cube, x cube plus one, x cube plus two, okay, x cube minus one says that if you take any x point on it let's say if I take any x point. Okay, and I sketch the gradient or sketch a tangent the gradient of the tangent will all be the same. Okay, that means all these curve have the same slope function at that point. Okay, so because there is no definite curve, there's a family of curves, that's why we call this particular topic as indefinite integrals. Now the first part of this chapter will be slightly you can say a repetitive part where I will be talking about integrals of certain standard function from the virtue of the fact that we know its derivatives. Okay, so the next part of the topic will be finding the antiderivative or finding the antiderivative of derivative or you can say primitive of primitive of standard function, standard functions, not all the standard functions by the way. Okay, by the virtue that we know the derivative of those functions. Okay, so by the process that differentiation and integration are inverse processes will be trying to find out from our knowledge of derivatives, antiderivatives of certain functions. Okay, so let me begin with a simple. I think I've already given this in your bridge course also. So derivative of a constant is zero so integral of a constant with respect to x is a constant. Okay, derivative of x with respect to x is one so integral of one with respect to x is going to be x plus a C. Please note that while differentiation is a, I can say one to one correspondence here it is one to many correspondence. Okay, so you have to write a C here. Now many times I've seen students, I know not using the proper notations over here see let me explain you first of all what this notation actually means, maybe you would have been and I explained this in school. This symbol, this symbol is basically a stressed s. Okay, as for summation. The symbol was devised by the German mathematician Godfrey and Wilhelm Leibniz. Okay, and this this expression itself is basically summing up something which we call as a differential area this is the area actually. So if you write an expression like this. Okay, this is nothing but a differential area and you're trying to sum up a differential area to get a area function out of it. Okay, so when you write an area function you need let's say this is my function f of x. Okay, this is my graph of f of x. So at a distance x. Okay, if you take a differential. Change in the x, let's say this is your DX. This is the height, and this is the width of this differential area. So this da is formed by the height into width, which is f of x into dfx. Okay, so f of x DX, that is your differential area, when you sum up this differential area, when you sum up this differential area basically you're integrating it. And this process gives you the primitive which is also called the area function. So the answer that we get we as an area function. So if you don't write this DX, please note that you're trying to find the area of a rectangle without multiplying it with the width, or maybe the length. Okay, so people who write such kind of an expression in the school exams and all, I definitely give them zero, at least for my natural students, because this is not an area function. Right, so please do not do such kind of mistakes when you have a subjective paper happening in, let's say in your board exam or in your school, of course in J and all nobody's going to bother too much about it. By the way, I hope you have all got to know about our J results, I think the best rank in the J main was around 900 by trippin. This is from NPS Raja Jean again 900 rank in J main exam is a very, very, you can say a good achievement actually because it is written by it was written by almost 10 and a half lakh students. Okay, so out of 10 and a half lacks being under 1000 is a very big achievement. Okay, so most of them I think every fifth student is a 99% and nobody knows some of them who have already scored 100% tile in physics and and mathematics. By the way, next turn is yours, so you people will be now facing the music in couple of months time and I'm sure you are well prepared. Please do not give it give up at this point of time because here you need to pull up your socks and work even harder. Okay, don't get burned out too many exams are going to come school exams. Pre boards maybe will happen very soon. And then you will have the first term. First time will continue for one month. And then you will have a crash course happening from our side. And then maybe who knows the J notification comes out for January so a lot of exams will be up. So, some of you called me in the last week session and said you're already tired. You're planning to, you know, slow down a bit. This is not a time to slow down. Let me tell you. Right, it's like slowing down in the last lap of your race, right. So please ensure that you are at your performing best for the couple of, I would say hardly six to seven months more. Okay, and then you will see the, you know, outcome of it. I think the notification has not come out only the result has come out for the previous day, just two days ago. So we are not very sure when the application will be starting but maybe one month at least you have to wait. Okay, and the problem with your batch would be now that there are two board exams so what will happen one month each will go in that. I don't want to give that one month time subconsciously it will be running in your mind that my boards are going on right, even if you want to you cannot prepare for competitive exams. So that problem was not there with the last batch, they had the exam toward the end of their, you know, completion of syllabus and you know whatever time it says it did not matter. Here, your board exams may, you know, continue till December mid so full one month will go in that and again one month one month four will go in the second semester exam. So one month here is like, you know, it is trapped. Okay, so you need to be further more judicious in time management. Are you getting one because in this one month probably you will not try to experiment too much. My suggestion would be if you get your schedule, which I think I'm told that you haven't got it yet from the board side. You just know it is going to happen in November, you haven't got the schedule right mid November right so ISC is happening in mid November so I think CBC will also be happening around the same time. So for ISC it is going till December 22nd. So maybe yours also will go to December 22nd maybe. And then I think New Year will happen and everything will you know take place during that time. So please ensure that you don't overprepare at the same time you don't underprepare also are you getting what I'm trying to say. Like for a 50% syllabus don't give three months of your time I mean you can cover up and this is an objective based exam you're anyways very good at it. So don't overprepare that is what is my concern. Okay, because later on nowadays what is happening that people are not paying attention to the semester two topics. As you can see the attendance in today's class is only 27, whereas it used to be around 38 or something. That means these 11 students think that semester two topic is going on why to pay attention. And when they will come in semester two this would have already been covered and that time they will be doing a self study of this topic and you know, doing a very shoddy understanding of the concept. So don't overdo it that is what is my request to you. If I were you I will first address the topics which has very less time or less time gap, maybe in some exam you only get one day or two day gap. I will do those topics, or I'll start doing those topics for those which for those topics which I have or those subjects which I have enough time let's say one one week time or one week gap is there. I will probably do the revision in the one week time one. Okay. And yes it is going to overlap with KVP as well. That's that's all the more reason why I'm requesting you don't overdo things. Okay, don't be like okay my life is now semester one topics I will not look at anything else. No that in that case you will not do well in KVP why and you'll have to you know you'll settle in semester two topics concepts. Anyways, I'll keep giving you all those guns little later on. So we already know derivative of x square by two is x. So integral of x dx will be x square by two plus C. And now this is good enough for us to figure out that if you want to differentiate x to the power n plus one by n plus one. The derivative is x to the power and so integral of x to the power and with respect to x is x to the power and plus one by n plus one and cannot be minus one here. So this is what we call as the power rule of integration, you are only aware of it. So you might be finding me slightly fast over here. Okay, because you have been using it since your physics first chapter. Okay, maths you are officially learning it right now. So please note this down. Derivative of derivative of. So integral of mod x with respect to x is x. Okay, so integral of x. Sorry, one by x integral of one by x with respect to x integral of one by x with respect to x is going to be ln mod x plus C. And which is important, many people asked that we were told that only derivative of ln x is one by x, how come ln mod x derivative is one by x. See it doesn't matter. ln mod x if you redefine it, it is ln x when x is positive and ln of negative x when x is negative right. Now it doesn't matter whichever form you differentiate let's say if you differentiate the first one. This result is already x, one by x you know, even if you differentiate this, you're still going to get one by x. Isn't it, so still gives you one by x. So it doesn't matter. So when ln x or ln mod x the derivative is one by x only. Okay, so integral of integral of x to the power minus one with respect to x is ln mod x per C. So this basically fills the gap where your n is minus one. So this is a place where your n is actually minus one. Okay. Let's also discuss the derivative of exponential functions, a to the power x by ln a derivative with respect to x is a to the power x. So integral of a to the power x with respect to x is going to be a to the power x by ln a plus C. So in the special case of it, when your a is e, it becomes something like this. This results are already known so I'm just going a little fast over there. Can I go to the next slide. And one more insider I have got to know from the training set I'm attending with respect to all India. Mass Teacher Association that your objective questions are going to be framed out of the subjective questions of your past year. I have come to know. So there has been already a team who is working to convert past year subjective questions into objective questions. Okay, so this is a I would say an insider information but mostly many of you would have already speculated it also. So please ensure that past year papers of CVSC at least last 10 years paper, you don't leave any paper unsolved. And now I think the market has already been flooded with MCQ type chapter wise questions and all you can easily find it in SAPNA bookstore. You can get it from there. I think you can also try sites like. They have soft copy of these Oswal, you know textbooks, but he has to pay some, you know, minor amount 150 200 degrees depending upon which book you need. Okay. You can check it out after the class. Okay. All right, so we'll be talking now about trig functions. Sinex derivative is already known with respect to x which is cos x and cos x integration with respect to x will be sine x plus C, sine x plus C. Derivative of negative cos x with respect to x is sine x so integral of sine x with respect to x is negative cos x was C. Okay. Derivative of tan x is known to us as secant square x. So integral of secant square x with respect to x is tan x to C. These results are already well known to us. I'm just putting it in a form of a note so that you can later on access to it. If you want to. So integral of C kegs is C kegs tan x. Sorry, derivative of C kegs is C kegs tan x so integral of C kegs tan x is C kegs plus C derivative of minus cos c kegs is cos c kegs cortex. So integral of minus integral of cos c kegs cortex with respect to x is minus cos c kegs plus C. Derivative of negative cortex is cosecant square x. So integral of cosecant square x with respect to x is minus cortex per se. And apart from this we have also learned our inverse trig functions. This is going to be sine inverse x per C. We can also write it as negative cos inverse x per C. Doesn't make much of a difference in both the results. By the way, in integration this is a well seen, you can say problem, or you can say commonly seen problem that your answer yet you get and the answer at the back of the book might not match. Okay. They may differ in constants, number one, or they may differ in the forms. For example, when we deal with trig functions, or many times we do substitutions to solve questions. So depending upon what substitution we have made your answer will be, you know, coming in form of that. Somebody has solved the same question by using some alternative substitution. He will get mostly the same form but yes, in a more, you know, you can say in a more simplified manner or in a more, you can say different format. The answer at the back of back of the book and the answer that you get may differ either in constants or they may differ in this simplification. Maybe some trigonometric identities have been applied to simplify it. So please watch out for those things. In fact, in comparative exams, including your first semester exam, or second semester exam if it happens in objective format, please be driven by the options. Okay. Don't solve this, don't solve any