 Okay guys everybody is there so I can see Ananya, Arka, Preeti, Shahrak, Shiramya, Shiram, Shugos, Tejas, first of all good afternoon to all of you, alright, all of you can see my screen as well, yeah great fine. So continuing with the last class we had talked about the concept of properties of definite integrals, right, so let's begin with that, properties of definite integral, properties of definite integrals. So I think three properties which we had already done last class but for the benefit of all of you I'm going to repeat those properties once again. The first property was there's nothing in the name that means if you integrate a function which has been written in a variable x from a to b or you write the function in a variable t from a to b it doesn't matter, so there's nothing in the name of the variable used in the function this was the first property. Second property was if you swap the upper and the lower limits of the integration your result becomes negative of the previous one, okay this was what we had done, no one more I had done I remember one more I had done which was splitting the limits. So if you have a function f of x, okay from a to b and this function experiences a change in the definition at some point c or experiences discontinuity at some point c, okay, so two things here either it may experience a change of definition at c or it may experience a discontinuity at c, okay then what do we do is we break the limit of integration from a to c plus c to b, okay remember for a function to be integrable it must be continuous in the interval a to b it must be continuous in the you can say close interval a to b is that fine. Now why I'm saying this is because there is a very important type of concept which is going to come later on but I'm going to give you a question related to it right now itself because I see a lot of mistakes being made by students, okay so I'll give you a question let's say this this concept is important because of this question let's say I give you a scenario where you need to integrate let's say dx by cos square x plus 3 sin square x, okay and the limit of integration is 0 to pi limit of integration is from 0 to pi, okay now there is a student x and not the variable x there's a student by the name of x what he does he first divides the numerator and denominator by cos square x, okay fine no harm in doing that so if you divide numerator is 1 over here so if you divide 1 by cos square you'll get a secant square, alright and down in the denominator you'll get a 1 over here and a 3 tan square x, correct limit of integration is not going to change because you have not made any substitution you have just done a simple division of the numerator and denominator, okay now what this guy does he makes a substitution of tan x st okay I think this is the normal process that we follow while integrating such kind of problems remember we had done a type in the class dx by a cos square x plus b sin square x plus c it belongs to that type only, correct now this person he does a substitution of tan x st, okay that means secant square x dx is equal to dt, correct so he converts this integral as dt by 1 plus 3t square, okay and to his surprise he notices that the limit of integration will be 0 lower limit remember when you put x as 0 t will also be 0 because when x is 0 tan of 0 is also 0 and when you put x as pi t will be tan of pi tan of pi is also 0, correct now to his surprise he sees that the limit of integration is from 0 to 0 okay and he writes this answer as 0, correct and of course he gets a 0, 0 as marks because this is wrong, right now what mistake has happened what mistake has happened here you may type or let me just unmute you guys so that you can speak okay anybody who wants to speak can speak what mistake has this person done over here anyone who wants to answer this there are 17 of you let me show this on a graph and I'll show you that from 0 to pi the area under this graph is not 0, okay but this guy gets the answer as 0, okay let's see it from a graph let's see it from the graph so let me just open up let me open GeoGebra for you hope you can see the GeoGebra screen so let me type a function y is equal to 1 over cos square x you can say cos x square okay plus 3 sin square x 3 sin square x okay as you can see this is the function and you are integrating from 0 to pi so 0 to pi okay so if you watch it out on the screen I'm going to shade the desired area the desired area is going to be sorry yeah the desired area is going to be this part this is the desired area correct right and you know that this result is not 0 right and you know this result is not 0 correct so why does this person get a result of 0 no problem away it's okay yeah guys any idea why this person is getting his answer as 0 when the answer is not 0 please type in your response so you can speak out okay if you don't have an idea let me share this with you he gets an answer of 0 incorrectly because this function very substituted tan x st tan x is actually facing discontinuity okay we all know that tan x very important information tan x is discontinuous at x equal to pi by 2 is discontinuous at x equal to pi by 2 correct and this is the reason why this is the reason why his limit of integration go from 0 to 0 right so if you had to ideally solve this problem he must break the limit of integration at pi by 2 okay so an ideal way to solve this is he should first break this from 0 to pi by 2 x square dx by 1 plus 3 tan square x plus pi by 2 to pi okay secant square x dx by 1 plus 3 tan square x now the reason he got a wrong answer is because he did not break the limit of integration at pi by 2 he did not break the limit of integration at pi by 2 and that's why this property becomes super important another reason why this property is important is because there are some functions which change definition okay there are some special functions with change definition special functions you can call it or you can call it as piecewise defined functions okay piecewise defined functions they also change definition at certain points in the interval and hence it becomes very important to break the limit of integration at those very points okay I would like to start the day with a problem solving on this property let's pick out a problem on this okay let's take this question let's take this question hope you can all see the question on your screen so we have to integrate under root of cos x minus cos cube x from minus pi by 2 to pi by 2 any idea anyone sure take your time so we have to integrate under root of cos x minus cos cube x from minus pi by 2 to pi by 2 just free field just feel free to type in your response in the chat box so let me help you out in this first of all we we have to analyze this term over here under root of cos x minus cos cube x okay under root of cos x minus cos cube x okay so when we take a cos x common it becomes 1 minus sine square x correct which is nothing but under root of cos x into under root of 1 minus sine square x okay under root of 1 minus sine square x is basically cos square x so under root of cos x into under root of cos square x now many people by mistake write the response for this as they write the response for this as cos x which is wrong okay under root of some something square is actually mod of that value okay so write the expression for this will be under root of cos x into mod of sine x okay this is the right way to express this so ultimately you are integrating minus pi by 2 to pi by 2 under root of cos x into mod of sine x sorry into mod of oh it is oh I am so sorry this is this is cos square cos square this is sine square so this is cos square cos square sine square and the result for this should be mod of sine x mod of sine now mod of sine x this nothing happens to this term but this term will change its definition right you know that sine x is positive from 0 to pi by 2 and negative from minus pi by 2 to 0 so sine x changes its definition as sine x from x line between 0 to pi by 2 correct and minus of sine x from minus pi by 2 to 0 is that fine so what I will do is I will change the definition of this function at 0 so minus pi by 2 to 0 I would end up integrating under root of cos x into negative sine x okay and from 0 to pi by 2 it would be under root of cos x into sine x now does it become a simple thing to integrate of course it does okay so let's evaluate this here if you take cos x as t in the first integral let me name this integral as i1 here so let's say this is your i1 integral and this is your i2 integral okay so in i1 in i1 you can write cos x as t in i1 you can say cos x as t so minus sine x dx becomes a dt minus sine x becomes sine x dx becomes a dt what about the limits of integration limit of integration here will be from 0 to 1 okay under root of t dt okay a similar substitution can be used here as well so this would be under root of t minus dt in this case so minus sine will come up okay from 1 to 0 from 1 to 0 yes I don't now we can use the swapping of the upper and the lower limits so if you swap the upper and the lower limits of the second integral it becomes plus 0 to 1 under root of t dt so what have I done I've just swapped the position of 0 and 1 and absorbed this minus sine correct and if you do that you end up getting 2 times 0 to 1 under root of t dt okay and that is nothing but 2 into 3 to t to the power 3 by 2 divided by 3 by 2 and limit of integration is from 0 to 1 which gives you the answer as 4 by 3 1 to the power 3 by 2 minus 0 to the power 3 by 2 which is 4 by 3 is your answer is that fine so a couple of things to be noted under root of a square is mod a it is not a okay and when mod comes it's a special function and when special function comes you have