So good morning, everybody. So let's start. So as many of you have joined in So today we are going to take up a very very important chapter for you in calculus In fact, this is the heaviest of all the calculus chapters that will come across in class 12 and before I begin with let me tell you this chapter requires a lot of practice and There are different types of you know questions asked under this chapter and for each type There is a different algorithm So this requires all the more practice of at least, you know 15 20 questions of every type So right now when we all have the summer vacation running on this is a good chapter to begin with because you will get ample time to practice and You realize that every every problem has a different approach. I mean you have to think laterally to solve a question Moreover in this chapter you would require that you have to be a very good observer You may have seen Differentiation chapter earlier. It was very conventional. You had some set of methods set of rules which you apply and you get your answer But this chapter is exactly opposite of that chapter Okay, not only in terms of the process, but also in terms of it being unconventional That means you have to think laterally. You have to think out of the box Okay, sometimes an approach may not click Okay, you have to go and change your approach and come back again at the problem So when I was preparing this chapter In my class 12th, I almost practice this chapter for one and a half months Okay, and I did more than thousand plus problems to get confidence Till date also this chapter surprises most of the students even if you know, they have practiced a lot Because different varieties of questions are formed every year Okay So the prerequisite of this chapter is only the concept of your basic differentiation You don't have to know much So I'll be giving you the overview of this chapter So, let me just give you an overview of entire calculus actually So in calculus, what all do you have? Okay, of course in the integration part will talk about Overview of integration. So under integration, you'll be studying two types of integration. One is called indefinite integrals Indefinite integrals or indefinite integration Okay This is going to be just the raw integration. You are going to learn the process over here Okay And the process is basically The process of doing anti-derivative. So we'll talk about it in the subsequent part of today's course So here you are learning the process Here you are learning to get the area function. What is the area function and all we'll talk about that Okay, this has got various types If I recall properly it there are more than 20 types of integrals that you will be coming across Okay, and I was telling the others before most of you joined in that This may take almost three classes three three and a half classes So nearly about 12 to 15 hours will go in indefinite integrals because this process involves so many types of integration Then you'll have definite integrals Then you have definite integrals indefinite integrals. You'll be basically using a lot of properties Lot of properties Right. It is primarily to do with the properties of graphs Okay, because under definite integrals, you are learning how to find the area under a function. Okay And it will be mostly governed by properties. You can say indefinite integral will only play, you know, 10 15 percent role in this Okay, mostly it will be governed by properties. So there are more than 14 properties, which will be told and This is the favorite of J. Okay So J main you will get more questions on definite integrals rather than on indefinite integrals But however, you need to know both the processes Why definite integral questions are asked a lot because it can be asked in the form of area under curves Okay, which is going to be the next topic, which is called application of integrals application of integration Under the application, you are going to just study one application. Okay area under the curves and In few instances, we'll be also talking about how to find the remansem remansem Okay, that is called Evaluating limit of a sum by use of definite integrals. So evaluating limit of a sum by use of Using definite integrals. Is that fine any questions so far? No, there will be no solids of revolution There will be no finding the length of arc and all in your application of integration But however, if you are a student of AP calculus, you need to study all those things Okay, washer method and disk method and all those things that is there There's no application of integration in kinematics also Okay So CBC curriculum also has to say Jay curriculum is very dry when it comes to application of integrals and Finally, there is a chapter which is You can say a summarizing chapter for all the concepts you have learned which is solving of differential equations Okay, so this chapter will use the concept of integration differentiation everything that you have learned Okay in the previous three chapters To solve equations which are made in differential coefficients. That is what we call as differential equations So we'll be learning how to solve first order differential equation primarily But however for people who are preparing for Jay advance, I'll give you an idea how to solve higher-order differential equations also Okay, so as you can see all these equations are linked So all these chapters are linked to integration. So integration becomes a very integral part of your calculus Okay, so it's good that you are starting this chapter in the summers. You'll get enough time to practice okay, so You know take a deep plunge in integration. It is addictive also sometimes Many times you'll feel like you're not doing integration every time right because if it challenges your gray cells You'll feel that it is addictive So this is the Overview of integration any question here. So I've told you about the content. I've told you about the time period Okay, DPPs have already been shared. So get started from today itself So we'll start with indefinite integrals and we'll first talk about why the name indefinite integral was at all given to it Okay Let me begin with a simple question to all of you If I say there is a curve Okay Whose dy by dx or the derivative of that curve with respect to x is 2x Then what is the curve? Then what is the curve? Huh 2x square plus C There's a curve whose derivative is 2x What could be the curve? So y is x square y is equal to x square very good Now Who was that? Sir Anurag Anurag said y equal to x square. Let's say Somebody else says Aditi says why only y equal to x square? It could be y is equal to x square plus 1 also right right this could also be a curve whose derivative will give you 2x Somebody will else somebody else will say why x square plus 1 it could be x square plus 2 also Right somebody could say why x square plus 2 it could be y is equal to x square minus 1 also So what I'm trying to say here is that there is a family of curves Basically, if you go you can do anything like this and still get the same answer so your curve is if If a curve is y is equal to x square plus C where C is some arbitrary constant Okay arbitrary constant sometimes also called as the constant of integration Sometimes also called the constant of integration So if you take any of these you know curves is derivative is going to still give you 2x So what I'm trying to say here is that there is no definite answer There is no definite answer as to what is that curve whose derivative gave you 2x So it gives you a family of curves Right why I'm calling it as family of curves because if you vary this you keep getting different different different family members satisfying the same Right so now these kind of curves will satisfy this Particular equation now boys and girls. I would like to highlight you here You are actually seeing a differential equation in front of you This is what we call as an ordinary differential equation By the way, we are only going to talk about ordinary differential equation throughout our course of study in class 12 There's something called partial differential equation and all which is not in the in a purview of our study Okay, that you'll study in undergraduate. Okay So what you see here is actually a first order first order ordinary differential equation So integration is a process by which you find solutions to differential equation Right so in some time you learn how to integrate 2x Correct. I'm sure most of you already know how to do this is just you know Connecting the dots kind of a thing when you're integrating 2x my dear you are actually solving this you're solving a first-order differential equation Okay, so you have already started you know solving differential equations So remember it was our fourth chapter of our discussion, but you have already started with it right in this very chapter So when you're when you are finding all these curves with satisfy this differential equation You are actually finding its general solution Okay, so integration is a process I'll repeat integration is a process by which you are finding general solutions to ordinary differential equation okay, and Your differential equation in this case when you are performing integration only once Okay, it is a first-order differential equation. Now. What is the example of a second-order differential equation? I'm sure you would have come across This equation in your life Where have you come across this equation in your life? S H M Yeah S H M is an example of a second-order differential equation. This is a second-order differential equation And you would realize that the order decides how many arbitrary constants should be there in the solution or in the general solution So if this is a second order, it will have two arbitrary constants, right? Remember used to write it like this a cos t b sin t Okay, so a and b will be two arbitrary constants because it's a second-order differential equation However, I'm not going to talk about, you know this as of now because any ways I'm going to spend a considerable considerable amount of time on this In the differential equation chapter But just to tell you why there is one arbitrary constant is because you are solving a first-order differential equation Okay, yeah solving a first-order De okay, but now it is clear why the name in definite was given because there is no definite answer It is going to give you a family of curve satisfying it Now in case of definite integral You are actually dealing with area. Okay, it's not that you're not dealing with area here. This is actually an area function Okay, so this function will be called as an area function Okay, it will be helpful for people who are doing a piece. They call it as the area function and Okay, so what is integration in a layman language? It is a accumulative Calculus process Right now now we are not no more into differential calculus versus we are into accumulative calculus versus Where we are trying to find out the area function From the idea of the slope of the tangent at every point of the curve Should I repeat it again? So integration is a Accumulative calculus or it is a part of an accumulative calculus where we are trying to get the area function From the idea of the slope of the tangent at every point on the curve Okay, so you're going to so you're going to construct back the function whose derivative is this okay by the way something very important that you know That you must all know as a nomenclature also when you Differentiate something let's say if you differentiate a function g Okay It gives you f Okay, then we say f of x is the derivative of g of x with respect to x So you must be thinking what is big in this we already know this okay, but when you integrate f of x Okay to get this g of x plus c then we call g of x in fact you can include c also g of x plus c is the primitive of f of x is the primitive of f of x Okay, so this term primitive is a new term to you Don't be scared if if the question reads finds the find the primitive of this so don't be surprised Okay primitive just means finding the integral Okay, so g of x which we call as the integral You can also call it as the primitive Okay And the function that you're normally integrating that is called the integrand Okay, just know these terms because you know while speaking to people you might need to use these termologies fine So a simple example is the coffee beans The coffee beans Are the primitive of coffee the coffee that you have in the morning? Okay, so coffee beans is basically the primitive of the cup of coffee.  