question in objective format without looking at the options. Look at the options first. Okay. Try all your, you can say MCQ solving, you know, strategies. For example, let's say if you have been given A and B values to find out and you realize that A values are same in all the option. Why will you face time finding A? Let's go for B first, because B will be the distinguisher. So like that you save your time in these comparative exams as well as your semester exams also. Okay. I mean, you're only experienced enough after one and a half years, you know your, okay, by the way, this itself, I mean, it comes from directly this isn't so which I don't have to write it again. So this is either you can say cos inverse x plus c or you can say negative sign inverse x plus c. Okay. By the way, for tan, I will write without mentioning cot inverse. So this is tan inverse x. I will not write cot inverse, cot inverse is directly driven by this. Next is a seek inverse. Seek inverse is one by x under root x square minus one. So integral of this term is going to be is going to be seek inverse x plus c. Okay. However, don't be surprised if you see minus cot inverse x plus c here and minus cos inverse x plus c here both are same thing because tan inverse plus cot inverse is pi by two. You can always write tan inverse as pi by two minus cot inverse pi by two plus c will become another constant of integration. Now something very optional over here if you are stuck with a problem where it involves hyperbolic functions. So these are optional results. I mean just for your extra knowledge. So hyperbolic functions, many of you would be not aware of I'll just talk about it in brief. So hyperbolic functions, sine hyperbolic x. By the way, this is pronounced as shine. I mean the English word shine, the sun shines. Okay, shine x. So when you say shine x, it is to be treated as sine hyperbolic x and sine hyperbolic x expression in terms of exponential function is this. This is kosh kosh. I'm so sorry. This is, this is minus. This is plus. Yeah. Then Dan, and hyperbolic x Dan is nothing but shine by kosh, which is e to the power x minus e to the power minus x by e to the power x plus e to the power minus x. Note that shine is an odd function. Okay, so this is our odd function. Kosh, just like cause is a even function. Okay, even fun. It is just a name given to it. See the name hyperbolic functions were given to this because it was observed that cause hyperbolic x minus sine hyperbolic x was actually a one. You can try it out by using these two expressions. And this very much resembles something like x square minus y square is equal to one which happens to be a rectangular hyperbola. Correct. And because of this it was given the name of a hyperbolic functions. So this term that you see this term that you see this term that you see they're actually given a special name. Okay. Sine hyperbolic x cos hyperbolic x tan hyperbolic x just for your information, Anusha. Many times, such expressions are not used but in case they are used, you should not be like surprised. Okay. A few things that I would like to share over here is that sine hyperbolic x derivative is cos hyperbolic x that you can easily see over here. If you differentiate this, you get the cos hyperbolic x function. Similarly, the derivative of cos hyperbolic x or cos is shine. Okay. Anyways, and a few things that we will be requiring over here is these functions have inverses. So inverse of hyperbolic functions. You already know how to find inverses because you have already done it in the function chapter. So please note down the inverses also. Sine hyperbolic inverse x is given by in the language of Lawn it is ln of x plus under root of x square plus one. Okay. This is sine hyperbolic x. Now those who have done integration in school. What do they represent they have a graph. I'll just show you the graph in some time cos hyperbolic inverse x is expressed as this. Have you seen this expression in integration concept in school. Do you realize that this this came in some special standard integrals right right right right. Okay. So please note this down these will be you know important tan hyperbolic inverse x is half ln. One plus x by one minus x. Okay. So, later on when we do our standard integrals will come to know that integral of integral of under root of one plus x square or x square plus one is given by sine hyperbolic inverse x plus C. Or what do you know as ln of x plus under root x square plus one. Okay. Similarly integration of under root of x square minus one is cos hyperbolic inverse x plus C, which is ln of x plus under root x square minus one. In school we don't tell that this is actually hyperbolic inverse or sine hyperbolic inverse or cos hyperbolic inverse. We only use the ln part of it but this is just for your info that if you come across such an expression you should not be surprised. Similarly this term is one by two ln or you can say one. Let me write it in terms of hyperbolic first so it is written as tan hyperbolic inverse x plus C. What do you know in school is half ln mod one plus x by one minus x but both are the same things, both are the same things. Yes, so I'll show you quickly the hyperbolic functions how do they look like. Normal curves only not something outside out of the box. Okay. So when you write s ln it will prompt you for this. Okay. Yeah. So this curve is same as y is equal to e to the power x minus e to the power minus x whole divided by two. Okay. As you can see both the curves have are overlapping each other. I'll just switch one on. Okay. Both are the same thing. Okay. So this is a sine hyperbolic function and it's a mirror image about y equal to x line is your sine hyperbolic inverse. Let me write first and then I explain you plus one to the power of point five. Yeah. So this is your inverse of this function as you can see their mirror image about y equal to x line. Okay. So this is what I was discussing with you. And I would also like you to see the graph of cost hyperbolic inverse and cost hyperbolic x both of them. So this is your course. It's like a parabola only. I had shown you in the bridge course maybe. So this is actually called a catenary. Okay. And it's inverse also. Let's see the graph. So y is equal to ln x plus x square minus one to the power of half right now in the present shape it will not be at lonely we partly warrior x which is positive so let's do it first. Okay. So it is only the mirror image of this part. Okay. When your x is positive because right now it's not a one month function. So it is not going to be invertible in the present ship. But if you just draw half the part of it just this part only. Okay. So this part inverse when x is greater than x is greater than zero. Okay. I think you can. Yeah. X is greater than zero you'll be able to find out the universe of it. Okay. Anyways. So this is the concept related to hyperbolic functions. So if you see your response in the competitive exam in terms of hyperbolic inverse. Don't be surprised. So many a time they will write this as sine hyperbolic inverse cos hyperbolic inverse. I would not use the word many a times, but it may be a random case. It may be a random scenario. Okay. So yeah. I'll wait on this slide. No worries. This should be greater than one actually greater than equal to one. Then a new show. Okay. Let's now let's now look into the rules of integration. Or you can say algebra of integration. Now I will not disclose all the rules here because we have certain rules which we'll learn a little later on like integration by parts, et cetera. So the basic rules I will definitely discuss with you. The first rule is if you have a function multiplied to a constant, then you just take the constant out and multiply it with the primitive and just put a C at the end. Okay. Of course, these results are also been already used by you. So this is not a new thing for us. We have already been seeing this. So if you want to find the first rule of integration, it's not a new thing for us. We have already been seeing this. So if you want to find the integral of, by the way, the function whose integral we are finding, it is called as the integrand. For example, this is your integrand. Okay. And here, please note that the C that you will be putting will be only toward the end of the process. Okay. Don't put a C every time that you integrate. Just combine all the C's and put it towards the end. So why should we put a C out? We just, I didn't get it. Taking the cons out means. Yeah. I mean, whatever you're doing, I mean, even if you don't write it, it's fine. I mean, the constant of integration will automatically come out from it. Okay. Next is, by the way, this rule I will write it in red for the simple reason because such a rule doesn't exist. And many of you who have learned this chapter in the bridge course, while I was seeing the assignment, I saw that most of you had done a very, very, you can say outrageous extrapolation of this particular formula. Applying your product rule of differentiation into your product rule of integration. So please note such a rule if you are thinking. Okay. As of now, nobody is thinking of this. I'm sure. But such a rule doesn't exist. Please do not apply such rule ever. Similarly, one more rule, which I would again write it in red because this rule doesn't exist. So if you have something like this, please do not do this. I'm just writing it in a very, you can say. Mockery sense, but many people in the bridge course. I mean, brisk or severe too young actually. So you're too early for you to learn this concept. These things are absolutely rubbish. Okay. They don't work. Okay. There is no quotient rule. There is no product rule like the way you have in your differentiation. So no, no extrapolation of any rules, please. No extrapolation of any rules. In fact, later on also this will be applicable. Okay. I was taking a session on integration in my school and some of the students. They thought that integration, you know, is just the opposite of what or just the same way as what they use in differentiation. And this is integration is very easy, very easy. And I started giving them few questions, basic ones. And I was shocked to see that they were, you know, flouting these rules left, right and center. Okay. So don't do these mistakes. Don't do these extrapolations just to, you know, apply differentiation concept. See both are not only inverses in the process, but also inverses in the approaches also one is very conventional while other is not conventional. I don't want to scare you. However, we have something called integration by parts. So that is something which you will see later on integration by parts. So integration by parts helps us to find out the integral of product of two functions. But this is to be used only when it is required because if you see the rule. This rule comes with a cost cost of you. This is in the form of another integral. Okay, so one integral gets converted to another integral. So please note that this process is not to be misused because you may end up getting an integral which is even more complicated. Or you may not be able to finish up this process it may enter into a you can say infinite loop kind of a scenario. Okay. So this will see later on not to worry about it. I think maybe the next class will be talking about it, not as early as today. And one of the most important rule that we are going to discuss here is the reverse chain rule. So reverse chain rule is basically a very you can say mild version of substitution. Okay. So why I call it as a mild version of substitution. If let's say a integral is known to you. As G of X plus C and you want to evaluate integral of a linear extension of X. So let's say X is linearly extended to X plus B and B could be any real numbers. Okay. Then you can also write this as G of X plus B by a plus C. This comes directly from your substitution concept which will be taking up in some time. So how it works basically is like you take your A X plus B as a T. So a DX will become a dT. So DX becomes one by a dT. So when you substitute it in this expression it becomes f of t one by a dT. And this integral are the primitive here is G of T so you can write it like this and just add a constant of integration. Of course, don't leave your answer in terms of new variable. Okay. So please do not misuse this rule. For example, if you know Cossack's integration is sine X and somebody asked you what is the integration of costs to X plus three. Okay, so let's say Cossack's integration is known, which is sine X. And somebody says what is the integration of cost to X plus three. Then you can write this answer as sine to X plus three by two. Okay. But let's say somebody asked you what is the integration of costs of X square. Then no, then this method is not going to work. Okay, don't start dividing it. Don't do sine X square divided by X and all. Okay. So this is not going to work in such cases as I told you no extrapolation of formula. By the way, there are certain integrals which are not integrable in the sense that they cannot be integrated by the methods that you will be learning. For them, you need to use some numerical techniques which is called the numerical solution of integrals. That's beyond our scope right now in fact, some of the international boards do teach them in their class 11th and 12th. For example, I have been taking classes for al curriculum a levels. So there they teach them some of the numerical methods to evaluate integrals. But many a times you will not be exactly able to evaluate an integral, you know, because it may be not be driven by certain procedures. So then you need a numerical solution of integrals, which you will learn later on maybe in undergrad so in a ICT syllabus it is basically taught in your undergrad curriculum. In undergrad