to take care of the breaking of the limits of integration depending upon where that special function is changing its definition cool good no doubt about this okay let's move on to the next one clear great let's move on to the next question let's have this one I'm sure you can read the question question is evaluate integral of 0 to infinity of gif gif this is the gif function of 2 e to the power negative x dx now I would like all of you to give it a try for one minute so let me see who all are attending ananya and rio arka away groove gaurav hey what's gaurav doing here okay ishika krishna laxia pranjwal peeti shashank shaurak shiramya shiram again once again good afternoon to you guys hello prajwal hope you are a great sports day today no problem no problem yes I can see arka I can see your messages I'll tell you what all messages you send so far it should be cause you send I got it okay I can see all of your messages don't worry ln2 is what laxia says okay laxia I'll not say right or wrong to it because I want everybody to try not to get influenced by the answer the best way to solve such kind of problems is plot them plot it it's not difficult graph it's a very simple graph 2 e to the power negative x just simple graphical transformation is required I have disabled chat actually no attendees can chat okay guys enough time was given let me tell you laxia your answer is absolutely correct very good so just a quick plotting of the graph of 2 e to the power minus x because from the graph we'll get a lot of information so 2 e to the power minus x graph would be the one which is dying like this okay remember as your x increases the graph is going to have lesser and lesser value and for negative values of x it'll have higher and higher values remember it will cut the x-axis at 0 comma 2 okay now just as a matter of information for all of you if you want to plot the graph of gif of any function okay number one step is first plot the graph of the normal function so first plot the graph of the normal function which we have already done over here okay second step is start drawing parallel lines okay parallel lines when I say parallel lines parallel to the x-axis at unit distance apart okay at unit distance apart okay so let me just draw them so let's say I start drawing the lines unit distance first line is like this okay second line will exactly cross through this okay third line will cross like this now the next step that we are going to do here is that the part of the graph which is sandwiched between the two lines will fall on the floor below it for example if you see this part okay let me just change the color over here if you see this part of the graph okay as you can see I'm shading it with blue okay this part of the graph will fall on the floor like this it will fall on this floor like this okay see the pink line okay hope you can see your pink line okay now this pink line of course will have a hole over here okay okay and that's why these functions are called floor functions because the part of the graph which is between two floor two c you can say two levels like this will fall on the floor okay this part of the graph again I'm using blue color no sorry let me use green color this part of the graph let me tell you will go on and on forever this part of the graph will again fall on the floor like this will again fall on the floor like this I'm shading it with gray hope you can all see it and there would be a hole over here are you getting it okay this part of the graph will again fall like this let me use some other color let me use orange in this case so this graph will be like this are you getting it so just start forming steps like this okay now my question was from zero to infinity so my playing area is from zero to infinity so from zero to infinity I want to find out what is the area underneath this graph correct so as you can see there's only one area that I can see so the area under this step okay rest this entire step has collapsed on the ground there's no area which has been you know covered by this step okay so the only answer is this answer okay by the way what is this point itself this is that point where two e to the power minus x actually becomes a one okay that means e to the power minus x will become a half that means e to the power x will become a two that means x will become an ln two okay so this graph this part is actually ln two okay x is ln two here okay and this thickness is one correct so obviously what is the area of this rectangle anybody can say okay length into breadth okay so area of this rectangle is one into ln two correct that's nothing but ln two okay so the answer for this question becomes ln two okay now how will you actually solve it in a school exam if it comes it's very easy in school exam will say from zero to ln two the function will behave as one okay and from ln two to infinity your function will actually be zero and therefore this result will be zero anyhow and this will begin will give you x from ln zero to ln two which is nothing but ln two no problem Adavath hope you are able to follow from here okay I'll show you on geogebra how this graph actually when you floor a function how does it make a difference to the graph okay let's check it out from the geogebra so let's let's go here and plot y is equal to two e to the power of negative x okay as you can see this is the graph okay hope all of you can see this graph fine okay moment I have flowed this function y is equal to flow of g okay see what is happening you can all see that this part of the graph from here on has fallen on the floor this part of the graph has fallen over here this part of the graph has fallen over here so blue lines are the part where they have fallen down getting the point okay and since we're integrating from zero to infinity only area that we are concerned is with this area zero to ln two from ln two onwards till infinity the area actually become zero because the line is collapsed on the x axis itself is that fine great so let me check if I have some more problem on this okay let's take this question okay hope you can all see the question clearly yeah so there is a function which itself is an integral okay it's an integral of mod 2t minus 3 from zero to x the question is asking us to discuss the continuity and differentiability of this function at 3 by 2 so at this point 3 by 2 we have to discuss the continuity and differentiability of this function okay can you all give it a try for some time no screen is proper hope you can all see the screen and hear me properly but well is it fine now for you so this function has been presented itself as an integral okay zero to x zero to x and they are they are interested in knowing its continuity and differentiability at x equal to 3 by 2 please feel free to type in your response or speak out the answer okay okay so pp is commenting that the function is continuous at 3 by 2 okay what about differentiability pretty not sure okay can you give it a try please okay sure is saying continuous but not differentiable okay okay I just wait for one more response to come in and then I'll start solving this others please participate Vinda Dhruv Kavya Krishna Laxia Nikhil Shashank Sitchinta would appreciate your response guys I think it's continuous okay Arka shallow continuous but not differentiable fine so thank you for your responses so let us now see what is the solution actually so let me first redefine this function right whenever there is a mod involved always it is advisable that you redefine the function before you work on it okay never work with mod gif a fractional part functions into operation always give them a new definition without these special functions in place and then start working on it so if I have to start working on this if my x is less than 3 by 2 because the contention point is 3 by 2 okay so I can have two intervals one is x less than 3 by 2 other is x greater than equal to 3 by 2 correct so when your x is less than 3 by 2 of course you are integrating from 0 to x and you know x is less than 3 by 2 remember this t cannot go beyond 0 to x correct so this t variable that you see over here this t variable this 0 to x defines from where to where it can actually go right so when your t is between 0 to 3 by 2 you know that 2t minus 3 will be a negative quantity okay especially when your t lies between 0 to 3 by 2 am I correct yes so mod of 2t minus 3 would actually return a negative of 2t minus 3 to us isn't it right in other words you are actually integrating 3 minus 2t dt from 0 to x if you want you can evaluate it but I will leave it for later time okay let me first write it in a in a form where I don't have any modulus now when x is greater than 3 by 2 it can be any value greater than 3 by 2 I don't know which value is it okay so what I'll do is I'll break the limit of integration first at 3 by 2 because I'm exceeding 3 by 2 already but 3 by 2 happens to be a critical point okay remember 3 by 2 x equal to 3 by 2 happens to be a critical point for us so I have to break the integral from 0 to 3 by 2 okay so what are you integrating from 0 to 3 by 2 of course the same thing that we had discussed a little while ago this thing okay and from 3 by 2 to x you don't know where you are going till you just know it is greater than 3 by 2 so let's say it goes to any value greater than 3 by 2 so keep it as x in that interval it will behave as 2t minus 3 in that interval it will behave as 2t minus 3 so this is how you redefine the function first without any mod into picture now what I'll do I'll give it a proper shape okay we all know 3 integration of 3 minus 2t is going to be integral of 0 0 to x 3 minus 2t dt will be nothing but 3t minus t square okay your limits of