So we'll do some integration of some basic functions by the virtue that we know its anti differentiation So let's go back to our differentiation we have learned that if you differentiate any constant you get a zero right it means Integral of zero with respect to x now. This is what this is how you need to write This is a symbol which is a stretched version of s So if you stretch s it becomes this okay s stands for sum because integration is a process of summing up continuously changing quantities this symbol was given to us by Godfried Leibniz Okay, given to us by Leibniz. Okay, Leibniz as you know, he was a co-inventor of calculus independently of Newton Newton was working on differential calculus and Leibniz was working on integral calculus So integral of zero with respect to x that is what we write integral zero DX Please note my dear students. I'm telling you right now here. This is a mistake Which 90% of you will make you will forget to write DX most of the time If you do that the symbol this is makes no sense to me Okay So if you don't mention the variable with respect to which you are integrating then merely writing this makes no sense to me And if you miss it out, let me tell you you will definitely lose half a marks for every miss you make while solving integration question Even board will deduct marks. So two things while doing integration problem never miss out DX Never miss out putting a C at the end So integration of you know zero is going to just give you a constant Next just recalling all our formula back Differentiation of x was one So integration of one would be X plus C don't forget this C Okay derivative of X square by two with respect to X it was X So integral of X DX will be X square by two If you continue this process differentiation of X to the power n plus one by n plus one is X to the power n It means Integration of X to the power n with respect to X will be X to the power n plus one by n plus one plus C But provided your n should not be minus one So this formula will not work When your n is minus one Okay, so for integrating one by X we have a different rule altogether So integral of one by X now try to recall whose derivative gave you one by X Lawn X absolutely, so Derivative of Lawn X Lawn X means log X to the base he gave me one by X So integral of one by X with respect to X will give you Lawn mod X plus C Now, why do we put a mod is because we don't want any No negative value of X to come in because log always accepts a positive argument Okay These formulas which I am giving you out of this this formula is very important. This is called the power rule of integration This is called the power rule of integration Okay, so please remember this Okay, very very important moving on any questions so far Any questions, okay Now, let's talk about some tignometric functions. We know that the derivative of sine of X Is given as cos of X. So this implies integral of cos X with respect to X is sine X plus C Okay Derivative of negative cos X What is it? Sine X so integral of sine X Would be negative cos of X. So we are getting to this formula just by the virtue that we know What are the derivatives? In the same way derivative of tan of X was secant square X So I'll be going slightly faster over here because it is just a repetition of your class 11th differentiation formula So it will be tan X plus C Okay, see many a times. No, let me tell you I'll tell you when the time comes just Let me complete this Are the derivative of secant X. What is that? C kicks an X So integral of C kicks an X is going to be C kicks plus C Okay, what are the derivative of minus cos X cos X Cortex? So we say integral of cos X Cortex With respect to X is minus cos C kicks plus C see no where I'm missing out a DX No, where I'm missing out the C that is the constant of integration So while you're practicing make sure you are, you know conscious about them. Okay. What are the derivative of? minus of Cortex Cosicant square X Okay, so integral of cos square X with respect to X is going to be minus Cortex plus C Is that fine any questions here? Okay Apart from that we'll also talk about the exponential functions Okay, now you will you must be wondering if you see this list In this list, I have nowhere mentioned the integral of tan X Okay, the reason being I don't know any function whose derivative gives you tan X In fact, it is not very commonly used function Neither I have given you integration of Cortex here Neither I have given you integration of C kicks here. Neither I have given you integration of C kicks here So to direct trigonometric basic trigonometric function that I've talked about is only sine and cosine There's no tan. No cot. No C no C Right. That doesn't mean their derivative doesn't exist. We'll talk about it when time comes. Okay, so we'll take all those things Don't worry. So now we are going to talk about Derivative of exponential functions So we all know if you differentiate a to the power X by LNA it gives it gives you a to the power X correct Okay, oh my god Constraint with the space so integration of a to the power X with respect to X will give you a to the power X by LNA plus C okay a Special case of this is e to the power X So if you put your a as e it becomes e to the power X by LNA which is one So nothing changes with respect to e to the power X So e to the power X is a function which on differentiation or on integration gives you e to the power X Okay, so this function is unperturbed by integration or differentiation Okay Apart from this Apart from this you must have learned about the derivative of inverse trig functions, right? Haven't you? Yes, yeah, so what are the derivative of? sine inverse 1 by X 1 by root 1 minus x square Yeah, sorry. What is it? 1 by root 1 minus x 1 by root 1 minus x square very good. Okay, so we say integral of 1 by under root of 1 minus x square is sine inverse x plus C Okay, and don't be surprised if you are a solved by the integration problem and go back and check the result And you see that instead of sine inverse x plus C the answer given is minus cos inverse x plus C Okay, so don't get surprised here. It's because many a times they will work with trigonometric identities to Give the answer in a different form So later on we learn that sine inverse x plus cos inverse x is pi by 2 So the term sine inverse x could be written as pi by 2 minus cos inverse x Correct. So when you add a C to this that means you add a C to this Okay, just give me a second So when you add a constant of integration, please note that pi by 2 plus C will be called as another constant of integration Right, so the same result that you had got could also be written as this see constant plus Constant plus another constant is another arbitrary constant Okay, so don't be surprised if you see the result to be minus of cos inverse x plus C Many a times you would realize that when you see the answer it is very different from what you have got It is not because your result is wrong It is because they may have started with a different approach and Ultimately, if you simplify your answer and their answer you will end up getting a similar expression only differing in constants only differing in constants Getting my point So don't start doubting yourself. Oh my god. What has happened? You know, it has gone wrong and all okay Nothing has gone wrong. It's just that they have taken a different approach or probably they have simplified it to a different form Okay. Now, many people ask me sir in board exams. What if they have the different form and we have done it in a different form? Don't worry about it. They have all the different types of the answers that can possibly arise from that question. Okay So continuing with this list, I think I've got slightly muddled up over here But okay, let me go to the next slide Yeah derivative of Derivative of tan inverse x. What is it? Very good one by one plus x square so we say integral of one by one plus x square with respect to x is going to be tan inverse of x plus c Fine I'm not talking about you know cos inverse x because you know how cos inverse and sine inverse are related so you can You know talk about their formula Similarly, I'm not going to talk about you know caught inverse x because you know caught inverse x as people who don't know this Caught inverse x and tan inverse x are again related by distillation Okay for all x belonging to real numbers. Don't worry. I'll talk about this in the chapter itf What is itf? Yeah, very good inverse signometric functions Okay, next is the derivative of seek inverse x. What is seek inverse x derivative? One by mod x this but I'll just write it like that without a mod so integral of this is Going to be now. This is one of the most deceptive in you know integrals many a times A term like this will appear and we are clueless. We are applying all our algorithms that we know So we failed to appreciate that this is a direct integral I know the direct integral of this is seek inverse x. So there will be a question in today's session Okay, which will be based on this Okay, apart from that just for your information. It is nothing to do with your J or your school level. This is an optional thing that you should know Okay, how many of you are introduced to hyperbolic functions? Okay, so I'll introduce you to some hyperbolic functions and their derivatives etc. I Sorry and their integrals etc. So there's something called sign hyperbolic x. Okay, it is pronounced as shine Shine not sign shine Okay, and this is basically e to the power x minus e to the power minus x by 2 Okay, there's another hyperbolic function Kosh Kosh Okay, cos hyperbolic x which is given by this formula. Okay, we ease your you know Oilers constant Now why they are called hyperbolic function is because they satisfy this you know identity They satisfy this identity Okay Now people will ask why they were given cos and sign terms because it resembles very much like your Trignometric identity or that Pythagorean identity cos square x plus sine square x equal to 1 the only difference is there's a minus here Okay Now if you see it basically fits into the equation of a rectangular hyperbola That's why it gets its name as hyperbolic functions Okay Now this is sign. Sorry shine Kosh and then there is Dan Okay, Dan X. Okay, Dan X if you put an edge it become Dan X. Okay, Dan X is nothing but shine by Kosh So it becomes 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 Okay, the last class you did limits with me, right? Okay, tell me what happens if I take limit of than as x tends to plus infinity Okay, and limit of than as x tends to minus infinity fast I'm testing you on your concept of limits Okay, you did this last class with me Tell me what is the limit of than as x tends to infinity and what are the limit of than when x tends to minus infinity? Is it so difficult? Okay, so Aditya is saying that as x tends to infinity it is going to be one Okay, is he correct? Is Aditya correct? Okay, and as x tends to minus infinity Pratham tells me that it's minus one. Okay. Yes, of course both of you are correct By the way, if you're doubtful how it comes so when you're evaluating the limit as x tends to infinity Okay, remember we take e to the power x common something like this Because it is right now in the infinity by infinity form Okay, cancel this out as x tends to infinity these two gentlemen will go to zero So as a result your answer will be one that means this this graph will die on x equal to sorry die on y equal to one line So y equal to one is asymptotic to this Okay And if I take the limit as x tends to minus infinity Okay, then basically what do we do is sorry minus infinity? I have to write Okay, then what do we do is you take e to the power minus x common Correct. So it is e to the part 2x minus one by e to the part 2x plus one So now as x tends to minus infinity e to the part 2x will become zero Zero, so it'll give you minus one by one, which is minus one. So y equal to minus one is also asymptote to these graphs Let me show these graphs to you if you have not seen them before So on your GeoGibra, there is a ready-made function The moment you type sign it will show you sign Hx Okay, put an x. This is the graph of shine very much like the graph of x cube Okay, gosh Gosh Okay, this structure is called a catenary This structure is called a catenary It is basically the structure taken by your necklace when you are wearing it around your neck Okay, this is the structure taken by your necklace Or you can say it is the structure taken by any wire of uniform mass per unit length Hung between two points whose length is lesser than the or whose distance is lesser than the length of the wire So if you see that wires of electric wires and all hanging in the Around that will be in the shape of a catenary. Okay showing you now Dan Dan So then as we had discussed y equal to one and y equal to minus one will be like asymptote Okay, as you can see it is limited between these two fine Okay, now Why I'm introducing you to this function. In fact, I'll introduce you to their inverses as well Okay, this is very important because many a times will use our online calculators to solve something And you will see that answer they give will be in terms of you know Sometimes in terms of hyperbolic function, so you should not be surprised that okay. What is this answer actually? So there is inverse of sign or shine which is called sign hyperbolic inverse X and it is given by LN X plus under root X square plus one People who have done integration to a slightly you know advanced level would know that they see this term later on So we'll come across this term later on I'm not sure whether we'll see this today or in the next class Then there's something called cos hyperbolic inverse X which is the same thing, but with a minus sign yeah, and Finally tan hyperbolic inverse X Which is half LN one plus X by one minus X okay Please note that the domain of this function will always be a value which is between minus one to one Because the range is between minus one to one so domain of the inverse I have not done inverse of a function with you officially, but when we do that You will understand that the range of a function is the domain of its inverse and vice versa Okay, so now these concepts will be used in in our some of the formulas which I'm going to give you Oh, I'm so sorry. Yeah, sorry This is what you wanted to see Done Yes, sir, okay, so now there are three important integrals that You know normally when we integrate this Okay Many books you will see that the answer that they would write is this In fact all yours, you know school level books all the je books will write the answer like this But don't be surprised if you see sign hyperbolic inverse X plus C given as the answer It is because this term itself is signed hyperbolic inverse X So you may use some you know online calculators wolf ram or some kind of You know calculator online. They may give you the final result like this Okay, so it is just for your awareness. It is not like, you know, you have to know this So normally the answer to this is given as ln mod X plus under root X square minus one But don't be surprised if you see Cos hyperbolic inverse X given to you as the answer This integral is given as half ln One plus X by one minus X plus C And don't feel surprised if they if at the back of the book they say tan hyperbolic inverse X per C Not not your books, but probably it will be given in your online calculators. Okay So this is just for your information Now we are going to talk about the rules of integration. Hope you have copied this down Any questions so far Is that fine? Can we go on to the next page? Aditya Okay, so now Remember when we did differentiation, we followed it up with the methods of differentiation or the rules It's like when you know how to drive a car, you should also now obey the traffic rules So they are the rules of integration that we are going to learn In fact, there are many many rules for different different types of questions But these are the broad level rules that you must follow Some of the rules I will introduce only when the time comes Okay If you're integrating a function f of x multiplied to a constant You can bring the constant outside and multiply it to the result Okay And of course put a constant of integration, which anyways will come from this process Okay, so for example, if I'm integrating cos of two cos of x It is just like integrating two times cos of x So constant can be pulled out and you can write this as two sin x Okay, plus c now another question that people ask me Sir, do we have to write c in the bracket of two? That means do we write two c here? No, not required two c c log c e to the power c everything is just c Okay, I don't know what you guys are thinking right now Okay, so whatever is the c Okay, don't put any kind of just keep it as c nothing else Okay Secondly when you're integrating the sum or difference of You know multiple functions it is as good as Adding or subtracting their respective integrals Okay, and just put one c at the end. Don't put c for this c for this separately and don't add it for what's sake Okay, just one c at the end of the solution is sufficient Okay Other than this other than this If you feel if you feel that there is a rule like this, okay, you know why I'm writing this in red Because such a rule does not exist Okay, so if you feel there is some rule that you had done in differentiation and there's a rule similar to that in integration I'm very very sorry. There is no such rule like this Right So the product rule that you have learned in your differentiation chapter that doesn't work in integration the same way. Okay, so avoid extra polluting any formula Avoid extra polluting any formula means do not try to invent your own formula Okay, that's one of the gray mistakes which people have been doing. So there is nothing like quotient rule As you have learned in your integration. There is nothing like this.  