curriculum for four semesters you will be doing maths out of eight semesters of your BTEC. If at all you're trying to do BTEC from any engineering college for four semesters you will have maths. So in each of the semesters you will be given some different mathematical tools. So those numerical techniques are definitely a part of your I think I learned it in my third semester and I don't exactly remember third or second semester. So those things will not be covering in our class 12th or JEE preparation. Alright now time to test you on whatever you have done so far. I know most of you have already done basic integration in school. So we'll do some general problems just as a warm up. Okay just to check where do you actually understand in your understanding of these rules and your formulas. So let me begin with the simple question. Integrate secant square x into cosecant square x with respect to x. On your chat box just say done. Done very good. Now three done okay. Can we discuss it out? Okay brother let's discuss it out. Anusha done okay. Alright so let me ask Anusha how did you do this question? If you're going to mute and talk that would be great. So you had given us a rule in the previous few slides. Which rule? The integral of f of x g of x dx is equal to f of x integral of g of x dx minus integral of d by dx of f of x that rule. Oh you use that rule? Yes sir. Okay. I mean wasn't it complicated using that? I realized after I almost finished it. So please do not use the integration of parts very so early because definitely you will be using it later on. But only for problems which require integration by parts to be used. See first of all I wrote this guy as 1 by cos square sin square and I wrote that one as sin square plus cos square. Individually divide each one of them by the denominator. So if you divide this will become sec square. And if you divide this by this it will become a cosec square. Okay. And these two are basically your functions whose standard integrals are already known. So this is tan x and this is minus of cortex. So your answer is tan x minus cortex plus c. Is this fine? How many of you got this answer? I don't think so that will give that answer for them. Okay. All right let's take another one. Let's do x square plus cos square x multiplied with cosecant square x. Whole divided by 1 plus x square. My problems may seem to be slightly challenging because why I'm purposely giving you challenging because you've already done it in school. Okay. Is there any person who is attending this session who has not done integration in school? I don't think so there's anybody like that. Mostly all of you have done it in school. Okay. So I'm just, I'm giving you only them handpicked problems which are slightly main look complicated to you. This topic is practice driven. So even if you are, you know, preparing very hard for your semester one topics. Please save some time for doing integration in between. But that's not, that's not what you gave me, right? You gave two cortex to me. How is your answer same as this? Just write it done if you're done. Okay. See this doesn't, the solution of this doesn't require any prior knowledge just whatever we have discussed in the previous seven to eight slides. That is more than enough to solve this question. Don't start applying higher versions of your knowledge of integration. Maybe we'll be saving that for our upcoming concepts. Okay. Yes. Any success anybody having no quotient tool etc is basically a challenge actually so you have lesser tools. The complexity of the problem is almost the same as what we used to get in our differentiation. But we have a lesser tool to basically get our results. So integration becomes more challenging. Okay, so let's solve this question. So what I'm going to do is I'm going to add a one here and subtract a one. Okay. Just because I want to create a one plus x square term. How will that help us? And how will that cosecant square be taken care of? Let us try to have a look at it. So first of all, I'll write it as x square plus one minus one minus cos square. The whole thing multiplied with cosecant square. And one plus x square in the denominator. I will just simply write it below these terms. This is anyways a one. This is sine square by one plus x square. And there's a cosecant square waiting outside. Multiply with this cosecant square throughout. Now if you see this form is pretty much integrable because cosecant square x integration is minus cortex. And one by one plus x square is tan inverse x. So integration of one by one plus x square is a tan inverse x. We have already seen these formulas in our standard results. Is it fine? Aditya, now it is clear. Yes. So you have to think, you know, unconventionally while solving these questions. Any question, any concerns here? I think only Siddish could solve it. Can I go to the next slide? Okay. Next slide. I'll give you a question. Which is actually very commonly asked question in board exam also. And there's a twin of this also. There's a twin question of this and that is tan x by C kicks plus tan x. Okay. So please try this out. So very simple question. Again, let me know by saying it done on the chat box. Done. If you just have a good command on your trigonometry, you can solve this question within 20 seconds. Siddharth is also done. Siddharth is also done. Done, done, done. Gaurav is done. Gayatri is done. Great. So you just have to recall that one by C kicks plus tan x is actually C kicks minus tan x. That's it. I'm sure you would have done this property million times in class 10 itself. Right. So this question is as good as C kicks times C kicks minus tan x. Which is actually nothing but integration of sequence square x minus C kicks tan x. And this is only a tan x. This is only a C kicks plus C. Okay. Same goes with here as well. The twin problem. Okay. So here one extra step you might have to do. You get a minus tan square. Right. Minus tan square. You just write it as sequence square minus one. Or you just do one thing. Add a C kicks a practice. You'll get a one minus this integral. Okay. Anyways, we'll do it here also. It doesn't take much of a time. So this will give you C kicks minus tan x integration of this is going to give you a plus x. Sign cause conversion also will give you the same result. But yes, maybe with slightly more time involved. Isn't it? Time factor will be slightly more next. Any idea anybody? Okay. Okay. Yes. So basically you have to write this expression as a square as a try to convert it to a perfect square. Not a perfect square. Try to make a square out of it. So this expression is x square minus seven express to a right. Can I write it as this x minus seven by two the whole square. I'll give you minus 49 by four and you want to so minus one by four. In short, it gives you two x minus seven square by minus one by four. Right. Okay. So this term over here that you have within the under root sign within the under root sign this term can we not write it as two x minus seven the whole square minus one by four that four will go on the top and become a two. Okay. Now let us try to recall one of the formulas that we have done in the beginning part of today's chapter, which was integral of x and the root x square minus one. What was it? Seeking inverse express. Right. Now you can see there's a linear extension of x as two x minus seven. So the same result will now become twice off seeking inverse two x minus seven by two. Of course this two and two will get cancelled off. Giving you the result like this. Okay. Many times people ignore this formula. Okay. They don't remember seek inverse x the integral of one by x and the root x x square minus one seek inverse x. Is it fine. See there are many ways to do it also. You can substitute two x plus one is one by T. But that process will discuss later on in fact much later on in our chapter when you do irrational function integration with you. Yes, you can do this by that method also. That is a more robust way. This is this was like by chance this came out to be like this and okay. Then you realize, oh, I could apply seek inverse x answer over here. So the other method which I mean I'll be going to talk about later on will be helping you to integrate something like linear under root quadratic. So this form I will take up a little later on in the integration chapter and I will tell you how to integrate the very same by this process here. Actually you will you're not aware here. The substitution is the linear term should be substituted as one by T. But anyways, we'll talk about it later on. Officially, I have not even started talking about substitution to be very precise. None of the question that we did required substitution. Okay. I'm just checking your basic integration skills. That is why I wrote general problems on integration. I'm not even started any kind of substitution so far. Okay. Next one. Maybe I should switch on the poll for this. If you're done, you can give a response on the poll. Okay. Got off. That was quick. I think one and a half minutes only. And two people have already responded. That was quick. Good. Good. CT like is all about speed. Concepts are pretty simple. It's all about speed. Three people. Three answers. Okay. What is this? Okay. Finally, one option is leading. This is a CT based question. Okay. Last 30 seconds because almost two and a half minutes have gone. So I joined in now. What happened? So. The internet was out with whom? So. This excuse is very old. So. Give some new excuse. So. Normally when internet's go out. There's some kids. There's some cable damage or something that normally takes. At least half a day to get repaired. Sir. Very frankly speaking, sir, I had a lot of rice in the afternoon and I went to bed and I. Went into deep sleep. And now I woke up. Just kidding. I trust you don't worry. Okay guys. I will call it off now. Five. Four. Three. Two. One. Two. One. Take a guess. Take a guess. Take a guess. Okay. People are refusing from guessing. Sir. J main exam said negative marks. Okay. So most of you have said B. B for Bangalore. I know why people. Be supposed to be the safest option when you don't know it. Okay. Now the very first thing that will come in my mind. In fact, it should come to the mind of any student is cause a plus cause B formula. Which is to cause a plus B by two. Into cause a minus B by two. Okay. Now. Now, now what to do now what to do. Now, can I use half angles here also. So instead of cost three X, I can write to cause square three X by two minus one. When you simplify this, you would realize in the denominator, at least I can see a three minus four cause square three X by two coming. Now this basically rings a bell in my mind. What bell it rings is that had it had I provided it with a cause X by two, this fellow would have easily become the formula for cause nine X by two, which is sitting on the top. So let's do that. Let's provide it with a cause X by two. Okay. So when you provide it with a cause X by two in the denominator, let's see what miracle happens because of that. So down in the down in the. Oh, there's a three X by two. Oh, my bad. Sorry. This is three X by two cause X by two cause three X by two. Okay. Now down in the denominator, this becomes three cause three X by two minus four cause cube three X by two. This is actually a formula for cause nine X by two. Isn't it? So cause nine X by two gets cancelled from here and here. Okay. Leaving you finally with leaving you finally with two cause X by two cause three X by two. Now basically use your transformation formula to convert product into some. This is as good as saying. Yeah. There was a negative sign also. I believe it. Yeah. There will be a negative sign also coming because this is a formula for negative cause three. Yeah. Negative sign. Yeah. So this will give you, this is going to give you nothing but negative. Just combine it back costs to X. Plus cause X. Isn't it? Isn't it? Any questions? Any concerns? Yeah. So this is nothing but negative sign to X by two. Negative sign X. In fact, you can take a negative common plus a C. Is this an option for us? Yes. Option D. Now this was just a play with your trigonometric identities. If you know your trigonometric identity as well. Then this is basically a simple. Is it fine? Next we'll start taking up. A question which will basically, you know, require us to start thinking in some different angle as well. Let's take this, this question. I put the poll on also in case you want to. Give a response. Okay. I've got a response. Okay. Those who want to respond, please do so within the next 30 seconds. Okay. Five, four, three, two, one. Okay. Not many of you have responded, but again, I think just a guesswork has gone in. Most of you have again said B as your preferred option. Anyways, we'll check whether B is correct or not. See, here you realize that the new denominator term. It's an expanded format. Okay. Just try to differentiate this term. If you differentiate this term, we'll get three X square plus e to the power X plus X e to the power X. Okay. Now. To a certain extent in the numerator that you have X e to the power X minus two e to the power X. Let us try to generate that term using this derivative. Okay. So this I want to make something very interesting, you know, very important outcome from this particular approach. So if you write this term, you'll see that this term is already there. So I'll just write down this term exactly as it is. And then I'll do some omission and addition if at all it is required. So I don't need a three X square because there's no three X square in the right side. I need a minus two excuse. So I'll do minus three to the power X. So as you can see, this term is already there. Minus two e to the power X. This will give you minus two e to the power X. And there's no