integration are from 0 to x so it becomes 3x minus x square okay I hope nobody has any doubt about this correct let's talk about the second one so this is covered let's talk about the second one so 0 to 3 by 2 3 minus 2t integral will be nothing but 3t minus t square and your upper limit is 3 by 2 lower limit is 0 so it becomes 9 by 2 minus 9 by 4 minus 0 which is nothing but 9 by 4 right this is fine guys hope I have not missed out on anything if I have please highlight it okay and I'm sorry yeah and 3 by 2 to x I have to integrate 2t minus 3 okay that's nothing but t square minus 3t from 3 by 2 to x so this will be x square minus 3x minus 9 by 4 minus 9 by 2 am I correct so this becomes x square minus 3x plus 9 by 4 okay remember we have to add these two okay because our this our integral is the combined of these two integrals okay so I'm combining it here combining these two results I'm getting the function value as x square minus 3x plus 9 by 2 fine now let us redefine my function over here again now that very complicated looking function I have redefined it in a simple and simple words which is actually easy for you to work on so when x is less than 3 by 2 the function is behaving as 3x minus x square and the function is behaving as x square minus 3x plus 9 by 2 if your x is greater than equal to 3 by 2 correct now will you like to take in a minute and now change your answer if you want to given that these are the two functions is it continuous at 3 by 2 first of all everybody please check continuous at x equal to 3 by 2 let's do a continuity test test for continuity at 3 by 2 so left hand limit that is limit extending to 3 by 2 minus for this function would be giving me 9 by 2 minus 9 by 4 which is 9 by 4 and the right hand limit as x tends to 3 by 2 would be x square minus 3x plus 3 by 2 which is 9 by 4 minus 9 by 2 plus 9 by 2 which is again 9 by 4 okay so yes it is continuous okay so continuous for sure okay continuous at x equal to 3 by 2 no doubt about that now let us check its differentiability so test for differentiability test for differentiability at x equal to 3 by 2 for J and all exams as I already told you you can actually directly differentiate and put the value of 3 by 2 so f dash or you can say L f dash 3 by 2 would be nothing but 3 minus 2x and you put the value of x s 3 by 2 it's going to give you 0 I guess it gives you 0 right and r f dash 3 by 2 would be 2x minus 3 and again when you put xs 3 by 2 back to 0 yes so left hand limit right hand limit both match they are same which means the function is differentiable also at x equal to 3 by 2 okay both continuous and differentiable yes that is the answer for this question continuous as well as differentiable at 3 by 2 not a very difficult question with respect to the continuity and differentiability part but yes there would be some mistakes done by you in redefining the function so please be careful any question any concerns please highlight it guys if it is clear please type C L R on your screen please type C L R if this concept is clear okay great this is also to test how many of you are still in the column okay great so we'll now move on to the next property I think property number four property number four somebody please keep a track of the numbers because I tend to forget the numbers which I am at okay property number four is the one of the most important of all the properties okay and this is called the king's property this is called the king's property I've already done this with the napplers okay what is this property this property says that if a function f of x which is continuous from a to b okay so f of x is a continuous function from a to b then the integral of f of x from a to b could be written as integral of f of a plus b minus x dx okay very important property it says that integral of f of x from a to b is as good as integral of f of a plus b minus x from a to b now I am going to prove this okay I am going to prove this non geometrically as well as geometrically let me prove it first non geometrically okay let me start with the left hand side left hand side is a to b f of x dx correct now what I'll do is I'll put a substitution of x as a plus b minus t what I did I made a substitution of x as a plus b minus t okay that means dx would be negative of dt dx would be negative of dt undoubtedly a and b are some constants here correct now let me change this integral so if you change this integral x can be replaced with a plus b minus t dx can be replaced with a minus of dt and what about the limits of integration remember guys one very important thing whenever you are making a substitution to solve any integral and it is a case of definite integral make sure you change your limits of integration as well okay if you don't change your limit of integration you have to go back and substitute it in terms of x which of course you would not like to correct so in indefinite integral that was compulsory but in definite integral you don't have to go back to your old variable once you have changed to a new variable treat this as a starting point of a new question and forget the old question all together so for that it is very imperative very important that you change the limit of integration as well okay and it will happen a lot of time that you will forget to change the limit of integration okay so be watchful about it right so how do i change the limit of integration very simple put your x as a since your lower limit of x is a put your x as a over here so if you put x as a then see what is your t coming out from here your t will come out to be b right so lower limit here will be b correct then put your upper limit b into in place of x so if you put up b here let's see what we get t as so b b gets cancelled t here will become a so upper limit is a in this case are you getting this point this is how you change the limit of integration now i never like a minus sign a stray minus sign like this because sometimes i may forget to you know put this minus sign so i always observe it if i have to observe it i have to swap the upper and the lower limits correct now treat this as a new problem okay forget about whatever has happened before okay so treat this as a new problem altogether now the first property that we had discussed there is nothing in the name i put back this t as x and there is no change in my answer right i know this may surprise you because earlier i started by saying x is a plus b minus c and i somehow put t as x back again i know this is going to surprise you but that is how this works there is nothing in the name i just changed the name from t to x correct so i started with this expression and ended up this expression that means king's property is proven okay hence proved which is nothing but your right hand side okay so treat this as a new question itself any doubt regarding the proof person okay anyone shiramya shiriram sucinta shugosh tejas guys please speak up please feel free to stop me in case no doubts clear oh yeah aved that's that's what i was telling you i am not relating it to a previous x okay i'm just changing the name from t to x that was not the original x so from here think as if this is a new question altogether this is a new question right in a new question can i not write t as x here i can i'm never referring to the previous x i'm just changing the name no no no fragile we'll not need the original x that requirement is there if you have not changed the limit of integration clear okay all right now i'll give you a geometrical proof for this because that's what j e and other exams are looking out for whether you understand the property geometrical okay so when you say integral of let me write the name over here geometrical proof geometrical proof so when you say integral of f of x from a to b okay what are you looking at you're looking at the area between the f of x and the x axis from x equal to a till x equal to b okay so let me sketch a graph like this so let's say this is a graph okay and this is a graph between x equal to a till x equal to b okay this is my function function y equal to f of x and when you say i'm looking out for integral from a to b you are actually looking out for this area okay let me call this as a for the timing okay now the property says this area is should be same as the area between f of a plus b minus x and the x axis from a to b correct it claims that this should also give you a let us check does it actually give us a okay so now i'm going to first plot the graph of let us first step number one let me plot the graph of let's plot y is equal to f of a plus b minus x okay how do we plot this for this we go by sequential steps first we plot the graph of y is equal to f of minus x okay how do i do that how do i plot the graph of y is equal to f of minus x please type in your screen i'm drawing a miniature graph over here okay just i would need your inputs to help me to plot the graph of y is equal to f of minus x flip around the y axis absolutely correct so if you flip around the y axis correct me guys won't it look like this okay that means this end will now come at minus a and this end will now come at minus b yeah reflect about y axis absolutely correct next step what i'm going to do is i'm going to change my x with x minus a plus b okay what i'm doing now i'm at this stage i'm replacing my x with x minus a plus b what effect does it have on the graph you will say it will shift the graph a plus b units to the right yes or no so if this gentleman gets shifted a plus b units to the right how would it look like won't it look like this won't it look like this correct where this end will be at a and this end will be at b yes or no correct and you are actually trying to find the area underneath this graph