There is no quotient rule as you know them in differentiation Yes, there is a rule called integration by parts. There's a rule called integration by parts Okay Though I'm giving you the formula right now, but do not use it until unless you have been officially taught this particular concept So in today's class, I'll be giving you a lot of you know questions which will involve product of two functions But do not use them. It is just like, you know, let me give you an analogy.  Okay, because it comes with a lot of You can say, um, what do you call? negative things Okay, what do you call that in medicine terms? lot of Side effects. How can I miss that word? Yeah, it comes with a lot of side effects. The side effect is you may end up integrating, you know, which is uh integrating something which will go on and on forever. Okay, so don't misuse this Okay, even though I'm giving you the formula, don't misuse it. The formula says If you're integrating product of two functions It is given as f integral g Minus integral of Derivative of f into integral of g Whole integral Okay, again as of now be blind about this formula. We'll talk about it later Okay, when I deal with this topic now, how do we choose f and g? That everything will be, you know, told to you later on as of now just ignore this formula Okay The next rule that we are going to discuss is something which we call as the reverse chain rule A reverse chain rule Okay, short form RCR What is the reverse chain rule? It says that If the integral of any function Is g of x plus c Okay, then Then if you are changing your, you know input of x by let's say ax plus b That means you are linearly extending x to some ax plus b for example Let's say you in you know how to integrate cos x. Let's say cos x integration We know that it is sine of x plus c right If instead of cos x you were given cos 2x plus 3 Then this formula will help you So this gives you The same answer but again x replaced with ax plus b divided by the coefficient of x Divided by the coefficient of x, which is a okay very much similar to the chain rule which you had learned in differentiation And that's why here we call it as reverse chain rule because in chain rule used to multiply that by that constant, isn't it? Here you are going to divide by that constant I'll give you a simple example for this Yes, any question anybody No, no, no, that's why I was very, you know, uh Stressing on the fact that it could only be a linear extension of x no quadratic extension Okay, only linear extension it will work So instead of x if you have let's say x square plus 2x plus 5 Sorry, this formula is not going to work for that. You have to apply some other methodology Okay, it is only for those minor changes where x has been replaced with some ax plus b form So now we all know integration of secant square x which is tan x plus c So if I asked you what is the integration of secant square, uh, let's say minus 3x Uh, plus 4 Okay, so for this you can directly use your reverse chain rule and say it is going to be Uh, tan of minus 3x plus 4 Divided by the coefficient of x over here, which is minus 3. Okay. Just add a c at the end. That's it Okay, this is called the reverse chain rule. Okay Other than that, yes, there are different types of algorithms that we follow for different varieties of questions. Okay And We'll introduce to you when the time comes. So we'll start solving questions in this uh topic. So let me begin with the first question So here comes the first question for you For every question, I'll give you around 90 seconds because they're not worth more than that Uh, don't don't need to type out the answer. Just write done if you're done. Okay, so exactly after 90 seconds We'll discuss these questions Probably some question will take you lesser also. Okay, so the answer is done Sudha also very good shankin done Triple also done. Okay All right, so let's discuss it, uh, dear students since we do not know any kind of a quotient rule The only option that is left with us is to expand the numerator And you know manually divide Manually divide each term by under root of x Okay, so when I do that I get 1 by root x 3 root x 3 x to the power 3 by 2 And x to the power 5 by 2 Okay, so as you know that from the rule of integration, we can individually evaluate their integrals and add them up Okay, now each of them will follow the power rule. So this is x to the power minus half Correct. So x to the power minus half integral will be what x to the power minus half plus 1 by minus half plus 1 Okay, don't put a c because c will put at the last So this is going to become 2 x to the power half Okay, similarly, this will be 3 Integral of x to the power half will be x to the power half plus 1 divided by half plus 1 Okay, this term will become 3 into x to the power 3 by 2 plus 1 by 3 by 2 plus 1 And this will become x to the power 5 by 2 plus 1 which is 7 by 2 divided by 7 by 2 Okay, just put a c at the end of the process Okay, you may further simplify this and write it as 2 root x This will be 2 x to the power 3 by 2 This will be 6 by 5 x to the power 5 by 2 And this will be 2 by 7 x to the power 7 by 2 Okay, first question very simple any problem in doing it any questions here Okay Next question Integral of 2 to the power x plus 1 minus 5 x to the power x minus 1 By 10 to the power x Just type done once you're done. Good It seems you people are practicing also integration on your own. Okay, very good So same people are saying done again. Okay, Shristi Aditya very good Nice Okay, again If you have to solve this we can write the numerator as 2 into 2 to the power x minus 1 5th into 5 to the power x And individually divide by 10 to the power x 10 to the power x Okay, so this is what we are integrating If you open this up it becomes 2 integral of 1 by 5 to the power x Okay minus 1 5th integral of half to the power x And I've split it this up as two separate integrals any questions here any doubts any question from Any question on conversion from this step to this step Now the reason for doing so is because now I have converted it to integral of this form. So please recall Integral of a to the power x is given as a to the power x by lawn. Okay Don't put a c and all c should be put at the end. So two times this will be 1 5th to the power x by ln 1 5th or lawn 1 5th minus 1 5th half to the power x by lawn half Okay, just put a c at the end If you want to further give it a nice touch you can say minus 2 into 5 to the power minus x by minus ln 5 And minus 1 5th 2 to the power minus x by minus ln 2 So it becomes 2 5 to the power minus x by ln 5 plus 2 to the power minus x by 5 Okay, any questions here any doubt So I'm again starting with very basic problems so that you know, everybody is able to understand those basic rules Next secant square x times cosecant square x again It's a request to all of you. Do not use product. Do not use Integration by parts ibp. Please do not use it right now. I'll tell you when is the right time to use it probably not in this class I think next class will be talking about questions related to ibp as of now. These can be solved without using ibp Without using integration by parts Done. Oh, wonderful. Wonderful Ish, what is that x? Oh, Dhan has also written the answer. Thank you. Very good All right, let's discuss this minus 1 No, no, no This will give you a function of x only minus one and all will not come out to be the answer if it is a minus one It will be c Okay Yeah, and I'll also try to like to remind all of you that don't miss out on a c. Okay Now this term I could write it as cos square x into sin square x Okay So what I'm going to do I don't know any kind of a quotient tool product rule. So what I'm going to do is See, that's what I said you have to think laterally many times. It's an unconventional way of problem solving Integration is one of the chapters where you have to run your mind like anything. You have to be a good observer Now this one you have on the numerator.  So what I have done I have converted it to Seekin square x plus cosecant square x And thankfully I'm aware of integral of both of these expressions. So integration of secant square x is tan x Integration of cosecant square x is minus of cortex Okay, don't forget to put a c. So this should be your answer tan x minus cortex plus a c Is that fine any question anywhere? Please also let me know if I'm going little fast Okay, again sunken there could be so many approaches Ultimately the answer that you should get should simplify to what we have got Okay, so this question is x square plus cos square Whole multiplied to cosec square. This is whole multiplied to you can you can put a bracket here if you want Divided by one plus x square any progress anybody Sure, sure. Okay, just he is done very good. Just see Okay, Kirtana We'll see we'll see Kirtana whether that's correct or not Okay, so enough time has been allotted Now here we have to do some manipulations I see a one plus x square down and I see an x square over here So obvious feeling that will come in my mind is Let's put it as x square plus one and subtract a one Okay, so basically I like it like this So this is what I have done to this bracketed term. Okay And of course outside we have cosecant square x waiting for us Okay, all divided by one plus x square now If you multiple if you see this term This term over here this term is negative of sine square x So I can write the entire expression as something of this nature multiplied with cosecant square x And divided by one plus x square. So one plus x square. I'll separately divide here. Okay So this gets cancelled Now when you open the bracket, you'll end up getting a cosecant square x Okay, let's have it separate minus sine square cosine cosec square will cancel themselves out and you'll be and ending up with this expression Okay, now these are two simple expressions, which we can integrate. This is going to be minus of cortex And this is going to be minus of tan inverse x remember we had done this particular Integration in the formula list Okay, so this finally becomes your answer for this Any question here? Can I move on to the next one? Anybody who wants to copy something or ask something? Please let me know Let's do this one Integral of one by one plus sine x So all these questions you must be realizing that with minor modification You are able to convert or break the function into Basic functions whose integral is already known to you. Nice. Nice. Everyone. Okay. People are giving the answer. Okay So for this question, what you'll do is uh, the moment we see one plus sine x an obvious feeling comes in our mind that Let's multiply and divide with one minus sine x. Okay this has been an You know very old process that we have been following since class 10th identity stigmatistic identities So this breaks the given expression to this and individually when you divide it one by cos square x and sine x by cos square x Okay, that leads to that leads to secant square x And this is nothing but tan x ckx Again, these are two basic functions whose integral is known to us. So that's going to be tan x minus ckx plus c So this is going to be your answer tan x minus ckx plus c aditya I think sign mistake has happened Okay, any question any doubt any concern? Okay, so these are like the warm-up questions that I would like you to attempt Uh, they might they might come in your school exams or your uh uts Let's do this ckx by ckx plus tan x And there is a twin problem of this which is tan x by tan x ckx plus tan x So approach is the same. You don't have to worry about the approach No, no same answer is not that it is a twin problem means a similar approach is there So darta, it's not the same answer Similar approach is there for both You're saying the same answer as the previous question. Okay, okay Yeah, because if you multiply with the cos x it results into the same expression. Okay an alternative way to do this is Okay, yes, you can convert it to sign in cos the alternative way to do it is you multiply with ckx minus tan x When you do that you end up seeing a one in the denominator Okay, so this guy becomes a one So that ends up giving you the same expression you are absolutely correct And that is nothing but tan x minus ckx plus c Okay If you have to solve this question all you need to do is This is one minus ckx by ckx plus tan x Okay, so instead of integrating this you can choose to integrate this integration of x one is x Integration this is just now we figured it out tan x minus ckx plus c So don't be like, you know Confused to see this question instead of this Next these are all confidence boosters so that you know the basic process at least One thing I have observed from my experience. In fact, it's it was my own experience that When you're practicing at a certain stage you will realize that Okay, after this I can do the question after this I can do the question. I know how to do it after this and you'll stop there Okay What will happen in due course? What will happen? You will start forgetting your basic formulas So for the initial one month While you are practicing integration chapter, please take your problem to completion until unless you write that plus c Don't stop anywhere. Don't think that okay after this I'll be able to do it Because while you're you know doing the problem completely you are also recalling all the formulas that you have learned Okay, so you're building your muscle memory done Very good. Okay. So in order to solve this Try to convert numerator in terms of the denominator So I can see 4x plus 7 under the root