e to the power. There's no three X square terms. So I subtracted it. So basically I've written the numerator in a very, you can say interesting manner over here. Now, why did I choose to write it like this? Let us now divide this term by the denominator itself. Both sites. The problem statement is integrating this site. That means you have to integrate this. So how does this breakup help us to integrate it? Let us look at the terms written in white first divided separately by the denominator. Okay. X cube X e to the power X. Any way you can write it doesn't matter. Here if you take a minus three common, you have X square e to the power X divided by X X square e to the power X. So you are basically willing to integrate this. So first thing that you'll observe that this and this will get canceled off and you can write it separately as two integrals. So what you see here in the first integral is that your numerator contains the derivative of the denominator. Isn't it? Because this term was actually the derivative of the denominator. So this whole term that you see is your differential of this integral. Okay. Now this sum is already well known to us. It is three L and mod X plus C. Now please understand here that whenever you get such kind of an integral, it is basically hinting towards the fact that if I start calling this as another variable, let's say a T. Okay. Let's say I start calling this as a T. Then your numerator is going to be a DT. Okay. So here is you can say the evolution of substitution method, right? And substitution method is one of the, I call it as a heart and soul of integration because in almost 70 to 80% of the question, you would require substitution. Maybe once, maybe twice, maybe multiple times also to solve the question. So substitution is basically a method in integration which converts a complicated integrand into an expression which is easy to integrate. So it just converts one integrand to a simpler integrand whose result is easy to be found out or easily available to us or easily figured out. Okay. So this expression itself would have been difficult one until unless you realize that, oh, if I call my denominator as a T, numerator will become a DT. And now one by DDT integration, anybody can do it because this is one of the standard results you have already seen. So your answer becomes ln mod, whatever is the denominator term. Okay. Minus, you may write it in a fancy way like this. Okay. And club up your, your ln terms together. So it is going to be one plus e to the power x by x square plus a c. In fact, all the terms are positive. So you can just remove the mod and put normal brackets away. So is this in the options? Yes. Option number A. Great. So here is now the approach of substitution that is being introduced. So now we'll talk about method of substitution from here on. And the very first thing that I would like to, you know, discuss over here is how to find integration of certain functions which we had not seen in our formula list. If I go back to the formula list, you would realize that in the formula list, I had not given you the integration of many trigonometric functions. If you see this list, there is no integration. If you see the right side, there's no integration of tan. There's no integration of cot. There's no integration of seek. There's no integration of cosec. So all these results which have skipped out can now easily be found out by method of substitution, which I'm going to take up now and we're going to spend a considerable amount of time on this. Tell you are all confident about this approach method of substitution. So this is this method is not a, you can say a conventional step by step mechanism. Okay. Do this, do this, do this. And you get your answer. It is just so you can say a ballpark approach which helps you to solve a problem. It doesn't give you a precise, I know this thing that, okay, whenever this comes, you have to do this substitution. I mean, of course, we have made some rules because of so much of practice, but this is mostly driven by your art of substitution. So it's an art, I would say, just like coding is an art. Okay. So in this part, you'll have to be observant. Okay. If I do substitute this, probably it'll convert the problem to a simpler integral. Okay. Why did you choose to use substitution here? Because without that, how will you solve it? Partial fractions. You know what is partial fractions? Partial fraction is an approach to convert a proper rational functions into simpler rational functions. We don't apply partial fractions to trigonometric functions and exponential functions directly. We have to use, you have to convert it by some substitution to a polynomial based function. Do you call this as partial fraction? This is not called partial fraction. Maybe your intention is to call it as breaking up into fractions, but this is not called partial fractions. Okay. And see, the thing here is that if you don't use substitution, how else will you solve it? That is my question. There is no, you can say quotient rule. There's nothing like you will break it out from it to get a simpler function. Even if you do, what will you do with that? Are you getting my point? So the thing is, you have to learn to think out of the box sometimes and mostly that will come from exposing yourself to more and more questions. So I'll start my discussion with, I'm sorry, I should start my discussion with tan x. So when it comes to method of substitution, the very first problem that we'll be targeting to solve is integration of tan x. Okay. So in tan x, we have sine x by cos x. Now this is a plain simple expression. What do you do with this? Or are you aware of any function whose derivative is tan x? Then that is good enough. That will help us to solve our problem. But unfortunately, I am not aware of any function directly. Or even if I have done in the past, I don't remember having seen a function whose derivative is going to be tan x. So how else are you going to solve this question? What if it is inconsistent equation? What if it doesn't give you any A and B? If it does, then well, I'm good. Replacing x with tan inverse x, then what will happen? Then you have to change your dx as 1 by 1 plus x squared. No, I'm not saying that is wrong. But ideas, you have to substitute at any stage. Whether you do, see, what you're saying is absolutely, you know, you can apply it. You're saying let's substitute x as tan inverse of let's say t. Okay. So dx will become 1 by 1 plus t squared dt. Okay. When you put it over here, it becomes tan of tan inverse t. Okay. Into 1 by 1 plus t squared dt. Correct. Which is actually, I mean, even if you consider this to be this expression. Here you have to use a substitution again. And the substitution is you have to put 1 plus t squared as that's another variable. Let's say you. Right. And differentiate both sides with respect to t again. You'll end up getting t dt as du by two or half do. So it becomes half du by you. And that's nothing but half ln mod u plus C. So substitution was required, whether you go for, you know, this kind of a substitution or any other substitution. So it's half ln 1 plus t squared. Okay. And T was, I think tan x. Okay. That is nothing but half ln C square. And that is, ellen's more C K expressly. Absolutely. This is a right answer. Correct. But the point is substitution, substitution was required to solve it. Okay. Directly. Why do you do so much things when you can do it? Let's say I call this as approach number one. You could do approach number two where you directly substitute cos x as your t. Okay. Why do you do so much of things? So when you do that, you get negative sin x dx as your dt. So your numerator. By the way, I'll just rewrite this step once again. When we teach a person a new, I mean a novice person integration, we do it separately. We say we differentiate both sides with respect to x and we write it like this. And then we shift the dx to the other side. I hope nobody has a problem. I'm going to jump from this step to this step because I don't have time to write all these stuff because we are up against, we are up against, you know, time crunch in the examination hall. Okay. So your numerator is directly negative dt. Okay. Denominator was a t. So it gives you directly ln mod t plus c, which is directly minus ln mod cos x plus c, which is directly ln ckx plus c. Okay. Now I can still solve this problem if I take sin x as a t. But problem is we have to find out a way which will give us a faster result. So that's what I was saying. Multiple substitutions can lead up to the answer, but the challenge is this substitution will save my time. Right. So I could also take sin x is t and get my problem done. So I'll give you a third approach. Third way to solve it. Okay. So what I will do, I'll take sin x as a t. Okay. So cos x dx will become a dt. So I will manually generate a cos x here by multiplying and dividing it. So this could be written as sin x cos x dx. You can club it as a dt term. This you can write again as a sign. Okay. So this whole thing becomes t by one minus t squared dt. But here I have to waste my time doing one more substitution. Let's say I call it as a K. So minus two t dt is dk. So this can be written as a minus dk by K, which is again negative ln mod k plus C, which is again negative ln one minus t squared plus C, which is again negative ln one minus sin square x plus C, which is again ln. Sorry, there was a half factor also. Sorry about that. Yeah. So this is again ln mod ckx plus C. Okay. So so many ways to get to the same result. Three ways to get to the same answer. Okay. So going forward, please put this in your formula list. Integration of tan x, integration of tan x, no doubt is ln mod ckx plus C. Before I move on to further questions. Some, you know, mistakes which people do, and I would like to warn you against those mistakes. Number one, and that's a very commonly seen mistake is forgetting to substitute dx in terms of the differential of the new variable. Okay. That's number one mistake. Okay. Many people have seen they will substitute the entire function in terms of the new variable, but they would not replace the dx term. Okay. That's number one. Number two mistake is keeping two variables into the same integral. So they will have a, you can say, kitschy kind of an expression where there will be a t also. There will be an x also. Okay. And many of them, I don't know with what smartness they will treat one of them as a constant with respect to the other. Okay. And they'll start solving the question. Okay. You may laugh at these mistakes, but in my teaching career, I've seen these kinds of mistakes coming up. Are you getting my point? Okay. There was a student I would not like to very long ago. I think in 2015. Okay. The question was something like this. Hmm. Question was integrate x minus two x minus three. Okay. So the question, the student, this is just to tickle your funny bones. The question, the student, you know, read it as root of this. So he said, okay, root of this quadratic is two and three. Correct. So he integrated two, he integrated three, and he gave those two as the answers. Can you imagine people doing these kind of stuffs also? Okay. CBC is doing this kind of stuff. Okay. So please don't, you know, give these kind of shocks to your teacher in the examination hall. Root means literally she found the root, the root of that quadratic, not the square root, the root of the equation. Since both have the word root inside it. So she confused the root of that expression with finding the root of that quadratic. And she got the roots and just integrated the roots. Okay. Anyways, Anusha, what was your doubt? Which step you did not understand? Everything is evident now here. So from here, t dt became minus half dk. So this term became a minus half dk. This term was already at k. So one by k integration with respect to k is ln mod k, minus half is already there. k was one minus t square. t was your sin x. So this became, this became minus half ln cos square x. Let's see. Minus half means reciprocate it and take an under root, which is going to give you a seek. No, not necessarily. Aditya. Not necessarily a denominator has to be taken as t. I have seen question where your numerator is also taken as a t. I'll give you a question on that side of that sort. Okay. Please do not make those kind of extrapolation that always denominator term is substituted as a t. No. See, I will not give you any steps over here. I'll just tell you the rationale behind substitution. Why do we do substitution? Whatever part of the expression you substitute, what makes you do that substitution, that is only what is important. And that rationale is converting it to a simpler integral. So any substitution which makes your problem simpler than the previous one or a one step closer to you getting it solved can be used. Now it needed to be always of a term in the denominator. It could also be a term on the numerator. Okay. All right. So having continued with this, we'll also talk about integration of cortex. Okay. So for cortex also, I will be saving your and my time by doing it directly in this. Unfortunately, the approach is the same. Hence you will have a feeling that always the denominator is taken as a t. And not only that, you will realize that every time it's get converted to Ellen only, but it is all coincidence and it is happening and it will be happening in all the four integrals that we are going to take in a few minutes time. Okay. So it's going to be Ellen mod sine X plus C. Please note this down. Very important result integration of cortex with respect to X. Added to your formula list is Ellen mod sine X plus C integration of C kicks. This also has a good story. By the way, I would request you to attempt this question. What is integration