you realize that it's the very same curve with the extremities flipped yes or no this point a and b let me call it has now come like this so this is your b and this is your a so you have just flipped the position of a and b on the graph so this b has occupied the coordinates b has come to the position of a a has come to the position of b correct when you flip such a graph the area under the graph will not undergo any change do you agree with me on this or not flipping off the graph doesn't change the area under it okay imagine as if you have a piece of handkerchief in your hand and you just you know swap your position of your right and the left hands okay you'll end up realizing that the area will not change and hence this will also have the same area correct that is your function will have the same area as what it had before that means the property here absolutely holds true this will also give you the same area as this one will give you okay so both area would be the same and hence kings property okay you'll say well good what kind of problems can I expect on this so let's solve few problems on this in fact I'm going to solve a lot of problems on this because I feel this property is the heart and soul of definite integrals so let me begin with a question okay let's start with this one this is an ncrt question okay so please feel comfortable solving it yes yes of course a very give the same area but that's not what the property says okay so I was actually targeting the property so you are trying to say a to b f of x okay will give you same result as f of minus x from minus b to minus a okay you're actually trying to state this yeah which is actually true if you see the properties if you just if you just you know sub x with minus x let's say if I take x as minus t or something dx will become minus dt correct so this will become f of minus t and dx will become a minus dt and your limits of integration is minus a to minus b so if you want to observe this external negative sign you can flip the provision of minus a and minus b and and this is what we obtain and there's nothing in the name and there's nothing in the name means you can flip back your t as x again okay that's what you're trying to say uh away does this answer you concern okay okay so I'm only getting responses p p uh at that they've all started giving the answers I'll just wait for one more response and then we'll discuss it it's not at all a difficult problem not at all a difficult problem very bad okay so I'm getting a lot of responses as five by two five sorry five by 12 okay let's check it out so as you can see this this question is very difficult to integrate if you don't incorporate things property guys let me tell you this straight away that if you don't use the limits if you don't use the properties of definite integral it nearly becomes impossible to integrate things okay so what I can do over here is that this is your function this is your function f of x okay this is your function f of x uh please exclude d of x out of it so let me redraw it let me redraw it so this is your function okay and this is your a and b so this is your a and this is your b this is your a and this is your b okay now according to this property first let me name this integral as i so this is your function f of x which is uh sine x by under root of sine x plus under root of cos x okay and if I use this property using king's property using king's property okay I can also be written as pi by 6 to pi by 3 under root of sine okay pi by sorry pi by 3 plus pi by 6 if you do that is b plus a this will give you pi by 2 okay so what I'm doing wherever there is an x okay don't test this don't don't put your x as a plus b minus x here okay so please exclude this okay only in the function you have to do it so wherever there is an x replace it with pi by 2 minus x by under root of sine pi by 2 minus x plus under root of cos pi by 2 minus x so this gives me pi by 6 to pi by 3 under root of cos x by under root of cos x plus under root of sine x is that fine okay now what I can do is I will add these two i's these two i's I will add them okay so adding those two i's i and i will give you a 2 i okay since the limit of integration is same in both of them I can actually directly add the functions so I will add under root of sine x by under root of sine x plus under root of cos x plus under root of cos x by under root of cos x plus under root of sine x hold together I can add it okay so 2 i is nothing but pi by 6 to pi by 3 under root of sine x plus under root of cos x by under root of sine x plus under root of cos x okay which will cancel out eventually and give you a 1 because both numerator and denominator are same so ultimately you are integrating 1 from pi by 6 to pi by 3 okay which is nothing but x put the limits of integration and this will give you pi by 3 minus pi by 6 which is pi by 6 so your 2 i is pi by 6 meaning i is going to be pi by 12 so absolutely correct those who answered as pi by 12 okay I think the first one to answer this was at that very good so the use of this property makes things so easy and convenient else by brute force solving it by using directly the concept of indefinite integral would have made this problem super complicated is that fine is that fine guys is this clear okay so there is a very special case of king's property a special case I would say 0 to a f of x dx is 0 to a f of a minus x dx so basically what has happened in the extended property where we had a to b okay this was the extended form which was the broader version which I discussed with you a little while ago what I did is I made this a as 0 and I made this b as a okay so this is how this property has come up okay so this is a special case of king's property this is the extended case you can call this as an extended case or extended version fine so be aware of this also okay let's take few more questions on this so basically what we are trying to do is by the use of king's property we are converting the integral into a form which is integrable by us okay so let's look into this question please try this out okay so I've already started getting response Shiram Lakshya and Shorak think it's 0 Kavya thinks it's 5 by 2 okay Shorak wants to change his answer 5 by 2 okay so guys enough response I've got for this let me begin with the discussion see again let's let's start using the king's property over here so this is your function f of x this is your function f of x from 0 to a so 0 to a f of x is your this function log of 3 plus sorry 4 plus 3 sin x over 4 plus 3 cos x okay now if I use king's property let me call this as i if I use king's property I have to replace my x with pi by 2 minus x correct so 4 plus 3 sin pi by 2 minus x by 4 plus 3 cos pi by 2 minus x is that fine and simple use of complementary ratio we can get this as 4 plus 3 cos x by 4 plus 3 sin x okay is that fine now if you see carefully it is actually the same term written in a reciprocal fashion right so if you reverse the position of the numerator and the denominator within the log you need to incorporate a negative sign outside so it becomes log of this is that fine yes or no isn't this negative i isn't this negative i so what conclusion have you reached you started with an i and the same i converted itself to negative i suggesting that 2 i is equal to 0 which implies i is equal to 0 okay so some of you who give the answer as 0 and then change it back to pi by 2 you earn yourself a negative marks okay you are correct in the first attempt okay no worries we'll take many more problems so the answer for this is 0 so guys let's try to analyze what has happened here through the use of king's property of course brute force will not work okay it may work but it may take a lot of time so even don't even think about using it so what we did was through the use of king's property we manipulated the integral and ultimately it gave the negative of itself okay now this is one way of solving it in many such situations you would realize that it will convert it to a form which you know how to integrate by using indefinite property also okay i will take up an example based on the same next okay i hope this is clear no questions about this great so let me pull out another one okay let's take this one let's take this one so hope you can read the question the question says evaluate x sin x y to the power x plus one integral from negative pi to pi please note here a and b both are negatives of each other so a plus b would actually become a zero okay i'm giving you a minute for this go to meeting pi people are giving me answer as pi guys let me tell you you are correct well done very good so again let's call this as i i from minus pi to pi x sin x by e to the power x plus one if you use king's property it becomes i remember a plus b minus x remember you are replacing f of x with f of a plus b minus x here a and b are negatives of each other so you are actually replacing the function with negative f of negative x okay so wherever there is an x you are putting a negative x so negative x sin of negative x by e to the power negative x plus one the x okay i remember minus x and sin minus x both would be same so it's minus pi to pi x sin x by e to the power negative x plus one or is that fine now what i'm going to do is i'm going to multiply and divide numerator and denominator with e to the power x i'm going to multiply numerator and denominator with e to the power x so this brings the integral uh integrand to uh x sin x e to the power x by one plus e to the power x correct now let's add these two guys let's add the i in yellow and the i in white okay so when we do that what do we get we get a two i since the limit of integration is same from minus pi to pi this is