symbol, right? So what i'm going to do is i'm going to write 8x plus 13 as twice 4x plus 7 minus 1 individually This term and this term i'm going to divide by the denominator So when i'm going to divide first of all I will get an under root of 4x plus 7 from here And let me separate out the integral also. I don't want to keep it in the same Okay, now both are basically cases of reverse chain rule Okay, so this is like a extended linear extension of under root x This is a linear extension of x to the power minus half Okay, so recall reverse chain rule. So the answer here would be 4x plus 7 to the power of 3 by 2 divided by 3 by 2 but also divided by 4 This is very important Okay, here 4x plus 7 And it's minus half plus 1 by minus half plus 1 But also divided by 4 are you getting this point? So don't forget to divide by 4 in this case So here you can see it's one third of 4x plus 7 to the power of 3 by 2 Okay, and this is going to be minus half under root of 4x plus 7 So this is going to be the answer for this Any questions here any questions? Okay, let me know if if you want some time to copy things because i'm just you know rushing through it sine qx Now don't start extrapolating any formula When I say extrapolating means if you think integration of sine x is minus cos x So integration of sine cube x will be minus cos x the whole cube plus c it doesn't work like that. Sorry Do not invent your own formula Works strictly according to the standard integrals or the basic rules that you know I have not given any such rule which will make you cube the result if there's a cube of the integrand Okay, pratham. Nice. Good R. Eman I really doubt known that's not correct R. Eman Okay, let me take this up quickly Now nothing to worry. We already know that we have a multiple angle formula to help us out over here Sine cube appears in the formula of sine 3x now Why I'm taking the help of this formula is because let me tell you boys and girls It's easy to integrate tignometric ratio of a multiple angle Whereas it is not easy or it is difficult to integrate a tignometric ratio raised to a higher power So what we do we always convert a higher power to a multiple angle Because converting it to multiple angle will just require the use of reverse chain rule So in the same effort I'm going to convert sine cube x as 3 sine x minus sine 3x Whole divided by 4 Okay, so here where your tignometric ratios identities will come into your action So instead of integrating sine cube I can integrate this entire expression Since 1 by 4 is a constant. I am keeping it outside the integral. Okay. Now you can easily integrate them integration of sine x is minus Cos x integration of this will be cos 3x by 3 Okay, just put a c if you want you can put your Terms of pretty also minus 3 by 4 cos x And plus cos of 3x by 12 Okay, this is going to be your answer One more confusion.  Okay, don't mess up with the sign Wait, wait, wait. I have not taught substitution yet. Aditya. Wait I've never mentioned any word like substitution so far Okay, but I'm soon going to do it. Don't worry As of now whatever resources we have we are dealing with those resources only. Okay, just the basic rules and reverse chain rule All those stuff Substitution I'm going to take up immediately after this question In fact more questions will take one more then we'll take substitutions Okay, Aditya. You want me to go to the 13th slide? Okay So everybody I am assuming that you have copied this question on your notebooks Okay, I'm just going to the 13th slide as of now. Yeah, Aditya. This is the 13th slide Do you have any questions on this? I just wanted to copy it All right again since we don't know any kind of a product rule and all in fact Even if I know I will not apply it to this Because as I've told you that that's like a nuclear weapon. Don't start misusing it when you can break an egg with a spoon Why to pick up a hammer?  I'll choose to integrate this fellow So that will give me half integration of cos x is sine x And this is sine 5x by by So that'll give me half sine x Minus One-tenth of sine 5x any question here. Trippan. I think you made a sign mistake again check Yeah, any questions here anybody? What's wrong with this? All right, let's take up this question So in the past 19 minutes, we have solved almost 10 questions. So that's a very good speed Of course, there are various other methods also to do it which we haven't yet learned But whatever you know so far that is sufficient to do this question. No clue. No idea Anybody anybody has any idea how to do this? Okay, let's look at this term which is within the under root sign this quadratic term. Okay, if you multiply it You see x square minus 7x Okay Now what I'm going to do is I'm going to multiply this term with a 4 That means I'm going to make it as 4x square minus 28x plus 48 now If you were a good observer you'd observe that this is nothing but 2x minus 7 the whole square minus 1 Do you realize that The same 2x minus 7 which is present over here also Right So this is what exactly I'm going to do I'm going to multiply a 4 inside the root That means to in order to compensate that I must have a 2 on the numerator Okay And inside I am going to write it as 2x minus 7 the whole square minus 1 Okay, if you want you can keep this 2 outside the purview Right now doesn't it remind you of the Integral of x and the root x square minus 1 Is it where the x-roll is being now played by 2x minus 7 correct Yes or no So the answer to this we can directly write to seek inverse 2x minus 7 But don't forget to divide by a 2 because this is a linear extension of x So 2 and 2 will get cancelled. So just put a C. This is what your answer is going to be Okay Let me see if anyone realized that. Oh, yeah, Kirtana realized that very good kids now Okay, so you have to be a good observer in that case Now moving on To another important rule which is called the method of substitution. In fact, it's not a rule It's an algorithm or a method. Okay This is what aditya was talking about a couple of minutes ago method of substitution I call this method as the heart and soul of integration heart and soul of integration Because almost in 80 percent of the questions you would read some kind of a substitution Now first of all, why do we need substitution? We need substitution to convert a complicated expression to a simpler integrand or a We need to convert it to a simpler form which we can easily integrate Right either to your basic formula or to your standard results that you will be seeing, you know in dukos Okay, so the whole and soul Aim for method of substitution is to convert your complex integrand To a simpler integrand whose integration is known to you The second question arises. How do we know how do we do these substitutions? Now, there is no fixed method for it. It all depends on question to question So for that you have to be a very good observer that that is what makes this chapter very unconventional There's no fixed rule that okay take this and substitute it. Okay There are in few cases But many a times you will only come to know them Through a lot of practice Okay, so practice and a good observation skill is going to sail you through this particular chapter Okay So i'm not going to give you any kind of a rule for how to substitute Of course in some cases I will give you but not in every case Because there can be millions of problems framed and you have to think millions in a millions in a different way Let me start with The integration of certain trigonometric terms which we haven't seen yet. So let's say if I want to integrate tan x, how will I integrate it? Okay Now, let me solve this for you so that you get an idea. How do we do substitutions? So we know tan x could be written as sin x by cos x Okay Now when you see such a term When you see such an expression Okay You can see there is a sin x term sitting on the numerator and there's a cos x term sitting in the denominator Okay So Let's say a student a What he does he substitutes cos x with t Okay When you substitute a function or a part of a function with t or when you substitute your uh, You know any particular term with a t, please note You have to convert the entire expression in terms of t when I say entire expression That means you also need to take care of this dx term also I've seen in my past many people do half hazard substitutions That means they will substitute the function But they will not substitute their dx As a result, they are not doing mistakes. They are doing blunder Everything will go for a toss Okay So please remember this That whenever you are making a substitution in that integrand Not only the integrand, but dx also needs to be converted in terms of the new variable And that's why it becomes so important not to miss out writing that dx because if you don't write that dx You'll probably forget to replace it Okay, so it's not only a matter of habit. It's a matter of compulsion that you need to write that dx So once you do that, let me differentiate both sides with respect to x So differentiation of cos x is minus sin x Differentiation of t with respect to x is dt by dx. Okay. Now what I'm going to do is I'm going to make the final, you know Shuffling over here. So let me put the dx term on top here Okay, what I can do is since there is a sin x into dx term I can directly obtain that from here itself in one shot as negative dt So I'm going to make a replacement of the numerator with a negative dt Denominator is of course a t Yes or no By doing such substitution, I converted it to a form whose integral is well known to me Remember, we had done one by x. Don't worry about this minus sign. Minus sign can be always pulled out So remember we had done integration of one by x, which was ln mod x Correct in the same way the integration here will become minus ln mod t plus c Okay, now don't forget to put your variable back in terms of x because t is something which you had introduced in the question So when I put back the t, I get this If I take a minus sign on top, it becomes ln ckx plus c Okay, so going forward, please remember this result because you're not going to derive this over and over again Okay, see I'm again repeating it I broke it up as sin by cos Into dx then aditya what I did I replace cos x with t Okay, I differentiated both sides with respect to x. So I differentiated both sides with respect to x Okay, I took the dx on the other side Okay, sin x dx will become negative dt So this becomes negative dt by t Which is negative ln mod t plus c. Okay, put your t back now as cos x Okay, this minus sign comes as a power on it. So cos x to the power minus one is c So ln ckx plus c is the answer Any questions here Now, let's say there's a student b Right Who thinks let me substitute sin x st Okay, see nothing will go wrong. Is this that His process might be elongated So if you take sin x is t Okay, differentiate both sides you get cos x Okay, by the way, uh, my dear students, let me tell you one thing here Are going forward I'm not going to write this step. I'm going to directly jump from here to here. Okay Is that fine? I'm not going to show you step by step. How did I reach this? I'm directly going to reach this Okay, any question with direct conversion from here to here? Okay now When you're solving the question Sin x by cos x into dx You need a cos x on top so I can create a cos x by multiplying numerator and denominator with a cos x Okay, so student b is taking this approach where he took sin x st Then he realized cos x dx is coming dt and there was no cos x on top So he manually created a cos x by multiplying and dividing this is called the manipulation You need to learn this because you'll have to do this in subsequent concepts Then what he did this cos square that he got down He converted that in terms of sin because he had substituted sin x st Now Replace your sin x with t replace your cos x dx with dt replace your denominator with 1 minus t square Okay, but still he's in a fix because we don't know how to integrate this also So what he will do next is he's again going to make one more substitution He's going to make this as some other variable. Let's say k Okay Now differentiate both sides with respect to t Okay And bring this dt on top even though I said I will not do it But I'm I'm doing it for the initial you know part of your learning Okay So I can say t dt the term which is present over here. You could write it as minus half dk Correct So this entire expression becomes minus half dk by k Now finally student b realizes that he has something which he already knows how to integrate So his result would be minus half ln mod k But now he has to go back from k to t and from t to x Now see how long the process has become But he will end up getting the same result Right. We know 1 minus sin square is cos square Okay And if you raise cos square to minus half power it becomes ln ck So ultimately the same answer he gets but with multiple substitutions Okay So why I told you the method be up in this approach of student b is because if you choose up taking a wrong function also as t You may end up getting the answer, but it's just that the process will be elongated. It will be a prolonged process Is that fine Is this example helping you to understand how method of substitution works? Now I would request all of you now to integrate in the similar way Caught of x Try this out Sir we can use this method for any problem or See again, uh, there is no rule like that. You can use it for any problem where it is required. You have to use it It's like technometric identities So for the previous problems you gave we can use it like it won't No, no, yeah, you can use it.  