of C kicks? First thing that you would write is one by costs. Okay. Then what do you do from here on? Multiply and divide with sign. No. Actually multiply and divide with costs. Okay. So that will make it cost kicks and the denominator is cost square cost square is one minus sign square. Okay. Take sine X as a T. Okay. So cost X DX will automatically become a DT. So this will become DT by one minus T square. Okay. Then what do you do with this? Okay. Now all of you, please pay attention. This term is actually one minus T into one plus T. Can I write it as something like this? Correct me if I'm wrong. One minus T by one plus T whole multiplied with half. Okay. This is basically breaking up into partial fractions. Okay. So once you have written it like this, you may choose to integrate this fellow. Let me take half outside. Okay. So this gives you half integration of one by T is LNT. So integration of one by a linear extrapolation will give you LN of the same term, but you have to divide by a minus one. So if you divide by a minus one, so something like this will be seen. Okay. Let's regroup it. It's a half LN one plus sine X by one minus sine X. Is it fine? Any questions? Okay. But the answer that normally is given to us is a slightly different format. What they will do with this term is they will multiply this with one plus sine X in the numerator and denominator. So that this becomes one minus sine square X. Okay. And this term is one plus sine X by cos X, the whole square. Now, please remember the power of two will come in front and that will get canceled with this too. Okay. And this term within the LN you divided by cost. It will become seek X plus 10 X. This is a more, you can say commonly seen result that we use. Right. We hardly use this result. And this will be what will be mentioned in your, you know, at the back of your exercises when you're solving a question. Okay. So please note down. This is the result for integration of seek X. By the way, there is a simpler method to achieve the whole thing. What is the simpler method? Multiply and divide with seek X plus 10 X. I'm, I'm writing this as method number two. Okay. Now take seek X plus 10 X as a T. Or another variable. And if you differentiate it, you get seek X 10 X. Plus seek in square X. DX as your DT. That is actually sitting on your numerator. See this whole term is sitting as in your numerator as a DT. Is it when you expand it, it gives you seek in square seek X 10 X. That is what is present over here. Of course DX is also there. Okay. So this whole term becomes DT by T. And this is a simpler way out, which directly gives you the result as Ellen mod seek X plus 10 X. Let's see. Okay. Anyhow you do it. I mean, you have to use substitution. No way out. So substitution is heart and soul of integration. Now there are other versions of the same result. I will show you those versions also. So first note this down. Another version of this result is. Where should I write it? Okay. Maybe I'll write it over here. No. One formula is already seen. Half Ellen. One plus sign X. My one minus sign X. Another version was half. I'm sorry. Not half without half. Ellen seek X plus 10 X. And seek X plus 10 X is reciprocal of seek X minus 10 X. So you can also write it as negative. And seek X minus 10 X. Okay. And there isn't half angles also, which is Ellen mod. Tan five by four plus X by two. Out of these four formulas, these two are very heavily seen. Okay. So at least these two should be in your mind. Now how do I get the last result? Very simple. Take this result. In this result, you write it. Everything in terms of sign costs. Okay. Use your. One plus sign X formula is cost X by two plus sign X by two, the whole square. Okay. And use your. One plus sign X formula is cost X by two plus sign X by two, the whole square. And cause X formula is cost square X by two minus sign square X by two. Okay. You may factorize the denominator as cost X by two plus sign X by two into cause X by two minus sign X by two. So can I do one thing? Can I be lazy enough. To just take care of that term here itself. Okay. Now divide throughout with cause X by two, it gives you one plus tan X by two, my one minus sign X by two. This is clearly a formula for, or this is clearly an expression, which can be written as tan of I by four plus X by two. Okay. That's how you end up getting the last formula. By the way, I'll just make some partition over here else. Everything will mix with everything else. Is it fine? Any questions? Okay. So four formulas already we have seen for C kicks. Similarly, without much waste of time, I'll also talk about C O C K X. The simple, the simplest approach to get integral of C O C K X is multiply it with C O C K X minus Cortex and take C O C K X minus Cortex as a T. So basically minus Cortex minus C O C K X Cortex plus C O C K and square X DX will become your DT. And this entire term is sitting on the numerator. So it's becomes DT by D. So this is L N mod T plus C. And this result is obtained L N mod. C kicks minus Cortex plus C. By the way, you could have also done it in second method by writing it as one by sign X. Multiply it with sign X and write the denominator as one minus cost square X. Take cost X as a T. So negative sign X DX will become a DT. So this whole thing gets converted to a DT T square minus one. So one by T square minus one is factorizable as this. And this is nothing but one by T minus one minus one by T plus one by two. Correct me if I'm wrong. Okay. So this is as good as integration of this term and this term subtracted. So if I'm not mistaken, it'll give you L N mod T minus one by T plus one. Okay. And that's nothing but one more result half L N. Sine. Sorry. Cost X minus one by cost X plus one within the mod. It doesn't actually matter if you switch their positions also. So even if you do L N mod one minus cost X by one plus cost X doesn't make a difference. Okay. So yet another result. In fact, I will write all the results together. No down in your formula list. The following results. We have already derived. Two of them. Which is this. Okay. And this is negative reciprocal. This is a reciprocal of course. C K X plus context will become negative L N. For C K X plus context. Okay. I hope that is clear to you. Just now we figured out you could write it as half L N one minus cost X by one plus cost X also. And one more result is there, which is L N mod tan X by two L N mod tan X by two. Okay. That is very obvious. When you simplify this in terms of two signs square X by two. This has two costs. And square square will be taken care by this half sitting outside. Okay. So before we go for a break, I will at least try to do a five to six questions with you where substitution is very, very important. And these are just the standard results which we obtain from substitution piece going forward incorporate these results in solving a complicated problem. And they have to be memorized. Okay. These results have to be memorized. Up there up to the left or up to the right. Which part do you want to see there? Okay. Done. Okay. Let's take some questions based on substitutions. In fact, this, this topic substitution itself is very long. I mean, there are a lot of types which are based on this. Anyways, let me start with this question. Ellen tan X by two divided by sin X. Integrate this with respect to X. No, no, no. Don't type your answer. Just say done. If you're done. Okay. So that is just done. Done. So here, if you see, you have to take your numerator as a T. Okay. So I did that you were asking, no, always the nominators taken as C. No. I'll take as a T. When you differentiate it, you will end up getting one by two sine X by two, plus X by two in the denominator, which is basically trying to say that DX by sine X is like your DT. So this term DX by sine X. This will behave as your DT term. Whereas this will behave as your T term. So you are integrating TDT. TDT integration is T square by two plus C. So your answer will be half of LN of tan X by two. The whole square process. Clear. Any questions? Okay. We'll show you some, you know, challenging questions also where substitution is, you know, very, very essential phenomena. Let's take this question. Read the question completely. I think it is asking for f of X plus G of X, not the integral directly. Okay. Anybody willing to give? Okay. So Aditya has given a response. Okay. So we'll see this. See if you look at the denominator term. Okay. I can see some terms like cost exiting one sitting. Of course, exponential term is not there. So let's start doing one thing. Let's try writing in the numerator terms associated with the derivative of this term. So derivative is this term, right? So let's see what we can do. So cos X minus sine X C. It's a leap of faith many a times. Integration is not conventional that if I apply proper product rule or proper quotient rule or chain rule, I will definitely get my answer. Please don't have that kind of a myth or a wrong notion in your mind. Okay. So integration is a chapter where many a times you will start with an approach and be prepared to fail. Okay. Be prepared to fail. And then again, take a fresh approach. Okay. So it's not conventional. So this term that you have on the numerator. First, let me write down this derivative term. Okay. Let's me make some add addition, subtraction. So e to the power. This term and this term is already there. E to the power X is not supposed to be there. So I'll remove it. Then I need to have a sign minus sign X and a minus X. So that also I will introduce it. So I wrote the same term in a slightly, you know, different fashion. Now let's see what happens when I divide it by the denominator and integrate it. So this time I'll separate it. In fact, minus sign, I can separate it out. Oh, sorry. This was sign. Okay. So if you see here, you have a very, very easy integral to perform. Okay. So in this first part of the integral in this part, if you take e to the power X plus sign X plus T is, as you can see, the numerator has already been framed in such a way that this will automatically come out to be the derivative. So numerator along with DX term will become the. DT term for you. And the second integral is just an integral of one with respect to X. So this is e to the power X plus sign X. Plus X mod Ellen. Okay. Now as per your question setter. He's asking for f of X plus G of X where f of X is the one which has been put as an argument to log and G of X is the one which is outside. So it is very clear that f of X is this term. And G of X is minus X. So if you add them. If you add them f of X plus G of X, you will end up getting option number B. Is it fine? Any questions? Any concerns? If you want, I can write it in a. Clearer way so that you can have a look at it. Is it okay? Any questions? See again, you have to be observant here. For example, I saw few related terms like sign X already there, X already there, cause it's coming from derivative of this. So these kinds of observations based, you know, a substitution I made. Now, and that's what I told I may have failed also. Okay. Then probably I would have thought of something else. What do you see is it clicked in one go, but many times it doesn't click in one go. So there may be several failures, which led to this approach. Okay. What we see is the AI rank. What we don't see is how many hours that person would have put in to get that AI rank, all India rank. So what do you see is like, okay, sir is able to solve it in one shot. What you don't see is how many times I would have failed when I was practicing it when I was of your age. So yes, it is something which is driven through a lot of exposure to the problems. Observation, you should be good at. Okay. And yes, as you keep solving more and more questions, you will start finding that, okay, this works. This will not work. That is why many types have been created because of it. When I was preparing for my J. We used to be given a very thin book Russian book called I am Aaron. You may ask your parents about it. Maybe they would have also referred to the same book. It's called basic calculus book on on by I am Aaron. And it had very less theory in it. It has mostly problems, problems, so we learned mostly by problem solving. So yes, you will be exposed to more questions later on, and then you will frame a theory out of it. And that theory also is not like completely correct. It may fail also sometimes. Next, let's take this question. Let's take. I mean, you would have done this problem in school also. So when you see this question, what runs in your mind? A simple statement will learn in your mind had the expressions been added or subtracted, then I would have easily done it, right? But right now it is multiplied. And of course, using integration by parts is not going to be a simple process. It may complicate the problem even more. I mean, in the present form, what will you do to get that? Whatever you're saying, what would be the starting process of this problem? But that was only for one tan x, right? Here there are three tan terms multiplied. What will you do? You will extrapolate it, multiply LN, CX, LN, C2X by two and LN, C3X by three, and multiply the result. That doesn't work. So, formula was only for one single tan x integration, right? Yes. Right, Gayathri. The starting point is first we have to break tan 2X as, sorry, tan 3X as tan 2X per 6. Use your basic trig identities. Okay. You yourself will see that things will get simplified. So, you will see that things will get simplified. You will see that things will get simplified. You will see that things will get simplified. You yourself will see that things will get simplified. And the integrant will appear in one of your terms. Okay, let's multiply the denominator to the left. By the way, let me use shorthand notation. I