you can take x sin x by one plus e to the power x and you can take one plus e to the power x common also correct yes or no correct uh prejuel your answer is correct okay now you can cancel this off now you can cancel this off correct and now you can see that you are trying to integrate x sin x from minus pi to pi okay now to integrate x sin x we know that i have to use integration by parts but let me tell you that is not required you can also integrate it without it also okay later on tell you how but as of now you can integrate it by parts so applying integration by parts for integrating x sin x let me use uh tic-tac-toe rule let me use tic-tac-toe rule over here also known as uh the di rule okay di method remember we used to make two columns d and i under d you write the function which you want to differentiate under i you write down the function which you want to integrate now differentiate x you get a one integrate sin x which gives you minus cos x now differentiate one you get a zero integrate minus cos x which will give you minus sin x okay put these signs plus minus plus and start multiplying in this fashion okay so your answer would be plus x minus cos x which is minus x cos x and minus one minus sin x is plus sin x is that fine this was the di rule that we had learned okay so now you can put the limits of integration now you put the limits of integration x cos x plus sin x from minus pi to pi okay if you put a pi you get yes of course we can we can also split the limit but there is a rule for that okay i'll i'll i'll talk about that also okay of course you can do minus pi to zero zero to pi absolutely no doubt about it okay so this will give you minus pi into minus one plus zero minus uh plus pi okay sure i'll i'll talk about it at that okay then you have a minus pi into again a minus one plus zero correct so thereby giving you uh let me check if everything is fine or this is plus pi yeah sorry yeah so this will give me pi minus of minus pi which is two pi correct and two i is equal to two pi let me write this result over here this is going to be uh two pi so two i is equal to two pi means i is equal to two i is equal to two pi means i is equal to pi sure otherwise i'll tell you the di method di method is basically those who have not attended the class uh on integration where i talked about the di method di method is uh important method by which you can actually do the integration by parts pretty quickly especially involving that u function which disappears on multiple differentiation okay how does this method work uh let me explain you with an example let's say i want to integrate x cube uh let's say e to the power x okay i want to integrate this and if you want to integrate this you have to use integration by parts where you take this as u and you take this as v now the problem with the normal method of integration by parts is you have to apply integration by parts multiple times isn't it because once you apply one single use of integration by parts will actually give you x cube e to the power x minus 3x square e to the power x okay again you have to apply integration by parts on this term okay so this makes the process slightly you know cumbersome and there is a lot of mistakes of making sign uh mistakes in putting the signs okay so this method says whatever function you are differentiating put it under d whatever function you are integrating put it under i okay start differentiating it so x cube derivative will be 3x square again 3x square derivative will be 6x again 6x derivative will be 6 again 6 derivative will be 0 understood adhavath and start integrating this e to the power x integral is e to the power x again e to the power x e to the power x e to the power x okay luckily all the terms here are e to the power x okay so just keep differentiating here just keep differentiating here and just keep integrating here okay now put a sign over here by default the sign is plus minus plus minus plus okay and start the multiplication by leaving this first just exclude this just exclude this so multiply this with this multiply this with this multiply this with this multiply this with this okay and write it along with their signs so read it like this plus x cube e to the power x so plus x cube e to the power x then minus 3x square e to the power x yes yes i'll solve it by splitting the limits also uh pretty just give me a minute then plus 6x e to the power x then minus 6 e to the power x okay so this is your result add a c and this is your final answer okay so this is much efficient and easy visibly the approach of implying integration by parts over and over again adhavath got it okay great yeah now coming to uh Priti's question can i solve it by splitting the limits let's do that also so Priti we are evaluating x sin x integral right and it was from minus pi to pi correct so first i can do one thing i can split the limit of integration from minus pi to zero okay and from zero to pi is that fine no problem with that so i just decided to split the limit at zero yeah now in the first integral here what i will do is i'll replace x with a minus t so dx is minus dt so this will become a minus t sin of minus t dx is minus dt limit of integration will be pi to zero if you see this closely it is actually minus pi to zero t sin t correct if you observe this negative sign your upper and the lower limit will switch their positions and again there is nothing in the name so i can write it back as x sin x dx i'm sorry i should not write a dx here this is dt my bad this is x sin x dx is that fine now this and this becomes same now okay so what we can do is this becomes zero to pi sin x sin x dx and this also becomes zero to pi x sin x dx so it is equivalent to two times zero to pi x sin x dx okay now next step is i'm again going to apply i'm again going to apply king's property on this so zero to pi zero to pi replace your x with pi minus x okay this is as good as saying two times zero to pi pi minus x sin x yes now let me call this integral i and this also i correct let's add them up if you add them up you get two i is equal to two times zero to pi x sin x plus pi minus x sin x correct adding it gives you zero to pi pi sin x correct by the way you can drop this two factor here itself so i becomes zero to pi integration of this okay isn't this a simple integral to perform okay you're integrating sin x whose answer is cos x okay and zero to pi minus one minus of minus one that will be your answer so i will be pi times oh i'm sorry integration of sin x is minus cos x okay so if you put the limits of integration it becomes minus of minus one okay and minus of one minus one minus one oh minus of there's a again a minus over here let me write it like this not to confuse anyone so minus pi cos x is minus one and again a minus one okay so that gives you minus pi into minus two which is two pi okay so this is what we had actually obtained in the previous slide also the integral of this was actually two pi correct yes or no and this actually came from two i actually let me check let's go back to the previous board there was a previous to previous board yeah so this was the result that we got now equal to two pi by splitting method is that fine no no no we got a two pi only away the pi is the actual answer of the final question this was what we evaluated right now the minus pi to pi x sin x is two pi let me go back to the final board board number 11 yeah see here two pi okay so this was what we're evaluating see here we started with this question and this is your answer for a two pi that pi was the answer to the final question which was e to the power x something was there right is that clear pretty how it works let's have another question on this okay let's work on this again an ncrt question hope you can read the question the question is evaluate dx over x plus under root a square minus x square okay or integrate one by one plus tan theta from zero to pi by two or anyone i mean one basically one can be obtained from the other okay how very simple you substitute x as sin theta okay so if you substitute x as sin theta dx is cos theta d theta correct so this becomes cos theta d theta x is sin theta under root of a square minus x square is under root of in fact i should write a here a yeah under root of this will give you under root of a square minus a square sin square theta which is nothing but a square cos square theta under root which is a cos theta so ultimately you are evaluating something like this okay drop the factor of a if you drop the factor of a it becomes cos theta d theta okay by the way limits of integration i was waiting for you to answer what would be the limit of integration remember when x is zero when x is zero theta will also be zero okay so limit of integration will be zero when x is a theta will be pi by two so upper limit will be pi by two okay so we are integrating from zero to pi by two zero to pi by two now if you divide by a cos theta it becomes zero to pi by two d theta by one plus tan theta so both integrals both these expressions are actually the same thing okay so by substitution i have shown that these two are same expressions evaluate it now pi by two okay everybody is saying pi by two only or pi by two is not correct i think a small mistake you all have done please find out that mistake sure sure take your time yeah i think now you have got the right answer kavya that's pi by four absolutely prithvi is not there in this call right prithvi and prithvi both okay anyways yes so how do i proceed with this it's better to go back to this step okay this step is more pertinent let me call this as i okay so in this step i'm going to use king's property let me use king's property at that step okay so when i use king's