Okay Is that fine ananya any question? Yes, sir. Thank you Okay, so here I'm sure most of you would have figured out that you could write it like this And by observation you can see that if you take sin x st Causes dx will behave as your dt. Okay. So the next step is taking sin x st As I told you I would directly jump to this step Okay So your numerator entirely will get replaced with dt So it becomes dt by t again. It's a you know integral which we know how to do it Okay, so the answer will be ln mod sin x plus c Please remember this result because please add this to your formula list because nobody is going to Derive this separately in the examination hall So why to waste time? Let me just cover up the integration of these Left out tignometric functions. So tan we have already seen cot. We have already seen let's talk about seek Now in case of seek Again, you have to think out of the box This will not help you out This will not help you out Okay So people who are trying to convert it to 1 by cos x remember it will not help you out There is another approach which again will come through a bit of practice for you So in order to integrate this what we do is we multiply and divide with We multiply and divide with ckx plus tan x Again, you will have to have this, you know, you can say foresight Okay vision. What are you going to achieve? See when you do that You realize you end up getting seek square x plus ckx tan x dx on top And in the denominator you have ckx plus tan x And you can see that this is the derivative of tan And this is the derivative of seek Right. So this will actually help you to substitute Ckx plus tan as a another variable t Correct. So when you differentiate both sides Automatically the term on the numerator will surface up So as you can see this term is already sitting on your numerator Isn't it? So that term can be replaced conveniently with a dt And denominator as we have already taken it to be t again, it comes out to be 1 by t So many of you would be thinking every time it comes out to be 1 by t. No, not every time But yes in most of the cases you will see that it gets converted to such kind of expressions Okay, this is basically actually covered under I'll tell you after I complete this question So it is going to be ln mod t plus c And put your t back your t was ckx plus tan x Okay Now let's say you are solving this problem And you turn your book back and see that The book has given the answer as ln tan pi by 4 plus x by 2 Now don't be surprised if you see this. Okay, don't start doubting your you know integration buses It is because the book probably has used a simplification of this term Okay, now I would like all of you to you know realize this on your own If you simplify ckx plus tan x Okay, we can write it as If I'm not wrong, we can write it like this correct And 1 plus sin x we can write it as cos x by 2 plus sin x by 2 the whole square I'm sure you are familiar with 1 plus sin x being broken up like this And cos x could be written as cos x square by 2 minus sin square x by 2 Okay If you split this up into two factors in the denominator You will end up getting a minus ba plus b formula Okay, this and this gets cancelled Okay in the remaining term you can substitute your or you can divide throughout with cos x by 2 So that will give you 1 plus tan x by 2 By 1 minus tan x by 2 correct So what I did the term which was remaining that is this term I divided my numerator and denominator By cos x by 2 Okay, and this clearly is the formula of tan pi by 4 plus x by 2 That's what you see over here Okay, so don't be surprised if you see this formula at the back of your book. It is one in the same thing Sometime they will also write it as ln ckx minus tan x Okay, do I need to tell you why? Do I need to tell you why they can write this term as this? 1 by ckx minus tan x Yeah, it is ckx for tan x is 1 by ckx minus tan x. That's Is this fine? So this is a term which probably needs a different approach altogether Of course, you are using substitution only but only after doing certain manipulations. So be ready with these manipulations And it may not come, you know Very naturally to most of us In the same way Hope you don't want don't have anything to copy in this can I move on to the next slide In the same way if you want to integrate cos ckx with respect to x We have a similar approach We multiply and divided with cos ckx minus cortex Why do we do that is because The moment you take cos ckx minus cortex as your t You realize that the derivative of this Okay, I'm just writing down the derivative of these terms So cos ckx derivative is minus cos ckx cortex and minus cortex derivative is cos square x So this into dx is dt And this term is Already sitting on the numerator check it out. So if you open the brackets you end up getting This curly bracketed term correct That means you are and again ending up integrating dt by t. So which will give you ln mod t plus c Which is nothing but ln mod cos ckx minus cortex plus c Please remember this formula very important And don't be surprised if you see ln tan x by 2 also in your answer Both mean the same you just have to reduce this now how to reduce it.  So they're all the same things Is that fine any questions here? So now in your formula list you have added integration of tan x Cortex ckx Cosy ckx as well and thanks to method of substitution By which we could do that So in your mind, please, you know recall all these formulas Integration of tan x close your eyes everybody Think integration of tan x ln mod ckx plus c Okay, done integration of cortex ln mod sin x plus c done integration of ckx ln mod ckx plus tan x plus c You can also remember it as ln mod tan pi by 4 plus x by 2 plus c done integration of cos ckx is ln mod cos ckx minus cortex So remember there is a minus here in seek when there was a positive So ln mod cos ckx minus cortex plus c You can also remember it as ln tan x by 2 plus c Okay As of now scribble it down somewhere because you would definitely need it in problem solving So let us start solving questions on method of substitution other than what we have done So let me take up a problem with you. Okay, let's take this up done very good So here you have to be a very good observer If you see this guy, it is 1 minus 1 by x square correct This actually comes as a derivative of x plus 1 by x Right and both of these terms are present Now this is an indication to you that we have to take this as t right Isn't it so you have to observe from the given expression What all could be substituted and what all will behave as you know, uh, the derivative of that It may be present in the question itself or you may have to manually Do some kind of tweaking with the question to generate that Okay, that's why this topic is very unconventional So you can see that 1 minus 1 by x square dx this entire term Will behave as a dt So this whole question gets converted to integral of t to the power 3 by 2 dt Correct. Yes or no So that will become t to the power 5 by 2 divided by 5 by 2 plus a c And put your t back. You are done. This is the answer. Okay. Does it make sense? Correct. Shishti pratham. Very good. Vibhav. Just check you're working Okay Let's take one more not one more many more. Sorry. Let's take this one. Okay simple So here you can see that tan x derivative Is secant square which is present on the top Okay, now many a times people take tan x If it is not wrong to take the whole thing also You can take the whole thing also st because constant will anyways not matter to you So if you differentiate both sides with respect to, you know, the respective variables you end up getting something like this And this term is already sitting proudly on the numerator. So that will be dt And 3 plus tan x is t again. Okay. So once again, you can see ln has Appeared. So your answer is ln mod 3 plus tan x plus t Okay So I think most of you are now feeling confident about, you know, how to substitute Okay, we are going to take more examples Try this one out Integral of 1 by 1 plus e to the power minus x Done One minute sir Now see here if you just take this as 1 by e to the power x and write it properly you realize you'll end up getting this Okay And e to the power x is the derivative of e to the power x itself So the obvious substitution that will come into your mind is let's substitute this entire denominator as a t So e to the power x dx Will automatically behave as dt. So this entire term will behave as a dt So again, it gets converted to dt by t form, which is ln mod t And t is ln mod e to the power x plus 1 plus c Let's take the another one and be here are constants. Okay a and b here are constants and you're integrating with respect to x Now again, if you see by observation that From the past you have seen that whenever you differentiated sine square x we used to get two sine x cos x Isn't it which is nothing but sine of 2x Okay And whenever you differentiate it cos square x We ended up getting minus two sine x cos x isn't it which is again minus sine 2x So The derivative of these terms is resulting into sine of 2x So what i'm going to do here is in light of this observation I'm going to put this entire thing as t So if you differentiate both sides you end up getting sine square 2x The derivative of this will be minus b square sine 2x Don't forget a dx over here, which was actually in the denominator of the right side But I've pulled it over here as I told you I will not write it like, you know Uh, this is equal to dt by dx and directly jump to this step Which clearly says that if you take sine 2x common Okay It gives you sine 2x dx as dt by a square minus b square So this term that is sitting over here this term that is sitting over here Okay, the one which I have shown with a yellow amoeba structure I could remove replace that with dt by a square minus b square So I end up getting dt by a square minus b square, which I'm keeping outside because it's a constant And the denominator is a t Okay, so this is what we end to get to integrate so we can write this as 1 by a square minus b square Okay, and this will become ln mod t plus c Okay, so replace your t back and this is what you should see This question is a potential question for you in your school exams So please mark it with a star and keep this may be asked in your school exam. Is that fine any questions? Let's take this question. Okay, so ln of tan x by 2 by sine x Yes, any idea how to do this? All right, so for a change now here you will take a numerator term to be t So this is a you know trend breaker many of us we feel that Only the denominator terms can be taken s t and the derivative will always be there on the numerator Don't be under that impression things may change. Okay So if you take this s t the derivative on both the sides if you take with respect to x you'll end up getting this Into secant square x by 2 into half This is going to be dt Please don't forget into half you are applying chain rule for differentiation So this will give you 1 by 2 if i'm not mistaken you'll end up getting sine x by 2 cos x by 2 isn't it so secant square x by 2 is Cos square x by 2 will come on the top and the cos of the denominator will get cancelled Thereby giving you dx by sine x as dt. Yes, sir. We already know that Integral of cos x is element and x by 2, right? Yeah So we very good very good. That's that's also a good good idea to do it So what he said, okay, let me complete this So You can take this entire term dx by sine x as dt. So it becomes t dt Correct. So your answer is half t square Plus a c that means half log tan x by 2 whole square plus c Yes, I know another important thing is You know that integral of cos x dx is ln tan x by 2 by fundamental principle or by fundamental theorem of calculus we can say derivative of ln tan x by 2 Is going to be cosec x Which means d ln tan x by 2 Is going to be cosec x dx Okay, so this entire term you can break it up like this x by 2 and dx by sine x which is cosec x dx you can write it as d ln tan x by 2 That means you have written something like t dt over here Okay, that'll again become t square by 2 plus c which is again square of ln tan x by 2 plus c very good You can apply that method also to do it Now i'm going to take a break over here. Okay, let's have a Let's have a break of 15 minutes Okay, we'll resume at 11 17 a.m Any questions? Please let me know Take a break right now All right. Welcome back everybody So, uh, let's start with this question. I hope you can see the question here Uh, you have to integrate dx by x square Times x to the power 4 plus 1 to the power of 3 by 4 Okay, let's have 90 seconds for this as well. Please type done once you're done Yes, any progress by anybody? This could be a potential question for your school or board exams also. Okay, let me help you out with this uh Just tell me one thing what will happen if I take x to the power 4 common out from this particular bracket term What will be what will it come out as? x cube Absolutely, and there's an x square already waiting outside. So can I say it will become x to the power 5? Inside the bracket. I'll end up seeing this now just try to relate this term with this term And of course I can say that this term is this term will appear or similar to this term will appear when you differentiate this term Remember 1 by x to the power 4 the derivative will generate something by x to the power 5 Of course a constant and all will come that I can easily take care of don't worry about the constant But what i'm trying to say here is that the derivative of this expression would result into such expressions Which will contain x by x to the power 5 in the denominator So what i'm going to do next is i'm going to exactly do the same and take this as a t Okay Differentiate both sides with respect to x you'll end up getting minus 4 x to the power 5 dx as dt Right in other words dx by x to the power 5 is minus one fourth dt Right, so this whole term. Uh, let me just wipe out this uh bracket here Okay, so this May remove this Okay, so this whole term over here That will behave as negative one fourth dt Okay, and this is t to the power of three by four So this entire integral gets converted to minus one fourth dt by t to the power three by four Which is a subject matter of our rule of integration which you can directly do And this will give you the result as t to the power this plus one Divided by this so that gives you minus t to the power one fourth That's nothing but negative One plus one by x to the power four to the power one fourth That's done Okay, so you have to do such kind of manipulations in order to get your work done Okay, it'll only come with practice practice and nothing else Okay, let's take up this question Evaluate integral of product of tan x tan 2x tan 3x Very good Siddhartha Now the problem here is that we can't integrate product of three functions Okay, even if you knew how to do them we will not apply it at at this present case So what i'm going to do here is This is just a hint for people who are trying it this tan of 3x Try writing it as a compound angle here Okay Then see whether this is going to help you out or not So i'm sure you would have made use of that hint. So what to do next you just multiply this with this term So when you expand it, I'm going to write it in short form Uh, you can use this uh expression normally I use uh short forms like this Okay For sin I write s for cos I write c for tan I write a t But don't do that. Don't do these all kind of things in your school exam. Okay Now what I need here I need I need tx T 2x and t 3x correct. So from here I can write it down as T 3x minus t 2x minus tx Okay So what do you have done finally or eventually you have converted product as Some more difference of tan terms Okay, and this becomes