don't want to write too many things. Tan tan tan tan, who will write? What you see is your, you have TX, T2X and T3X, correct? This is what we want. That can be easily be obtained by making this the subject of the formula. So, instead of integrating this, you can choose to integrate this, which is your tan 3X, minus tan 2X, minus tan X. Okay. So, this is all you have to think out, right? It's not a, okay, I know a product tool. I'll keep using it whenever I see a product. No, that doesn't work like that. You will end up complicating the scenario more. Okay. So, this is reverse chain rule. I hope everybody remembers the reverse chain rule. Instead of X, you have a linear extension of X, means instead of X, you have some AX plus B kind of a thing. So, you have to use a similar formula, but don't forget to divide it by, is it fine? Any questions, any concerns here? This problem is a very famous one. It comes in school as well. Let's say X4, that's one to the power of three by four. Just try to done if you're done with it. All these problems which I'm giving you, they are all the ones which have been asked in some of the other exams, that is why. These are not like the typical questions. I will be sending you after the class, a complete list of 400, 500 problems. You need not solve all the problems, but just do some chosen ones from there which you feel are tricky. Okay. Yes, how to solve this question? Very simple here. You take this X to the power of four outside this power of three by four. So, it'll come out as X cube. And there's already an X square waiting over here. So, that will become X to the power of five. Now, one by X to the power of five appears when you differentiate this term, isn't it? Yes, solution is attached here. Solution will be attached to that, don't worry. Yeah. So, in this, you can either go for one minus X to the power of four as your T or you could make your life simpler by choosing this whole thing as a T. Or many times I've seen people going for T to the power of four also. Why? Because if you take this as T to the power of four, this four and the four in the denominator of this power will be adjusted. So, there are multiple ways. Some may go just for one by X to the power of four as a T. Some may go with one plus one by X to the power of four as a T. Some may go with one plus one by X to the power of four as T to the power of four. See, ultimately it's your thought process. What makes your life easier? Go for it. Not differentiate both sides. Sorry. With respect to X. Or with respect to whatever variables they are in. So, this four and this four will be taken care of. So, as you can see, DX by X to the power of five is negative T cubed DT. So, this whole term, you can write it as negative T cubed DT and down in the denominator is T cubed. Okay. So, T cubed T cubed gone. And as you see a simple integral for us to perform, don't leave your answer in terms of the variable that you have used to solve the problem. Okay. At least in your semester two exams, when you have subjective paper, don't do that mistake. Is it fine? Any questions? Okay. Now a special type of question. I think you would have already seen this while practicing this. This is let's say A. They all have a similar approach. So, I'll just write them together. And maybe I will solve one of them or we will solve one of them. DX you can put. Okay. Out of ABC, let us first target A. Okay. I mean, others have a similar approach. So, if you do one of them properly, we will be able to do the others as well. So, how do you solve A? If you have solved it earlier, this problem is present in Adi Sharma and all those books. Then fine. If you have not solved it, let's try attempting it. Okay. So, in this kind of problem, the typical approach that we use is, if you have same type of term that is sign, sign or cost, cost. Okay. We multiply and divide in the numerator with sign A minus B. So, for both these approaches, this is the, for both these types, this is the approach. Okay. Now, see what I'm going to do with that sign A minus B. So, for both these approaches, this is the, for both these types, this is the approach. Now, see what I'm going to do with that sign A minus B, which I have written. So, that sign A minus B on the top, you can write it as sign X minus B minus X minus A. Okay. I've just introduced an X so that it rhymes with the angles present in the denominator. Now, what are the benefit of using this? The benefit of using this is, let me just treat this as one angle. Let me call it as angle A. Let me call this as angle B. So, we have sign A minus B formula, which is sign A cos B minus cos A sin B. And down in the denominator, you have sign A sin B. Okay. Just divide individually, both the numerator terms by the denominator. So, you will end up getting cot X minus A. And this will give you cot X minus B. Now, these expressions are simple for us to integrate because cot X integration, we already know, this is just a linear extrapolation of that. So, this is going to become ln mod sin X minus A minus ln mod sin X minus B. So, this is ln mod sin X minus A by sin X minus B. Is this fine? Any questions? Any concerns? While for the last one where you have a mix of sin and cos, in this case, we recommend multiplying with cos A minus B and then split this up again as cos X minus B minus cos X minus A. Okay. This is what we recommend for the last one. So, for sin sin cos cos, use sin. For sin cos, mix you use cos. Is it fine? Any questions? Any concerns? Now, we'll take a break here. On the other side of the break, we will be discussing integral of the type sin to the power mx cos to the power nx. Few of the types related to it after the break. Yeah, sure. Please do it for homework. Okay. This part you please do it for homework. Is that fine? Okay. So, we'll take a break now. The time right now is 6.19. We'll meet at 6.34. Sir, yes, yes, yes, yes, 35 only sir, 35 only. No, every minute is important for us. Okay. See you in 15 minutes time. See, this type that we're going to take up, it is no different for method of substitution. So, we are going to use substitution here also. Right. But because these kind of problems have been asked so many times in these competitive exams that we make a theory out of it. Right. Ultimately, it's nothing, but it's a concept which is based on your understanding or you can say your skill to substitute. That's it. Okay. Nothing unusual is going to be discussed over there. Okay. All right. Now, depending upon m and n, which is your powers of sin and cos, there are, we normally have broken down this theory into four cases. Okay. So, first case is where your m and n are odd natural numbers. Okay. M and n are odd natural numbers. Okay. Or odd numbers. Odd natural numbers. So, in such case, what do we do? The approach is, algorithm is, again, this algorithm has been derived because of the, I can say the normal pattern that we have seen to solve this question. Okay. So, what do we do? We take the term, we, or you can say substitute, substitute, that trigonometric ratio to be t, which is subjected to, which is subjected to the higher odd power, to a higher odd power. Now, in reality, you can substitute any one of the two as a t and solve it. Okay. So, once you substitute that, see what will happen. Let me not write it as a, because this is the only step that we do. Let me explain it by an example. Let's say somebody has given you to integrate sine to the power of five x into cos cube x. Okay. Out of the two, you can see five is the higher odd power. Correct. So, I will substitute sine x as a t. So, trigonometric ratio means sine or cos whichever. Okay. So, when you substitute sine x as a t, cos x dx will become a dt. So, what I'm going to do here is, I'm going to borrow a cos x from the cos cube term. Okay. And I'm now going to use my t consistently throughout. This will become t to the power of five. This will become one minus t squared. And this will become dt. This will become a dt term. Now, this is simple to integrate. We can easily integrate t to the power of five minus t to the power of seven. Correct. So, this is simple. t to the power of six by six minus t to the power of eight by eight. And put your t back in terms of, in terms of sine. So, sine to the power of six x by six, sine to the power of eight x by eight. Now, I could have also substituted cos x and solve this problem. Right. So, you can also substitute cos x into this problem. But in the other case, you'll, you'll end up getting one minus sine square square. So, ultimately, you'll have to, you know, expand a, expand a slightly, you can say, lend me a term to do. Okay. See, I could also substitute. I could also substitute cos x as a t. So, in that case, minus sine x dx will become a dt. So, what I will do is I will borrow a, borrow a sine term from it. In fact, borrow a negative sine x and put a negative sign outside. So, this term will become, this term will become one minus cos square square. So, it's one minus t square square. Okay. And this is a dt. So, ultimately, you'll have to first, you know, expand this, multiply with t cube and then write it, write a term and then find its integration. So, this also can be used, but slightly lending, slightly lending. So, this is why we, we recommend substituting the term which has got a higher odd power. Okay. As a t. Is it fine? Now, in my time, we were never told these kinds of theories. So, this theory has come because we have seen this type of problem and normally people do this and that's why it has become a theory. Next, case number two. Any questions that you have here, please do let me. Case number two is where one of them, one of m or n is odd natural number. Odd natural number. And other is even, even natural number. Okay. In such cases, the algo says substitute the trigonometric ratios subjected to the even power, subjected to even natural number power as t. Okay. Let's take an example. Let's say I take sine cube x and cost to the power six x. Okay. So, one of them is even another is odd. So, this is odd. This is even. So, the one which is subjected to even power that will be substituted as a t. So, here you borrow one's negative sine x. So, you will be left with something like this and take this expression in terms of t. Okay. So, this is a simple integration t to the power eight minus t to the power six. And this is a simple power rule of integration. We already know this result. And put your t back as your cost x. Is this fine? Even if this theory was not known, you would have come to know it by trial and error. Now, what will be the problem if you substitute sine x as a t, if you substitute sine x as a t, then you will be left with a cost to the power five x because it's going to be zero in dt. And then cost to the power five, if you want to express in terms of sine, you will have to use fractional powers and all and that will create a lot of issues. Okay. That will create a lot of issues. Okay. So, let's follow this to save our time and effort. So, the one which has got even natural number as its power, substitute that as a t. Whichever of the two, that depends on question to question. Can you move on to case number three now? Case number three is when both of them are even. When both M and N are even natural numbers. So, here there are two approaches. One is by the use of trigonometric ratios and identities. Okay. And another is by use of complex numbers. However, both of them achieve the same purpose. Okay. Complex number to be more precise by use of D more based by the use of D more based. D more based. Both approaches have the same objective. Now, what is the objective? The objective is to convert higher powers to multiple of the angles. What is the objective? I'll repeat once again. The objective in this kind of a problem is to convert a higher power into multiple of the angles. Okay. How it is achieved or how these two methods are, you know, implemented to achieve that. Let's start with a question. Maybe the question will be a good example to talk about it. Let's say I want to integrate sine square into cost to the powerful. Okay. Now you think, neutrally, you think, how do you integrate this? Will you substitute a sine x as a t? If you do that, you will get cost tube term and cost tube term will be, you know, creating a fractional power. Same will be true even if you put cost x as a t. Isn't it? So what will you do here? So for here, we will try to convert it to multiple of the angles. Yes. So how do I do that? Simple. All of you please pay attention. First of all, I'll club some sine square and cost square term and make it as, yeah. So I'm using my trig identities, TRI for it. Okay. Then I'll come to the complex number approach and we'll see which, which is more, you know, beneficial to us. This is as good as a sine square to x. Okay. Now we convert sine square to x into one minus cost four x by two. And this has one plus cost to x by two. So slowly you are converting it to higher multiples of angles. Okay. Power has been getting changed to multiple of angles. So this will be one by 16. Correct me if I'm wrong. This will give you one plus cost to x minus cost four x minus cost four x to x. Now this also you provided with a two. Okay. Use your transformation formula. If I'm not mistaken, this is cost six x plus cost two x. Okay. So overall, if you see, it will become something like this. One plus half cost two x minus cost four x minus half cost six x. Now you finally integrate this because now it is in the format where you can integrate. This will become sine two x by four. This will become sine four x by four. This will become sine six x by 12 plus a c. Okay. I'm not simplifying it any further. Okay. This is your answer. Now you realize that this takes in a bit of your time, especially why you are, you know, you are wasting