property i will become zero to pi by two cos of pi by two minus theta i can directly write sine theta and this will become cos theta and this will again become a sine theta let's add the two i's let's add the two i's so two i will become zero to pi by two you will end up getting cos theta plus sine theta both in the numerator and denominator cos theta plus sine theta both in the numerator and denominator that's going to be one so two i is nothing but theta's integration from zero to pi by two that's going to be pi by two itself so i is going to be pi by four i is going to be pi by four yeah i know i know that mistake happens is that fine any question is that clear please type clr if it is clear so that we can move on to the next question here comes the next question for you okay let's take this one show that integral of under root of sine two theta sine theta d theta is pi by four uh can you where you prove both the forms to be equal uh i don't know which which one are you talking about which forms zero to a f of x which one previous question okay one second guys hope you have copied this question on your notebooks and you have already started working on it let me go back to the previous slide yeah are you talking about this ananya is this the page is this the working that you're looking out for little up this is the top part of the question you mean a little down or here can i go back to the question okay thanks can you divide this into two integrals uh i don't think so you would uh you can use king's property first see king's property what does it give you is don't expect king's property to always you know give you something which is very you know like they add up to give you one or something like that they may add up to give you a integral which is easy to integrate by the use of indefinite integral so don't expect miracles to happen every time you know when you are putting uh uh king's property if you're lucky enough it'll give you the same uh you know um reciprocal of that or negative of that whatever you were getting in the previous questions this is a slightly different question so if you if you apply king's property over here let's say i take this as i no no no we can't use that property there's no such property like that sorry okay i'm not solving it i'm just rewriting the question for you all yeah i'll give you a minute or so okay let's discuss so if i use king's property so let's say using king's property i can write i as i can write i as integral from zero to pi by two under root of sine two pi by two minus theta into sine pi by two minus theta d theta correct that's nothing but zero to pi by two and i think this opening up of the bracket will make it sine pi minus two theta which is same as sine two theta again but this would become a cos theta for sure okay let's add these two so let's add these two integrals let's add these two integrals so when i add it i get two i is equal to zero to pi by two under root of sine two theta and sine theta cos theta will simply get added up sine theta and cos theta will simply get added up remember these two terms are common in both these integrals okay so you can take it common and sine theta here and cos theta here will get added up okay now pay attention over here whose derivative is sine theta plus cos theta whose derivative is sine theta plus cos theta ananya am i audible to everyone ananya can you hear me uh check your audio settings on your laptop or desktop whichever you're using a phone okay no sorry abhay my question was whose derivative is this or in other words what is the integral of this right integral of this is going to be sine theta minus cos theta so what i'll do is i'll take this as t i'll take this as t okay so if i take this as t then cos theta plus sine theta d theta will become a dt will become a dt correct if you substitute it back over here so two i becomes check whether ananya is able to hear me yeah so two i is going to become i'm not going to write the limits of integration right now i'm just going to make this as dt and i need to convert sine two theta in terms of t now guys we all know that if you square both the sides here if you square both the sides over here hello oh but others are able to hear me okay so if you square this it becomes one minus two sine theta cos theta yes please check your audio settings so can i say this this term is going to behave as dt and this term is going to be behave as under root of one minus t square uh if ananya can hear me please uh leave the meeting and join again that would i think that would make things proper okay while others please pay attention over here so now what i've done is i have expressed this guy also in terms of t okay this has already started playing the role of dt and this has played the role of under root one minus t square okay now this is an easy function to integrate because we have already done this in our indefinite integrals yes or no by the way let me first change the limit of integration now when your theta was zero when your theta was zero remember t will be minus one is that proper okay and when theta is pi by two your t will be one uh working now glad to hear that ananya fine so now we have to uh you know do this integration i think which is very simple guys uh if you would all recall please recall that integration of under root a square minus x square was what x by two under root of a square minus x square plus a square by two sine inverse x by right i'm seeing that many of you are not practicing integration and as a result of it you are missing you are forgetting these formulas they are very essential formulas okay and only by practice you can remember it okay so coming back to the question coming back to the question so two i would be i remember a role is being operate by one over here so it's t by two under root of one minus t square plus half sine inverse t okay let me put the limits of integration when you put a one you get a zero here and you get a pi by pi by two into half which is pi by four minus minus one will also give you a zero here and this will again give you a minus pi by four so this is actually pi by two so two i is pi by two so this implies i is going to be pi by four which is equal to your right hand side and hence shown hence shown is that fine beautiful question i would say this question is very very good question and it actually brings a mix and match of your indefinite and definite integral both properties okay so your indefinite actually the indefinite properties was actually used to finish off the problem whereas your definite properties help you to initiate the problem okay so they are both working in tandem to solve this problem okay we'll take one more of this type just to give you more confidence about it and then we'll move on to the next property the question here goes like this uh integrate i'm going to write this question i don't have a snapshot of it zero to one dx by five plus two x minus two x square one plus e to the power two minus four x or no problem arka it has been recorded on youtube i will definitely send you the link after today's class okay so you carry on no problem okay let me tell you this session is also been recorded on youtube okay and after the class i will send everybody the link on the group okay so others please try out this question integrate this from zero to one the answer i will actually write it down because it's too ugly because if you get it you should know you have got the right answer the answer of this is one by two root eleven log of root eleven plus one whole square by ten okay so pretty ugly answer but i just wrote the answer because if you get the answer you should know that you are on the right track you have done it correctly so try to prove this this is the answer anybody you got it please just tell that you have got it so that i know how many of you are done and i can start the discussion yes anyone any success so far advaith shiram so chinta okay sure sure sure take your time take your time okay anyone who got it completely i know it may involve a bit of simplification somewhere after you have evaluated the final integral but somewhere close to are you getting have you started getting root eleven or something in your answer okay so let's let's start with this let's use king's property again so using king's property i will be zero to one dx wherever you have written an x you have to replace it with one minus x one minus x square one plus e to the power two minus four times one minus x okay so it becomes zero to one dx by if you open the bracket it becomes five plus two minus two x and you get a minus two and you get a minus two x square and i think you get a four x okay and here you get one plus this will become e to the power minus two plus four x correct a little bit of further simplification a little bit of further simplification will give you zero to one dx this will become two and two will get cancelled so five plus two x minus two x square okay so in fact we have got the same expression back but here you get one plus e to the power minus two minus four x okay now let's multiply the numerator and denominator with e to the power two minus four x so let me multiply the numerator and denominator with e to the power two minus four x so this becomes zero to one oh you got your mistake virtual okay so this becomes five plus two x minus two x square and this into this will give you a one and this into one will give you e to the power two minus four x okay this is actually a good news for us because i have managed to make the denominator same again back okay so you can see the both the denominators are same now let me add the two i's let me add this i which i started with let me add this i with this i oh sorry yeah