quite easy for us to integrate Okay, so let me write it in a complete form So tan of 3x minus tan of 2x minus tan of x So instead of integrating the product you are choosing to integrate this because this is well within our reach This is well within our knowledge So integration of tan 3x will be ln mod seek 3x whole divided by 3 Minus ln seek 2x whole divided by Whole divided by 2 integration of tan x is ln of c kicks just put a c Okay, so this is going to be your final final answer for this integral Okay Another twin of this is where you have tan 2x and 3x and tan 5x Right in fact in all the problems where you have the sum of the 2 equal to the third you can apply the same approach Is that fine Any questions here? Yeah, this problem is another another important school level problem And I'm hundred percent sure this will come in one of your exams Whether it's a ut or a semester exam or board exams pre-board exams It will come How do you do this question try it out everybody? I'm going to give you a minute to think on it Done. Okay aditya. Yeah, that's a right right approach aditya proceed with that Okay, so I'll just tell you a hint how to do do this type of question. So here You multiply and divide with Sign of a minus b remember a and b are like constants. So you can play with these constants As you feel like to our advantage Okay, so first step is multiply with sign a minus b Okay, now this approach is also used Even if you had this integral 1 by cos x minus a cos x minus b Okay, in short with when you have two similar terms Okay, let's say sign sign or cos cos we normally multiply and divide with sign a minus b Okay Now this I will give you as a homework try this out as a homework Now what I'll do with the numerator sign a minus b. I'm going to write it as sign x minus b minus x minus a Now I don't want to write too much What I'll do is I'll just write it in the short form So 1 by a sign a minus b integral if you expand this it becomes sign x minus sign x minus b cos x minus a minus sign x minus a cos x minus b divided by sign x minus a sign x minus b Okay, this is what we get Now individually these two terms that we have over here divided by the denominator Okay, so I think sign x minus b will get cancelled and you'll get c c x minus a by s x minus a minus here you will get Cos x minus b divided by sign x minus b Okay So ultimately you end up seeing something like this sign a minus b Cos and by sign will be caught caught integration. We all know l enough sign x So it becomes sign a minus b ln mod sign x minus a minus ln mod sign x minus b We may combine the log as a single term by using our log properties So your final final answer should look like ln mod sign x minus a By sign x minus b Okay, just put a c at the end and that's it This is a sure short question for your school exam Any questions here any question any concern you want to copy something should I drag it up drag it down? Anybody wants to copy Okay, this is the main thing Let's do a twin of this Where you have one as a sign and other as a cos Let's see whether you're able to apply it Now here just a quick hint you have to multiply and divide with cos a minus b try it out similar approach Okay, so quickly doing this In fact, I'm also doing along with you Okay Sign x minus a cos x minus b So this can be expanded as cos x minus b cos x minus a plus sign x minus b sign x minus a By Sign x minus a cos x minus b.  Let me know if I missed out anything So this is eventually one by cos of a minus b integral of cot x minus a plus tan x minus b Right and we very well know how to integrate them. It becomes one by cos a minus b This will become ln mod sign x minus a This will become ln seek x minus a but you can write it as one by cos x minus b I'm doing this because I want to club these ellen terms So on clubbing them. This is the final result that you must all get By cos x minus b very good Dhan very good absolutely correct answer Is this fine Any questions here, please let me know let's take the next question Let's take this one two to the power two to the power two to the power x into two to the power two to the power x into two to the power x Any idea anybody Absolutely absolutely the first term itself s t will work wonders for you Okay, because when you differentiate it by using chain rule other terms will automatically surface out that's very good So if you differentiate this with respect to x we all have seen that two to the power anything derivative is two to the power same thing ln two And again that something is differentiated now. This is your something So the derivative of this will be again two to the power that into ln two Again this derivative of this term Which is again two to the power x ln two And this is going to be your final, you know derivative of the left hand side So you can see clearly that ln two whole cube And the very same expression that you are integrating is basically coming up. Yes or no Which means this entire term which is present in the question could be written as d t by ln two whole cube So ultimately When you are integrating this with respect to x You are actually integrating this with respect to t. Okay, so the answer will be t by ln two whole cube plus a c So answer will be two to the power two to the power two to the power x into two to the power two to the power x into two to the power x By ln two whole cube plus c Now this type of problem can come in various shape. They can put five to the power five to the power five to the power x Approach is the same for all those cases Okay Now next i'm going to tell you certain useful. Yes, sir in the previous question Isn't our t just two to the power two to the power two to the power x Oh, i'm so sorry. Did I write everything? Just Oh, i'm so sorry. So sorry. This is just two to the power. Yeah Uh, thanks shawmic by mistake. Uh In the flow of speaking I wrote everything. Yeah, sorry. So this is just t by this term. Correct. Okay Now we are going to talk about certain useful trigonometric substitutions certain useful trigonometric substitutions You may require this substitution even in your derivatives chapter. Okay, so please listen to this very very carefully Whenever you have an expression Like this present in your integrand Okay expression like this means it can be in the various forms of the same thing. For example, it could be root of this It could be let's say a square plus x square to the power of five by two In many of these cases you realize that such substitution prove very very handy Substituting x as a tan theta or substituting it as a cot theta Why it is very handy is because it converts this to a single trigonometric expression So the moment you put x as a tan theta it will convert it to a square c square theta Similarly for cot it will convert it to a square c square theta So it's easy to deal with those you know expressions in the present shape if you have Terms of the form a square minus x square or any term related to it Then x equal to a sin theta Or x is equal to a cos theta proves very helpful In a similar way Any expression related to x square minus a square then x is equal to ac theta or x is equal to Or x is equal to a cosec theta helps very very Much in this case. So it proves very very helpful if you have under root of A minus x by a plus x or you can say reciprocal of this In such cases x equal to a cos 2 theta is very very beneficial Now these substitutions are not only limited to this particular integration chapter. You can use it anywhere you feel it is required Okay, if you see x minus a by b minus x Or if you see x minus a times b minus x Then x equal to a cos square theta plus b sin square theta proves very very helpful Is that fine? We'll take a few questions where you would realize that such substitution is making our life very very easy Okay Let's go on To a question based on this We go back Sorry you want me to go back Just just let me put the question Then we can take back Okay, let's take this one People, please note this question down. We'll go back to the i'm going back to the previous slide Yeah, so i'm going back to the previous slide those who want to copy Let me know once you're done Yes, sir done One more thing I would like to highlight here that the substitution which I gave you was where the essential ones It is not an exhaustive list. Okay I mean Okay, after doing this follow, I'll give you more more such cases But it actually depends upon how you perceive the problem Now in this case Anybody has an idea how to do it any response anybody? So we can complete the square Yeah, then put x plus one is equal to eight and theta and then differentiate That's wonderful. That's wonderful. So as you can see in the denominator You can see x square plus two x plus two right So here we can say I can put x plus one as one tan theta or you can say tan theta correct That means dx would become secant square theta d theta So what I'm going to do is this one into dx I'm just going to replace it with secant square theta d theta And this will become one plus tan square theta the whole square One plus tan square theta the whole square Which actually gives you if I'm not mistaken cos square theta Because this also becomes secant square secant square square will become Secant to the power four and one of the secant square will get cancelled And this you can easily convert into double angles Like this And integrate it So that'll give you theta plus sine two theta by two Plus c Here a bit of challenge you may feel in converting it to x back So let's convert it to x back. So now From this expression, I can say theta is tan inverse of x plus one Okay And not only that We also know sine of two theta is two tan theta by one plus tan square theta correct So for this expression I Yeah, I can write it as one fourth into two tan theta By one plus tan square theta Right This also I'll replace it Tan inverse of x plus one. So your final final answer would look like this half Tan inverse of x plus one plus half x plus one by x square plus two x plus two Any questions here? Any questions here? Okay, I'm now going briefly to the previous slide where I would like to Oh, sorry. Sorry. Sorry. I already pressed the button. Can I go back to the previous question? Yes, sir. Okay, not question. Basically, I would like to complete the list by giving you a few more conversions but My description is see when I was learning this chapter We were never taught about these conversions But nowadays I have seen it's a world of readymade formula We were mostly left to our own to think about these substitutions So I'll complete the list with few more. So let's say when you see these kind of terms Okay, or let's say this this time of terms We preferably substitute x as a tan square theta Or x is equal to a cot square theta Okay If you see these terms x by Or you can say under root of x by a minus x Or let's say reciprocal of this Or let's say this Or let's say reciprocal of this We prefer substituting x as a sin square theta or a cos square theta Any questions here on the other hand? Sorry on the other hand if you have terms like this Under root x by x minus a Or under root of x minus a by x Or under root of x x minus a Or under root of You can say one by under root of x x minus a In such cases We prefer substituting x as a secant square theta or a cosecant square theta Okay, and again even if I have given you all these expansions Probably you have to exercise your own discretion while solving questions Okay, what substitution will fit the best And it's not necessary that you have to use substitution to solve questions Let's take few more questions On substitutions This we already okay, let's try this one. This is an ncrt question One by x to the power half plus x to the power one-third Yes, any idea what to do any any substitution that comes in your mind anybody? The one minute Yeah, sure Okay hint for you all Substitute x with such a term which has a power of lcm of two and three That means substitute it with t to the power six Okay This six is basically the lcm of two and three. Okay, correct Aditya So what is the benefit of doing so when you replace your dx it will be it will be replaced with Six t to the power five dt.  Okay So t cube plus one is factorizable as t plus one T square minus t plus one and if you just cancel out this term It will leave you with six times Integral of t square minus. Sorry. Sorry for missing out dt Let me do let me write dt over here Again, it's a matter of habit if you start forgetting it, you'll keep, you know That habit will stay with you even if you are attempting your school papers Okay, so it will be six times T cube by three minus t square by two plus t minus ln of mod t plus one Okay, put your t back t is uh sixth root of x So this will become one third x to the power half This will become half x to the power one third T will become x to the power one by six minus ln mod x to the power one by six plus one Okay, you may open the brackets and multiply with six throughout so that becomes your That becomes your final final answer. Is it fine? So again, this is a learning for you that whenever such kind of a terms are there We normally put x as t to the power the lcm of the denominators of these powers Let's try some few difficult ones also So yeah, let's try this one So in such questions, all you can do is take a leap of faith. Okay, because It doesn't give you observation may not be been a very helpful to you So all you can do is take a leap of faith So if I have to do anything with this problem, I would first try attempting This step. So let's say this is try to make it Can we try to make ckx plus cos x as In the form of tan x or because tan inverse of tan x will become x. No, sir Uh, okay as of now, yes, but not always Okay, will that help you in some case? So, how will you generate the other terms? That is what is the question Okay, probably you may try taking sin cos x as t or something Correct Yes or no, that's what you're trying to take But then there will be an x term whatever you are planning to get over here that will be unaddressed So as a leap of faith, let us try taking this If things work out, that's fine. If it doesn't we'll come back and change our approach Okay Now what are the derivative of tan inverse something we know it's one by that something's whole square Okay into derivative of that something that's going to be ckx tan x Okay minus sin x Okay Okay, like this Okay Now, let me simplify this even further Let me write this as One by cos x Plus cos x the whole square one more thing I'll do I'll pull out a sin x from here If I pull out a sin x I'll end up getting one by cos square Minus one any questions Now let me take the lcm over here and let me