a lot of time while you're writing sine, cost and et cetera. So three alphabets, you know, you have to write for every trigonometric ratio that takes away a bit of your time. Okay. The same thing can be achieved by using our demobilized theorem. Now, how? Let us try to see that. Okay. By the way, any question related to this approach, do let me know. Clear. Any questions? Any questions concerns? Do let me know. But demobilized theorem. First of all, let us recall the demobilized theorem. Let us say if you take a complex number cost six plus I sign X. Okay. Let's say I call it as a Z. We all know that Z to the power N, N being some integer will give you cost NX plus I sign NX. And if I raise the same to power of minus N, it'll give you cost NX minus I sign in X. Okay. I'm just revisiting a bit of theory before I start applying it. In other words, Z to the power N plus one by Z to the power N. We all know it's two cost NX and Z to the power N minus one by Z to the power N. In fact, one by two I, this is nothing but in fact, one by I, I should say. Or you can say to why you also know issues. This will be sign NX isn't it? Okay. Let me just take the two here as well. So as to maintain a consistency in the way I've written them. Yeah. Okay. So we are going to use this. How? Let's check. So let us look at the integrand first. Okay. Please note that this doesn't help you to integrate. This just simplifies it to a form which you can integrate. That means it converts it to multiple of angles. I'm repeating once again. Demover's theorem doesn't help you to integrate. It basically converts your integrand into a simpler expression. Or you can say a power or power is getting converted to multiple of angles, which is relatively easier for us to integrate. Please understand this. Okay. So now sign square X from here. Can I do one thing? Can I put N as a one over here? Okay. If you put N as one over here, I'll get sign X from the second term. So it'll be one by two Y. Z minus one by Z. Z minus one by Z. Whole square. And similarly, Cos X from here, I'll do one thing. I'll put N as one and I will just take a power of four. Is this expression clear to everybody? Any questions? Anybody has any issue with this expression? Do let me know. Okay. So this becomes one by 64. Z minus one by Z squared. Okay. And this becomes Z plus one by Z to the power four. Now, all of you, please pay attention. Your problem will be solved in very quick time. See, if you can write this as Z square minus one by Z square whole square. Can I write it like this? In fact, you have to write less in this form. Okay. So this is one by 64. This is Z to the power four one by Z to the power four minus two. This is Z square one by Z square plus two. Okay. Now this expression that you have and this expression that you have. If you read it in light of this term. If I put Z as a four. Can I say one by two Z to the power four one by Z to the power four is actually cost for X. That means Z to the power four. I'll write it here itself. That means Z to the power four one by Z to the power four is two costs for X. Okay. So in short, what we can do is we can write this term as two costs for X minus two will remain minus two. Similarly, this will become two costs to X. Okay. And you can just do away with this to two to two. One and that will give you a 16 over here. Okay. You can also switch the position of this one minus cost for X. If you see very closely, you have actually reached to this step, this step. Yeah, there you go. This one. Yeah. So one by 16, one minus cost for X one plus cost to X. But we're writing very less actually. Okay, you don't have to write cost, cost, cost, okay. After that, you can just, you know, follow your regular mechanism, the one which you have already done in these steps. Okay. So you have already, you have already reached this step. So I'll just directly jump to the final step. So from here, you can write it as one by 16. This will be nothing but one plus cost to X minus cost for X. And I think it was half cost six X plus cost to X. Okay. I hope I've written everything properly. Check it out. Or this was minus. Yeah. So, I mean, after this, you know how to integrate it. So please do so. So once you have done this, you can make your integration faster. You can just do your integration here itself. Now you will understand the efficacy of this method when the power is slightly higher where your trigonometric ratio will take much longer time. And this method will take a slightly lesser time. For example, let's take a question. You will realize the efficacy of this method. When I take a heavier power. Is this fine? Any question with respect to this? I'm not, I'm not integrating this. I hope you can integrate it from here on. So I've written proceed here. Let's take a question which has got a higher power. Let's say I want to integrate. Anybody wants to copy this? Please do so. Then I will move on to the next slide. You can skip that. Why do you write all those things? Directly write cos x as half of x plus 1 by x. I'm sorry. Half of z plus 1 by z. And sine x as 1 by 2 y, 1 by 2 y z minus 1 by z. No need to write everything down. See, let's say I want to do this question. Cost to the power 10x integration. Let's say. Okay. Now imagine you doing this problem by using trick substitution. How will you do it? Anybody has any idea how to do it? You will do cos square 5. Then you will write it as 1 plus cos 2x by 2 to the power 5. And then you will raise this to the power 5 and oh my God, six terms will come out. This will be a tough task. It will be a tough task to solve this by. Signometric ratios and identities. See, I'm not trying to showcase the importance of one over the other. Both have their own importance for lighter powers. Signometric ratio is good enough. No need to go for complex number of push. But for this case, you can directly substitute cos x as half of z plus 1 by z. Okay. And see how fast it will be. When you raise it to the power 10, it becomes 1 by 2 to the power 10. 1 by 2 to the power 10 is this. Now see how beautifully the terms will come over here when you raise this to the power of 10. All of you please pay attention. z to the power 10 will come and the last term will be 1 by z to the power 10. Check it out. First term will be z to the power 10 and the last term will also be 1 by z to the power 10. Okay. Then 10 c1. Next term will be z to the power 8 and the second last term will be also 10 by z to the power 8. So they will automatically pair up like this. That means always you will get a term of this nature. So that you can directly use this formula and convert it to multiple angles. This will always come out. Always guarantee. Okay. Next term will be 45 z to the power 6 and 45 divided by z to the power 6. So you can club it up like this. Okay. So this is 10 c1, 10 c2, 10 c3 is 120. I hope you all remember your 10 c values. Okay. Then 10 c, this is 10 c1, 10 c2, 10 c3, 10 c4. 10 c4 is 210 if I'm not mistaken. And 10 c5 is 252. Right. So see the benefit of this method. Only you write this step and the next step is your conversion into multiple angles. So this is 2 cos 10 x. Done. This is 20 cos 8x. Now this is already in the format which you can integrate. This is 90 cos 6x. This is 240 cos 4x. I know some heavy coefficients are coming, but that is not my concern right now. Okay. So once you have written cost to the part 10x like this, you can go for the integration. Right. So let's do it. I mean, since we have brought it to so far. So this will be 2 sin 10x by 10 20 sin 8x by 90 sin 6x by 6. Of course, you can do a lot of cancellation, which I'm not interested in doing because my purpose is not to teach you simplification. My purpose is just to highlight how efficient this method could be in such cases. Is it fine? This is your answer. Any questions? Any concerns? Do let me know. Yes. Yes, why not? Yes. If you picked up for that, you'll end up getting stuck. Right. Right out. You may try out using it. Let's see what implications it has when you use it for other forms. That's a good question. Actually asked. Okay. Finally, the fourth case, which I'll be discussing here when your m and n are rational numbers which add up to, which add up to a negative even integer. It happens mostly in the case where your powers m and n are, you know, fractions, negative fractions. I'll just take an example for the same and I'll tell you the algorithm also. Let's say I take this question signed to the power of minus five by three X and let's say cost to the power of minus one by three X. So in this type of question, first of all, any kind of a negative power, we take it to the denominator. So I'll go here is negative power terms to the denominator. And then let's say this negative integer is minus 2k. Then divide your numerator and denominator by cost to the power of 2k. Okay. And then everything will convert into tan. So take tan X as a t. Okay, let's, let's demonstrate this. I'll go on this expression. So here both are negative powers. So I'll write it like this. Oh, first of all, m is minus, let's say this and n is minus one by three. Do they add up to give us a negative integer? Yes. Negative even integer. Yes. Negative even integer is minus two in this case. Right. So your 2k is your 2k is your two. Okay. As per my expression. Okay. Again, these things have been, you know, unnecessarily theorized. I will use the word theorized. We never learned this when we were, we learned the hardware. Okay. When we used to do this. Okay. Then next step with this and let's try dividing it by some cost. And then I mean later on it was made into theory. So now you divide your numerator and denominator by cost square. Now this minus two means you have to divide by cost square. So both numerator and denominator, you have to divide by cost square. So this will become secant square on the top down in the denominator. If you see, you will always get tan of the same power as that is on sign. Check it out. See one by three and two that will compensate to give you cost five by three and sign five by three cost five by three. Now we have reached this step. We have divided the numerator and denominator by costs to the power of 2kx. Now take tan x as a T. So tan x as a T. If you take, you will end up getting secant square x dx dt. So this is dt by T to the power five by three. And this is a simple, I can say power rule of integration, which you can easily solve any questions. One more question we'll take on this type and then we'll move on to our standard integrals or special integrals. That is the important thing because most of your problems are based on that only. So one question only I will take it. Take care. See, I normally love to do a lot of integration question, but you know, we are having a time crunch also because the board has come in between all of a sudden. We want to cover as much as possible before the boards. You know why? Because till December you will not learn anything. You will not want to learn anything. So that will be a, you know, dead hours for us. Just say done if you are done, okay? Done? Done. So m plus n here is I think minus 20 by five, which is actually minus four, a negative even digit. So as I told you, if it is in the negative power, bring it to the denominator like this and divide both your numerator and denominator by cost to the power of whatever is this negative integer, remove the negative sign and raise it on cost. That is cost to the power four X. So this is seek to the power four X DX. Down in the denominator, you can save your time and directly write 10 to the power seven by five X. Okay. It'll always come like that. Trust me. It's obvious why it comes like that. I don't have to explain that. Okay. So now this one of the seek to the power four, you keep it as a seek square and other you write it as one plus 10 squared. Okay. Now take your tan X as a T. So secant square X DX will become your DT. So this will become one plus T square DT by T to the power seven by five. Correct. Now I don't think so. I have to, you know, tell you any further how to integrate this because I think all of you are experienced enough to integrate this. Okay. So please use power rule and proceed. Okay. Many times I will write proceed, proceed because I don't want to save the entire time doing the simple integration. But this is something which I would like to share with you my own story. When I was preparing for my J E after integration was over, I was given almost around 1000 questions by my coaching institute to solve. So 1000 question is like, you know, if you calculate almost three minutes also per question, that is 3000 minutes almost how many 50 hours of work was given to me and 50 hours of work. I had to complete it within let's say two to three weeks. So what I used to do is I used to solve a question and after I reached a stage where I knew how to solve it, I used to abundant the question that, okay, I have now reached this stage. I know how to do it from here on. Leave it. Let's move on to the next question. What happened in due course of time? I started forgetting the formula only. What formula? You'll come to see some formula later on in our today's discussion also. So don't do that mistake which I did. So please as far as possible, of course, when it is very simple step, you can skip it. You can go to the next question. Don't have to do it. But once you reach a step where you feel that there is a formula required over here, which I may forget in long run, then please do that problem to the end till you write that plus C. This is my sincere request to everybody. It happened with me and when I went to the examination room, I could bring it to the final form which the result was already given to us and I forgot the result because I was not writing it. I always used to like, okay, from here