let me add this i to this i so two i will become zero to one since the denominators are same that is five plus two x minus two x square one plus e to the power two minus four x numerator will simply get added up one plus this okay now this term and this term can be cancelled off okay and cancelling it gives you a very simple integral to evaluate which is which can be actually found out from your understanding of definite in the indefinite integrals as well this year all done completing the square method right isn't this a standard integral that we had learned in our indefinite so let's do that so i'll take a factor of two out so half zero to one dx this is going to be five by two plus x minus x square which is nothing but half zero to one dx this is going to be minus x minus half the whole square and i'm sure i'll get 11 by four over here correct till this step everyone is fine let me know if you have not understood anywhere is there any ambiguity about any step please stop me here itself clear everyone okay now we can complete a perfect square here as well we can write this as root 11 by two the whole square okay now this resembles the integral dx by a square minus x square correct we all know the result for this the result for this is one by two a ln of a plus x by a minus x correct so i can write this as half one by two a one by two a is root 11 ln of ln of a plus x a plus x by a minus x okay this is two i so i can say i is equal to this by four so i can change this with four now limit of integration is from zero to one limit of integration is from zero to one so let me just take it on top over here one by four root 11 when you put a one you get a root 11 plus by two plus half by root 11 by two minus half okay and when you put a zero you get minus ln root 11 by two minus half by root 11 by two plus half okay now two two factor you can remove off from everywhere so i'll remove it here itself so this two is gone this two is gone this two is also gone this two is also gone so everywhere there was a two now this term is actually the reciprocal of this so you can change the position and add it so you can make up plus here minus here and you can add it correct so it becomes two times so one by two root 11 under root of 11 plus one by under root 11 minus one correct now let me take this up so one by two root 11 you can rationalize the denominator by multiplying it with under root 11 plus one both numerator and denominator so this becomes two root 11 ln mod under root 11 plus one the whole square and it's i'm sure we get a 10 down correct isn't this the answer that you wanted to prove let's check is this the answer you wanted to prove yes exactly this is the answer we got is that fine so remembering your old indefinite for an integral formula is very very important okay so keep practicing it at least make it a target to solve 10 problems every day on indefinite integrals at least for a month great all right so now we are going to switch on to a new set of property which prop which number was that anybody can property number four or five oh it was probably number five okay fine property number five is also very important property which actually i call it as halving the limit property half the limit property okay what does this property say as you can know from the property itself why it has been named like that the property says integral of integral of 0 to 2 a okay of any continuous function okay so i'm assuming that my function is a continuous function from 0 to a this could be written as twice of 0 to a f of x if if with double f okay if your f of x is equal to f of 2 a minus x okay and this is equal to 0 if f of x is negative of f of 2 a minus x can you prove this now i would give you one minute to try out the proof at your own end i would appreciate anybody who can give me geometrical proof for this anybody who can come up with a geometrical proof for this i assure laksha i'll show the slide uh guys hope you have copied this down because i'm going to show the previous board to laksha okay so uh do you want me to drag up or drag down i think after this step you should be able to manage it on your own after this is just a question of indefinite integrals right laksha uh let me know when you are done with it so that i can go back to the last board okay i'll scroll it up a bit is that fine okay great okay so prajwal has found something uh absolutely correct prajwal the graph of the function would be symmetric about x equal to a line if f of x happens to be equal to f of 2 a minus x that's absolutely correct yes shiram correct you guys are on the right track awesome good what about in the second case prajwal and shiram what happens if your f of x is negative of f of 2 a minus x what can you comment about the nature of the graph in first case you are absolutely correct they are symmetrical about x equal to a line what about in the second case uh they are symmetrical about the x axis no no not exactly no no prajwal okay anyways let me help you out with this okay so uh first let me show you a geometrical proof let me show you a geometrical proof for this okay if you're claiming that your function is satisfying this functional equation okay all of you please try to understand this just try to understand this uh your function is continuous from zero to a okay so there's no break there's no jump in the function from zero to a now you're trying to say that the value that the function has at some x is same as what it has at 2 a minus x okay let's say i take zero x as zero let's say at zero its value is here then it'll have the same value at 2 a also correct so let's say 2 a is somewhere over here so exactly the same value will be here also correct if you put x as one it'll have the same value at 2 a minus one okay so let's say at one x equal to one if you not write x it'll block all my space let's write one at one let's say its value is here then at 2 a minus one it'll have the same value correct f of 2 i'm just taking some random value just to figure out what is happening okay at 2 let's say this is my 2 at 2 let's say its value is here then at 2 a minus 2 also the same value will be here correct can i say if i move on to a it'll actually become same that means this graph will meet somewhere at a okay suggesting the fact that suggesting the fact that this function this function would be symmetrical about x equal to a line so let me draw x equal to a line with these dotted okay so your function this is your x equal to a line will be symmetrical about x equal to a line okay so the meaning of this statement is your function f of x is symmetrical about x equal to a line okay now what is the requirement of the problem requirement of the problem was to find out the area under the function from 0 to 2 a that means these guys were asking me to in find this area isn't it till 2 a right isn't it equivalent to saying i find the area from x equal to 0 till x equal to a that is let me shade that area with different color let me use blue color isn't it like finding the area from 0 to a and doubling that up yes or no so i can say the yellow area is nothing but it is twice of twice of the blue area yes or no okay so this integral 0 to 2 a f of x is twice of 0 to a f of x and therefore this property so this property comes from there any question guys with respect to the first part of this property please type clr if it is clear type clr if it is clear okay awesome great great great now let us move on to the next part of the property so i have to drag this slightly up next part of the property says if f of x is equal to negative or you can say negative of f of x is f of 2 a minus x under this let us see and let us analyze graphically what is actually happening okay let's try some values put f as 0 so minus f 0 is f 2 a that means whatever it has at 0 exactly the negative of that value will be there at 2 a so let's say here okay whatever is happening at 1 exactly negative will happen at 2 a minus 1 okay so let's say one value is here then this value will be here correct if you continue doing this if you continue doing this and go till a minus f of a is equal to f of a that means f of a will be equal to 0 that means at a this will meet the graph okay so i'm just drawing a rough shape okay this will be the nature of the graph okay don't treat it to be straight line it can be like you know something curvy also right pressure let us mirrored about a comma 0 that's what i wanted to hear this graph will be mirror image about a comma 0 that means this area that is the white shaded area and the yellow shaded area will exactly cancel each other out correct so white area will cancel the yellow area getting my point if everything gets cancelled what is my answer my dear friends zero getting the point and that's how this property has actually come up and you get a zero over here see here okay this is what competitive exams want you to understand just mugging up these formulae or mugging up these properties is not going to help the cause okay they just wants to become a presenter why they just want to take further class no problem making you the presenter should i make you the presenter oh happened my mistake okay okay now i'll prove this non-geometrically also okay let's prove this non-geometrically also let me move to the next page so again i'll rewrite the property because i don't want to go back to the previous slide this is the property if f of x is equal to f of 2 a minus x and this is zero if f of x is negative or you can say negative of f of x is f of 2 a minus x correct let's prove this non-mathematically okay so let me start with the left hand side zero to a f of x and let me break the limit of integration from zero to a f of x a to 2 a a to 2 a f of x is that