just take the lcm here also so cos Multiply throughout with cos square that would be great So cos square this will also be cos square cos square will enter as a cos so it will give you one plus one plus cos square x the whole square, okay Here also if you take the lcm it will give you sin x one minus cos square x by cos square x My god, it's becoming very very huge But thankfully we are almost there Uh, cos square cos square will go off. This is sin x into one minus cos square, which is again a sin square So numerator will clearly become a sine cube If I'm not mistaken In the denominator if you open the brackets you get cos four You get two cos square and cos square which is three cos square And a one Now this is what is there in your numerator. Yeah, this term is what we have got You can include this term also in your purview. Okay, so that whole term has actually become The whole term has actually become a dt So actually you're dealing with dt by t Because this term was your t This term was your t Okay, so it worked in this case ellen tan inverse ck express cos x Plus c will be your answer Okay, and there's equally chance that it may not work In those cases we have to use higher level skills to solve it that I'll keep on, you know updating you as we go on Any questions here So you should not do any mistake in this application of chain rule while integrating it. Is that fine any questions? Let's try this out integration of e by x to the power x Plus x by e to the power x whole multiplied with ellen x try this out integration of e by x raise to the power x plus x by e raise to the power x whole multiplied with ellen x Okay, so nothing substantial is striking to us right now other than taking a leap of faith of putting this term as t Okay, let's try this out if it works then good and fine else will change our process Try this out everybody Let me attempt it along with you. So first thing that you would do is take Lock to the base e on both the sides So let me write it like this first Okay, let me differentiate with respect to x on both the sides So this will give you First time differentiating x so one into this Then i'm taking ellen x by sorry Then x into derivative of this will be reciprocal into one by e Okay, so this into dx Is your one by t dt Okay Now thankfully here ellen x by e we can write it as ellen x minus ellen e And this itself is one and ellen e and one will get cancelled out Because ellen e is also one. So these two will get cancelled out right And you can see that ellen x dx Which is there in their expression over here can be expressed as dt by t So this whole thing gets converted to This is by the way one by t This is t and this you got wrote it as dt by t Which actually becomes a very easy problem to crack because it is just the integration of this Which is nothing but t minus one by t plus s c So t is nothing but x by e to the power x One by t is e by x to the power x plus s c this becomes your answer Right, so I'm exposing you to a slightly difficult versions of the same concept any question here Any questions? Okay, let's take few more as confidence boosting Problems. Uh, yeah, let's do this one This I think you can do it in 30 seconds not more than that x squared times tan inverse of x cube By one plus x to the power six Anybody quickly tell me what substitution will you make here? Quick Absolutely This is a no no brainer. Okay. So the derivative of this will be this into derivative of x cube. So chain rule, please do not miss out on that So x squared dx by one plus x to the power six is one third dt So this guy Is one third dt and this is t itself. So This is your integral. So let's not waste time and quickly write down the result So your answer is one sixth of tan inverse of x cube whole square Plus c any questions Now many people would prefer taking first x cube s t You may do that, but you will require one more substitution So there are so many ways to approach But if you're going to do other ways, you're going to just take longer to achieve the final answer One second one second one second should have been said one second early. Good enough Yes, sir. Only when too many boards are there the software becomes slightly heavy and lagging That's why this one That's why this one quick. It's a 30 second problem not even 30 seconds 15 second problem Yeah, there you go. Done. Pratam is done already very good So you can see that this term that is sitting on the numerator again It it it requires a bit of you know Good observation skills If you differentiate this You realize that the term that I have circled out on the numerator makes its appearance Which clearly indicates the fact that this choice is correct Okay, so ultimately you see Something like this in your question That's nothing but integral of secant square t. That's nothing but tan of t plus c So your answer will be tan of xc to the power x plus c done Right They may give it in different forms also They may keep it in sine square cos square whatever so you should be able to deal with it The whole idea is this guy's derivative is sitting on the numerator any question here Now i'm going towards another type of questions Another type of integrals which is going to be heavily asked The type is integrals involving Integrants of this nature sine to the power mx cos to the power nx Under this there are various cases Let me talk about the first case here First case is the case where m and n belong to natural number And let's say both are odd Both m and n are odd Okay Let me give an example then it'll be good to explain it Let's say i take sine to the power 3x Cos to the power 5x Okay Now you tell me Let's say this theory was not known to you. Let's say I have not even told you the theory actually. Let's say this question was given to you How would you approach it? What would you substitute as your variable t? Any guesses anybody will you substitute sine xst or cos xst Okay to be very frank Any one of them will work Okay, so you can solve this question by substituting sine xs equal to t Or cos x equal to t also both of the approaches is going to work Okay But we we follow our algorithm over here. We put that expression As t which has a higher odd power Okay So put that trigonometric ratio as t That has A higher odd power A higher odd power preferably Preferably So as I told you it can work even if you put, you know the lower one also st Okay, I'll show you both the cases then you take your call why I chose this approach So right now I decided to put cos xst Okay, so Minus sine x dx will be dt Right that means sine x dx is negative dt So what i'm going to do here is i'm going to break this up as sine square x and i'm going to Pull out a sine x and give it along with x along with dx So sine cube I broke it up as sine square into sine Sorry, and I clubbed it with dx so that I could substitute sine x dx with a negative dt So this guy becomes negative dt This guy becomes t to the power five undoubtedly What will happen to this guy? One minus one minus t square correct So I can say this entire integral converts to negative of t to the power five one minus t square Okay, basically you're integrating t to the power seven minus t to the power five That's again power rule of integration that you can easily answer t to the power eight by eight minus t to the power six by six plus a c and your t is cos of x So this becomes your answer Okay Now even if you substitute sine x st it will work, but slightly more it will be longer just slightly more For example, if I take sine x st It will become cos x dx is equal to dt So what i'm going to do is From cos x i'm going to carve out a cos cos to the power five i'm going to carve out a cos x Okay So this will be behaving as dt for you This will be behaving as t cube for you But this fellow will be behaving as one minus t square square Okay, so slightly more time will take because you will end up You know opening this bracket again multiplying it with the term and then again integrating it. So this will be slightly longer Okay, but the idea is you can substitute any one of them to be t and do your work Okay Any questions here? So case one is understood no doubt about it Okay Let's talk about case two Case two is a case where m and n are natural numbers One of them is odd and the other is even Our example of this would be let's say a sine to the power of let's say four x Into cos cube x Okay Now, let's say if you are given this problem, what would you like to substitute st? Sine or cos? Cos the idea is you should substitute sine x st over here the one with the even power Okay, so the theory is substitute The tignometric ratio with the even power As t Okay, so if you put sine x st Then cos x dx will become a dt So borrow or cos x from here. So this will be like Sine to the power four x cos square x into cos x So this will behave as dt This will behave as one minus t square and this will behave as t to the power four Okay, now people who are saying who will substitute cos x st Remember when you borrow the sine x from here to make for your dt, you will be left with an odd power Odd power means you'll have to write it as one minus cos square to the power of three by two that will again create a trouble for you Right. So hence this rule, please keep it in mind Hence this rule Okay, so with this the problem will become a super easy question. That means you're integrating This and I'm sure after this you will be able to do it Any question here When one is odd other is even case number three When m and n are both natural numbers or you can say you can be whole numbers also Yeah, by the way instead of natural numbers you can put whole numbers everywhere. Okay And m and n both are even Both are even. Let me ask you a question Let I want to integrate sine square x into cos to the power four x Okay, what will you do in this case Now both are even So even if you decide to substitute one st the other will be subjected to an odd power And then using your trigonometric ratios will be difficult So can we substitute sine square x on st? Sine square x st. Okay. What will you achieve from there? You'll get sine 2x as your derivative Okay, now here there are two approaches one is by using your trigonometric double angle Or you can say multiple angle identities Sir Yes Can we convert that sine square x into cos square x then apply then substitute That's that's the problem. No, even if you convert it to cos square What what how will you how would you deal with cos to the power four and cos to the power six and all? Are you getting my problem here? Problem here is I'm getting a higher power of sine or cos. How will I deal with that? Higher powers have to be converted to higher angles or you can say multiple angles Okay, that's the only way you can deal with it So that's why the first approach says you can use your trigonometric multiple angle identities And the second approach is basically the use of complex numbers Okay By the use of something which we call as the de Moivre's theorem Okay Now second approach No, no, no, no, we can't use power rule on trig functions. That's the whole point never extrapolate these formulas Didn't I tell you the other time if sine x integration is known sine q we cannot write, you know We cannot extrapolate it Those are only meant for linear functions. Okay Now out of this The first approach is quite understandable. We'll do that first and then we'll apply the second approach But the second approach is not to be used for any school exams Okay, this is just for your own, you know for competitive exams. You can definitely use it so Let me first convert Sine square x into cos to the power 4x as a multiple or a higher angles rather than higher powers So the first thing that comes in our mind is converting sine square as 1 minus cos 2x by 2 Okay converting cos square as 1 minus sorry 1 plus cos 2x by 2 the whole square Any questions here? Okay Now let's take the constants outside so 1 by 8 And not only that 1 minus cos 2x and one of the 1 plus cos 2x will make 1 minus cos square 2x Like this Any doubt so far?  Now here also you can multiply and divide with the 2 Right and use the formula of 2 cos a cos b Oh, sorry So that will be cos a minus b plus cos a plus b Okay Right, you may further club sub terms if they are same for example minus half cos 2x and half cos 2x You can make it plus half cos 2x and you have minus cos 4x And minus half cos 6x So with so much of effort you have now converted This term This term over here As multiple angles And not only that you have converted the product as some more difference correct So instead of integrating this You can choose to integrate this Okay, and that will give you the answer as integration of 1 is x Integration of this will be sin Sorry sin 2x Sin 2x by 2 Integration of this will be minus sin 4x by 4 Integration of this will be minus sin minus sin 6x by 12 Okay, if you want you can take a factor of 4 out you can make it 64 it becomes 4x This will become 2 sin 2x this will become sin 4x And this will become minus sin 6x by 3 Plus is this fine So it requires a bit of effort from your side and this is what you have to follow in school Now, I'm sure you must be looking forward for the other method So now I'm you going to use the second method Which is the complex number method to the use of demyverist theorem Now just a quick recap of what is demyverist theorem for those who have those who were not there with us last year Demyverist theorem says that If there is a complex number like this Okay, and you raise it to the power of n n being a integer It gives you cos n theta plus i sin n theta Okay, this is the demyverist Demyverist theorem Any question with respect to this Okay, now I'm going to use this to solve this question. You must be wondering how Okay, all of you please Consider the following things. Let's say I call a complex number z to be Cos x plus i sin x, okay So what will you 1 by z? Tell me by the use of demyverist theorem 1 by z means I'm raising it to the power of minus 1 Correct, okay If you add them you will end up getting cos x 2 cos x is z plus 1 by z Isn't it? So cos x becomes half of z plus 1 by z and if you subtract them 2 i sin x will become z minus 1 by z. So sin of x will become 1 by 2 i z minus 1 by z any doubt so far Now for the first time that you're doing we need to you know, I need to show you all the steps But let's say you are applying this to a question You may directly substitute your cos x to be half z plus 1 by z And sign to be 1 by 2 i z minus 1 by z. You don't have to undergo the same process. Okay Now you must be wondering why did I choose to write sin x and cos x like that? Let's see. What is the benefit? So we had this term right if I'm not mistaken So