I could do it. So leave it. So I was not writing it. So I was not writing it. I ended up forgetting it. And of course I had to sit and derive it, but that wasted my time. No. Okay. So of course, you know how to do power rule of integration that you can leave it. But when you have a formula to be applied at, okay, after this, there is a formula already there to do it. There you do it to the completion. Okay. And those formulas are now going to start appearing for you as well. So now we are going to talk about some standard integrals. Also call as some special integrals. Okay. Whose results will be deriving also. Let us do it in this coming nine minutes. I'm sorry, 21 minutes. I should be done with that. So these standard six results that you should be well worse with the first one is this result. Second is this result. Some of them I've already taken in the beginning of our today's discussion. Fourth one is this. So there are six of them. These six of them, they will come in various shapes and sizes in the question that you will be solving. So be aware of them. I mean, know their results on your fingertips. Okay. So what are the results of this? So this also is actually under substitution only. Okay. I will be solving all these by using substitution. But yes. Tignometric substitution like the way you learned in your differentiation chapter. Okay. So what I'm going to do, I'm going to derive each one of these six results one by one in the next page. And I'm going to come back and put the result over here. I don't want to corrupt this page. Okay. So first one DX by X square plus a square. You will only suggest what integration or what substitution can I use here? Please suggest a substitution for me. Any trick substitution that will help you here. XS 8 and theta, right? So let's save our time and energy because you would have already done this in your school as well. Just for the purpose of completion of our understanding. So this will denominator will give you a square six square theta. Okay. And there will be nothing left other than one by. So this is one by a theta plus C. And what is theta theta is tan inverse of X by. So this result is there. Now many people will say sir, why are you a substitution and all? Okay. You can use this method also second method. You can take a one by a square common out and you can write it like this. Okay. Not recall that we had already done. We had already done one by one plus X square in our formula list and that was tan inverse X isn't it? So this is just a linear extension of that. So can I not write this answer as one by a square tan inverse of X by a divided by one by that means divided by the coefficient of X. Okay. Yes. This also can be done and you'll end up getting the same result. Okay. Another method to look at the same thing. Okay. It's cool. You all derived it. No. Just asking. I mean so that I can be fast. Okay. Shitish has confirmed. Yes. Yes. Yes. Yes. Okay. Next one. Second one. Second one. Let's do it here itself. I will do it. I don't have to use another page for it. Here again, you can go for a substitution X as a seek theta or cosec theta, but the best way to solve this question is by using of partial fractions. The way I did it in one of the questions before. The best way to solve it is by the use of partial fractions. So you can write it like this one by X minus a minus one by X plus a by one by two a. Okay. So instead of integrating this, you can integrate this. So this gives you one by. Two a. Ellen. Mod X minus a. Minus Ellen X plus a, but can I combine it like this and save my time and energy? This bracket was now not required. Okay. So there you go. This is the result for the second integral. I'll go and just jot it down over it. Okay. One by two a Ellen. One by two a Ellen. Mod X minus a by X plus. Okay. The second third result is actually negative of this. Right. So what we can do is we can put a negative and we can reciprocate this. But when you reciprocating it, we normally reciprocate it like this. Now many people say sir, why? You could have just written it as X plus a by X minus a. See, there's no other reason, but just the reason that it resembles one by a time hyperbolic inverse X by that's it. There's no other reason for writing it as a plus X by minus X. But because when you write it in this form, it resembles one by tan hyperbolic inverse X by that's it. Okay. That's the whole and sole purpose of writing it as a plus X by minus X. Within more, you can do any kind of changes. Okay. Coming to the fourth one. Fourth one also let us derive it. Let us derive it. This page itself is good enough for deriving the fourth one. Yeah. Let's have this place. So under root of X square plus a square. Here also you can go for the substitution X is equal to a tan theta. So numerator will become a secant square theta d theta denominator will become a seek theta. I hope I'm not fast here. Ultimately leaving you with seek theta. Seek theta. Okay. Seek theta is ln mod. Seek. Seek theta plus tan theta. Okay. And put your X back. So this will become tan theta will become X by a and this will become under root of one plus X square by a square. Right. Now here if you take the LCM of your a, you will end up getting X plus under root X square plus a square. Whole divided by a plus C. But normally what do we do is we write it like this. We write it as ln mod X plus under root X square plus a square. And minus ln a plus C is clubbed under another C. That means it's called a C. We call it as another constant of integration. Okay. So this is how you end up getting the result for the fourth one. By the way, if you remember this was actually sine hyperbolic inverse X by a. Okay. This is sine hyperbolic inverse X by a. This term. Okay. I think the other derivation also you can do very easily. Same result will come seek theta, but there you would substitute X as a seek theta. So please try this out. Only differences. You will end up getting minus sign here. The same expression will come over here. Okay. Sixth one is super easy. Sixth one is you can substitute X as, let me find a place. I may be appearing to do it very quickly because I want to save your time and my time also because you've already done it in school. Okay. This also will give you a cos theta only. So nothing is left other than theta. So this will give you theta plus C and theta is nothing but sine inverse of X by. Okay. Anybody needs more clarity. Please feel free to stop me. Okay. I will be more than happy to take this up. On a slightly lesser pace. Okay. This six result should be by hearted. Okay. By hearted because you will be using it. And what I used to do is let's say I reached this stage. I used to think, okay, I know the result. And ultimately I, in the exam, I was like confused. Oh, was it Ellen or was it one by two LN or was it what? So that kind of a confusion in order to avoid that, please ensure you are solving the problem to completion. At least the one which involves these. Okay. Let's take few problems based on the same. First of all, I will check your trigonometric substitution skills. Okay. I would like everybody to solve a question. This is the hyperbolic inverse X, right? This is cost hyper. This comes from cost hyperbolic inverse X by a plus C. This is not exactly cause until unless you have an X by a. Okay. Shiddish. Let's take a question where I would test you on your. Substitution skills. I would request everybody to solve this question. Then stuck. Come on. This was easy. Of course. Done. Okay. See, this is actually nothing but X plus one the whole square plus one. The whole square. Correct. Now for such cases, I will prefer taking X plus one as a tan theta. Correct. So DX is secant square theta d theta secant square theta d theta. And this is one plus tan square one plus tan square is six square whole square. Okay. So ultimately it gives you cos square theta to integrate. And cos square theta you can write it as one plus cost two theta by two. So your answer will be half theta plus sine two theta by two. Yes. Of course you have to convert things back in terms of X. So that's also not a big big ask actually because theta is tan inverse of X plus one and sine two theta is what sine two theta is two tan theta by one plus tan square theta. So half into this will give you tan theta which is X plus one by one plus X plus one the whole square. By the way, this term is actually X square plus two X plus one. So why to write it in a fancy way. Okay. So this is how trigonometric substitution can be useful not only to solve those standard integrals but also in any such cases which you feel that over here I can go for a trigonometric substitution. So trig substitution very very essential tool in your hand not only in differentiation chapter but also in integration. And of course you have already experienced it in your inverse trig functions. Okay now questions questions in general which is based on the standard integrals that we have done. First let me start with some basic ones. Let me start with some basic ones. Just to give everybody. Let's take a question. Let's take a question. Let's integrate this one by under root of one minus four X minus two X square shift ages done. Okay. How about others? Okay. Now here why I chose this problem because I've seen some people doing this blunder. I would not call that mistake. I'll call it as a blunder. Some people take minus two outside. Don't do that. I was a student who did this last year. He took it as a minus root two outside. Just to not do away with the negative sign. Please note that when it is under root only thing that you can take out is a positive quantity. Two if you want to take out as a root two you're welcome to do that. But don't take a minus two outside please. People do this in your school papers and then they'll come and say sir I lost mark. I did this a very small silly mistake that they will call it as not a silly mistake. It's a conceptual vendor. Okay. So if you want to take a two outside you can do it. I mean I'm not denying it. Even though that is also not required. But if you want to you can do that. Okay. So once you get this you need to convert it to a square inside the under root sign so that you can use one of the standard integrals that you have already seen. So this is X plus one the whole square. Okay. By the way this will generate a minus half. Okay. So make it as a three by any questions. This three by two you make it as root of three by two square. Right. Now what does it resemble? It basically resembles it is basically resembling DX by under root of a square minus X square. Correct. That is already known to us sign inverse X by. So I can write this as one by root two. Sign inverse X by a X by a is going to be this. Is this fine? Any questions? Any concerns? No, I will not be bothering you with too many questions of this. I will directly go to those questions where the question is not directly of this form but you can make it in this form by using some substitution. So let's take this question. Please try this out. C kicks minus one. If you're done, let me know with the done on the chat box. So what is going on in school nowadays? Anyone representative from Kormangla? Anyone from YPR? Anyone from DPS? Anyone from HSR? Revision. Okay. Integration continued. Okay. Gayatri, your teacher is smart. He is not wasting time. Finish differential equations. Are you a Ramakrishna? You started with undergraduate also. What so much hurry they are in? Any differential equation? What are they going to do for the next six months? Revision. See, there are two types of teachers. I'll tell you one who will do a chapter properly. I mean they'll take time but they'll do it properly. One who will complete it early and do multiple revisions, revisions, revisions, revisions, revisions. Okay. I feel the one who does multiple revisions, slowly people will stop coming to the classes only because they will be bored of those revisions. Okay. Done, Pranav. Finished probability. So integration and all is over Pranav in HSR. Differential equation, area under curves, not differential equation. I don't know, I'm speechless. Okay, let's solve this question. So first of all, let's write it as this. So this is under root of 1 minus cos x by cos x. Now see what I'm going to do here is multiply and divide with 1 plus cos x. Just to create a sin x. And facilitate substitution of cos x as t. There you go. Okay. So far so good. No problem. Now here, go for the substitution cos x as a t. So negative sin x dx will become a dt. So this will become negative dt by under root of t1 plus t. Okay. So this will become a negative dt t square plus t. Now we can proceed from here on by converting it to a perfect square. Again, please ensure you are solving a question completely. Don't do skipping of steps else you will forget the formula in the long run. Okay. So this is negative ln. Now here, here your x-roll is being played by t plus half. So it's x plus under root of. Now you could write the same expression or you could just write a t square plus t also because ultimately when you write t plus half the whole square minus half square, it's actually t square plus t only. So it has come from this term only. So don't waste your time writing and inverting it to t. Just save your time. And of course your t has to be substituted back as cos x. Cos of x. Okay. So we will stop here. And next class we'll be continuing with more such quick to urgent questions on this special or standard integrals. And I'll also be talking about several forms, several forms at least five, six forms. I will tell you how to, you know, and all of them will be based out of this only. Okay. More or less whatever we have done. And let's see whether we are able to start with integration by parts in the next session. I'm doubtful, but I will try my best.