this term is 1 by 2 i z minus 1 by z square and cos x is 1 by 2 z plus 1 by z to the power of 4 Okay, uh, if you collect all the constants, so here you will get minus 4 here you get 16 So minus 4 and 16 only minus 1 by 64 Okay, and not only that z z minus 1 by z square and z plus 1 by z square will make z square minus 1 by z square square And the remaining term will be like this Let me know if you have not understood from this step to this step. Is it clear everybody? Yes, okay If you expand this You are going to get z 4 1 by z 4 minus 2 z square 1 by z square plus 2 Okay Now i'm literally going to multiply these two terms Okay, now while multiplying some precautions has to be taken care So first term with this will be z to the power 6 This into the second term will be z square. So write it separately write it far Okay So this into this is z square right? So I wrote it far This into this will be 2 z to the power 4 so 2 z to the power 4 write it over here I can make some gap in between Okay Next is 1 by z to the power 4 with z square that'll give you 1 by z square. So club it with this Okay, this with this will give you 1 by z to the power 6. So club it with the first term Basically, you're clubbing it with the same term present in the numerator and same term in the reciprocal Then this with this will give you 2 by z to the power 4 Okay Next is minus 2 into z square will give you minus 2 z square and this into this will give you minus 2 by z square So what I can do is I can just switch the sign over here with a minus Okay, and minus 2 into minus 2 will be minus 4 Any doubt from this step to this step. Please let me know if you want you can club this term together with the two outside Any questions Okay, if no questions at this stage if everybody's happy, let me take you back again to demo of this theorem Now what will be z to the power 6 if z is this You say cross 6x plus i sine 6x What will be 1 by z to the power 6 Okay What will be the sum of these two Do you agree if you add these two you will end up getting this in a similar way Basically, you can substitute this guy as 2 cos 6x This guy as this guy as 2 cos 4x This guy as 2 cos 2x minus 4 and you're done So you have converted it to a sum of multiple angles So no need to undergo all these, you know sine cos sine cos cos sine all those sums here you're dealing with pure single number z Is this understood because after this all you have to do is integrate any questions here So just like z to the power 6 plus 1 by z to the power 6 became 2 cos 6x In the same way z to the power 4 plus 1 by z to the power 4 will become 2 cos 4x In the same way z square plus 1 by z square will become 2 cos 2x And rest of the terms are just copy for copy forward. Are you convinced with this? Any questions here? Okay, so once you have got this the only thing left off for us is to integrate So your final answer would look like this. So let's integrate this result integrate this So your final answer will look like this Integration of this will be 2 sin 6x by 6 This will be 4 4 sin 4x by 4 This will be minus 2 sin 2x by 2 And minus 4x. Okay And if you want you can check your answer is going to match with whatever we have got earlier I'm just making some final touch ups this 2 and this 6. I'm making it as 3 Okay This 4 and 4 gone off Okay Just match this answer. In fact, you can send the minus sign inside also Okay, if you can send the minus sign inside it will become minus here Minus here Plus here and plus here Isn't this the same answer that you had got here also? Just check it out this answer And this answer check whether they are the same or not and i'm giving you few seconds to analyze this once again and Ask any question if you have any okay, so if this is fine Use demoverish theorem to integrate cost to the power 8x Are do use complex numbers method to integrate cost to the power 8x Now in the first question I told you so many things But here you can directly jump with the fact that it cost to the power 8x could be written as 1 by 2 z plus 1 by z to the power 8 What is 1 by 2 to the power 8? z plus 1 by z to the power 8 see how will I write it z plus 1 by z to the power 8. How will I write it see first term is z to the power 8 last term Will also be 1 by z to the power 8 so i'm clubbing them together Second term will be 8c1 z to the power 6 and the second last term will also be 8 By z to the power 6 they'll automatically get clubbed up. You don't have to worry too much about clubbing them up Next will be 8c2 What is 8c2 28 z to the power 4 1 by z to the power 4 any questions 8c2 is 8 into 7 by 2 right that's 28 Then 8c3 8 into 7 into 6 by 6 that is 56 z2 plus 1 by z2 You must be wondering where is z to the power 7 and 1 by z they will not appear they will not appear only the even ones will appear And finally is your 8c4 8c4 will be 8 into 7 into 6 into 5 by 24 So that will make it 70 Okay Now the moment you have got this You are done with the question actually because that given expression could be written as Now this term is what? 2 cos 8x Okay, so this term is 2 cos 8x this is 2 cos 6x but multiplied with a 2 it will become 16 cos 6x This will become 56 cos 4x This will become 112 cos 2x Plus 70 and you are now ready to integrate See how fast it was if you apply your you know multiple angle by using your trigonometric ratios and identities It will take you almost five times this time Okay So integrating cos to the power 8x is as good as integrating this So which will give you 1 by 256 2 sin 8x by 8 16 sin 6x by 6 56 sin 4x by 4 112 sin 2x by 2 Plus 70x Is that fine any questions here? You can finally you know cancel out the terms which you feel that they're you know For example, this can be made as 1 by 4. Okay, this can be made as 8 by 3. This can be made as 14 right if I'm not wrong And this is 56 like that Is it fine any questions here? Let's go on to case number four Case number four is a case where your m and n are rational numbers But however they add up to give you a negative even integer. Okay, let's say minus k Let's say minus k then how do you deal with this question? Let me give you an example. Let's say I want to integrate 1 by sin to the power 5 by 3x And cos to the power 1 by 3x As you can see here the present problem is sin to the power minus 5 by 3 x cos to the power minus 1 by 3x right So m is minus 5 by 3 n is minus 1 by 3 And if you add them you will end up getting If you add these terms you'll end up getting a negative even integer. Okay, so what is the approach here? What is the algorithm here? So when such a thing arises be normally number one divide the numerator And denominator by Cos to the power of kx Right Remember your k is negative of m plus n. Okay the same k here Okay And second thing is we put tan x st Is the approach clear? So what do we do first? We divide the numerator and denominator by cos to the power kx where k is negative of m plus n And then we substitute tan x st. Let's see what benefit it has for us So Let's divide here 1 by cos square x And in the denominator also Let us divide by cos square x Okay So this will give you secant square x on the top Denominator if you see this one third will get cancelled leaving you with a five third right And sine five third and cos five third will become tan five third Something like this And now it's a very obvious thing to proceed from here on with tan x st So if tan x is t secant square x dx will become a dt Okay, so it becomes a dt by t to the power five by three Right, which is nothing but an application of power rule of integration. So that is t to the power minus two by three Is that fine? So this is what is your answer Any questions here, please ask Sure. Yes Okay, no question anybody we'll do some more problem practice I don't want to start with uh another uh type of thing at this point point of time because we just have 15 minutes of the class left So we'll do some questions Okay, let's take this question. This is an n crt question. Let me know once you're done Done. Okay. Shristi very good All right, this is not a big problem uh that you would be taking so much time see cos square x is what? Two cos square x minus one Okay, now theta is a constant. Okay, so don't don't get scared of theta But still you can use two cos square theta minus one for me Let the denominator be as such When you open the brackets on top you end up seeing cos square x minus cos square theta by cos x minus cos theta Now it's obvious that your numerator can be factorized and one of the factors can get cancelled So that will just leave you with cos x plus cos theta So your answer will be just sine x Cos theta is a constant. So just write x into that constant That's it done. I think most of you would have started applying your uh transformation formula, right? But I think that should also work Oh, I'm so sorry. So sorry. So sorry. Yeah, I meant to integrate by x, but I wrote d theta by mistake. Sorry. Sorry Okay, if you had written d theta, then your cos x will become a constant Then your cos x will become a constant try this one dx by Under root of e to the power 5x fourth root of e to the power 2x plus e to the power minus 2x cube So you may read this problem like this also e to the power 5x by 2 e to the power 2x e to the power minus 2x to the power of 3 by 4 5x by 2 Yeah Any success? Okay, let's do one thing from these two terms. Let me take e to the power 2x out Now, of course, it is Subjected to three fourth powers will come out as e to the power 3x by 2 why did I write a 4 here by mistake? I think Okay So inside here, you'll end up getting e to the power minus 4x raise to the power of 3 by 4 And if you observe that these two will collectively become e to the power 4x So e to the power minus 4x will go And in the denominator, you'll be left with e to the power minus 4x to the power of 3 by 4 Then I am sure you would know what to do next from here Take this as T So minus 4 e to the power minus 4x dx will become a dt That means e to the power minus 4x dx will become minus one fourth of dt So this entire term okay will become Minus one fourth dt So minus one fourth Dt and this is t to the power minus 3 by 4 So your answer will be minus 1 by 4 t to the power Minus 3 by 4 plus 1 will be 1 by 4 divided by 1 by 4 That gives you my okay plus t that gives you minus T T is this term Any questions try this one out Integrate sine to the power minus 7 by 5x Cos to the power Minus 3 by 5x The next class is going to be a very very critical class because we are going to talk about various types of other standard integrals that exist Whose use is very extensive in problem solving We'll also talk about Integration by parts We'll also talk about use of partial fractions for integration We'll also talk about integrating irrational functions Okay, if something is left we'll take it in the third class. So the third class would be on you know, uh More on integration by parts and on reduction formula. So three classes is definitely required There may be a fourth one half required After integration, which chapter do you want me to work with? Which chapter you want everybody after integration? Sir, you said you do ellipse and hyperbola Uh We'll do one thing. We'll keep this for some other day because right now when the school reopens Okay. Yes, I I wanted to be at par with the school at least matrices determinants is what you're doing right now, correct? See ellipse hyperbola you have done enough You have you have you have just left with five ten percent towards the end Which is not very great. We can cover that up in any time we want Okay, hardly take two classes. Okay But what will happen if I don't start doing 12 syllabus? It will be like, you know, too much will come at the end And we have to start with the bridge, uh, uh crash course session in the month of November For that remaining part will end up get enough time. Don't worry. So after integration, what do you want me to do? Inverse trigonometry Inverse trigonometry Not matrices determinants or you think matrices determinants is easy. I can manage it It's not easy chapter. It's it's a very very critical chapter Every year in j main and j advance two questions comes from that particular chapter Okay, meanwhile, let's complete this as you can see here the sum of the powers is negative integer Okay, so what I'm going to do is I'm going to write it like this first And I'm going to divide both numerator and denominator by cos square x Okay, so that will give you a secant square x on top And denominator you can directly say will give you tan to the power seven by five x For sure Check it out Okay So put your tan x st So secant square x dx will become a dt So this will become a dt by t to the power seven by five So it's t to the power minus seven by five dt That's nothing but t to the power minus two by five Divided by minus two by five So that will give you minus five by two Tan x to the power of minus two by five plus c Okay So Next class when we meet we are going to talk about some more standard integrals So next class agenda is standard Uh, some standard integrals basically the six types In fact, I would suggest you to read them up and keep it So one by x square plus a square type You may all solve this by using substitution So let me give this as a homework to you all of you for homework You would prove the result of these integrals which I'm going to give you next So there are six integrals like this All you need to do is your substitutions remember I have told you some trigonometric substitutions apply them and do them So please derive the results of these six integrals which I'm giving to you right now This is one by eight tan inverse x by This is one by two a ln mod x minus a by x plus a This is one by two a ln a plus x by a minus x Which is also called as one by eight tan hyperbolic inverse x by This is one. This is ln x plus under root x square plus a square Okay, which is sine hyperbolic inverse x by a also This is ln x plus under root x square minus a square Which is cos hyperbolic inverse x by a and finally This is sine inverse x by a So please prove them by using Use trigonometric substitutions Is that fine? This is the homework for you for the next class Okay, because we will directly launch ourselves with this concept. So that will save save a lot of time for us Okay Okay, never mind what whatever we are supposed to do after this chapter We'll discuss it in the last class because but I feel that we should start with what school is doing because your uts And all will come and then you will be like, you know, we have not covered the syllabus Sir, you blew definite integration Definite integration after this Or What is going on in school siddhartha? What is going on in school? Sir, we stopped that