fine i can break the limit at a right now this first integral i will not disturb this don't disturb this okay work on this let me call this as i or i1 so i1 which is nothing but a to 2 a f of x what i will do here is that all of you please see here i'll substitute x as 2 a minus t i will substitute x as 2 a minus t that means dx is negative dt okay so f 2 a minus t dx is negative dt what about the limits of integration what about the limits of integration all of you tell me put x as a here then t will be t will be also a so this will be a am i correct and if you put x as 2 a t will be zero so your limit of this will be a to zero correct now in order to observe this outside negative sign can i switch the position of the upper and the lower limits so i can say zero to a f of 2 a minus t dt and let me tell you there is nothing in the name nothing in the name means i can put my t back as x again if this surprises you that means you have not understood the previous approaches at any stage you can consider the problem to be a new problem and make a change in the name of the variable doesn't matter forget the previous axis okay so this i1 can be substituted back over here so if you're substituting your i1 back over here so this fellow i'm substituting it back over here okay then it becomes zero to a f of x dx is zero to a f of 2 a minus x now this thing i call it as a parent result this is a parent result for this okay now there are two children of this parent okay so from here i can get so this integral zero to a f of x can be written like this so this is the parent result okay now from this two parent two conclusions have been drawn conclusion one conclusion one is if your f of 2 a minus x is f of x that is your graph is symmetric about x equal to a line then this will get converted to twice of zero to a f of x dx as you can see the second integral this fellow will become a f of x isn't it okay conclusion number two conclusion number two if your f of 2 a minus x is negative of f of x then this result will cancel off the previous result so it becomes like this okay and this gives you the answer as zero yeah so yeah prajwal is saying he thinks he has an easier proof f of a is or prajwal can is your mic operating i mean can you unmute yourself and speak are you able to speak on the mic yeah can you unmute yourself and speak yeah i have yeah sorry yeah tell me so i mean that the integral f of x is uh capital f right integral off integral off yeah the primitive okay it's capital it's capital okay fine we'll continue with this i think i was yeah i think i was not able to understand that proof can can you just uh share with me offline okay okay and in next class i'll talk about it with others as well okay prajwal okay so we'll now take up quick questions on this concept let's start with this question let's start with this question please type in your response once you're done okay kavya has already come up with an answer let's wait for others to answer kavya uh preti got pi but she thinks it's wrong shiran also got a zero okay fine so let's discuss this now so this is your function f of x correct okay and this is your 2a why can't we solve this problem without using the halving the limit property see the problem that will arise if let's say a student tries to attempt this question without halving the upper limit i'm sure you all of you remember this property which i had done with you in definite integrals that if you have signed to the power mx cos to the power nx okay where m and n are natural numbers okay normally if one is even another is odd we take that even term to be t right correct so here we'll take sin x to be t so if we take sin x to be t which means cos x dx will be dt right and you can write this as sin to the power 100 i'm sorry sin to the power 100 x cos to the power 98 x into cos x dx so substituting it will become t to the power 100 and this is 1 minus t square to the power 49 correct into a dt okay but what will happen to the limit of integration what will happen to the limit of integration in this case when you put x as 0 t will also be 0 and when you put it as 2 pi t will still be 0 so you are ultimately integrating it from 0 to 0 right now when such a thing arises when you have limit from 0 to 0 it is slightly concerning okay there we have to be careful many occasion you will get this scenario okay don't blindly write down the answer to be 0 it may be 0 i'm not denying that the answer may not be may will never be 0 it may be 0 also but this is like a point where you should become slightly cautious okay when your limits of integration is becoming same it means that there is a scope of splitting the limits correct and hence you should try out for splitting the limit property see again shiram and laksha yeah answer may be 0 i'm not denying it in fact in this case your answer is actually 0 but this is something which should ring a bell in your mind that okay probably this is not the right way to solve the problem it may not always give you 0 as the answer let me tell you in fact in the beginning of the session today itself i had shown you a question like this where the answer was 4 by 3 right if i'm not wrong and the answer you got was 0 if you put 0 to 0 correct okay so be careful that's what my request is so don't take this approach in this case correct so let us try to see whether a halving the limit actually helps us in this case or not okay so what i'll do is i'll take the approach of halving the limit so let me erase this so this is your 2a and let me check whether f of x is equal to f of 2a minus x that means let me check if f of x is f of 2 pi minus x so when you do f of 2 pi minus x that means sine 100 2 pi minus x cos 99 2 pi minus x yes actually it becomes a point but it is not a point are you getting my point under the new transform substitution you end up getting from the same point to the same point which is not the case that you also know so it wrongly gives you an idea that your answer would be 0 see when you make a substitution of sine x as t correct you end up integrating from 0 to 0 that means you are not in that becomes an area under the point correct so you're trying to say that after substituting this curve collapses and becomes a point that is not correct right okay it may be a case that there is a positive part and there is a negative part which may cancel each other out but that doesn't mean it's actually an area within a point area underneath a point there are two different scenarios these are two different cases even though the end result for both of them is 0 you cannot wrongly you know conclude that it was area under a point okay so in this case you would see that your result is actually the same correct you get back your function even after doing this that means I can split this as I can split this integral as 2 times 0 to pi sine to the power 100 x cross to the power 99 x okay now here I can use my king's property let's use king's property so by king's property I can say it becomes 2 times 0 to pi sine this correct which is nothing but 2 times 0 to pi sine to the power 100 x and this becomes negative cos 99 x which is nothing but minus 2 0 to pi sine to the power 100 x cross to the power 99 x correct that means i becomes negative of i here so i becomes negative of i here please note that this was your i okay and this is negative of i this is negative of i so i becomes negative of i which means 2 y is 0 that means i is equal to 0 absolutely correct those who got the answer as 0 but such a conclusion should not be drawn at the stage which I gave you an example 0 to 0 are you getting this point yes yes absolutely correct Sira let's take another one integrate dx by cos square x times 2 plus tan square x from 0 to 4 pi again you already have a cx square created try to take tan x st you'll again get 0 to 0 so that 0 to 0 should you know alert you should ring a bell in your mind that something is not right about this try it out anyone who wants to answer pi by 2 no pi by 2 is not right okay let's discuss this so 0 to 4 pi you have a secant square x dx by 2 plus tan square x correct now substituting tan x st is not going to solve the purpose here because i'll end up getting 0 to 0 integral okay so let us try to see whether f of x is equal to f of 2 a minus x right now this is behaving as my 2 a correct so f of x and f of 2 a minus x so let us calculate 4 pi minus x which is actually secant square 4 pi minus x by 2 plus tan square 4 pi minus x oh yes they're both the same so basically f of x and f of 4 pi minus x are same so basically what do you conclude you conclude that f of x is same as f of 2 a minus x that means i can split this i as or i can break the this has 0 to 2 pi okay secant square x dx by 2 plus tan square x now again if i want to substitute tan x st it is not going to help me same 0 to 0 integrals will happen that means there is a further scope of breaking the limit so can i make it as and of course f of x and f of 2 pi minus x will be the same again that means this can be further in split up as 4 times 0 to pi again if i try substituting tan x st again same problem 0 to 0 so again i can do pi by 2 and i can again have a 2 over here so it becomes finally 8 times 0 to pi by 2 secant square x dx by 2 plus tan square x now at this stage you can substitute tan x st that means your secant square x dx will become a dt right so 8 times dt by 2 plus t square what happens to the limit of integration students when x is 0 t will be 0 when x is pi by 2 t will be infinity okay so don't hesitate in putting infinity if the tan function is coming as your answer so here your answer will be 1 by root 2 tan inverse t by root 2 of course 8 times 0 to infinity so when you put an infinity you know that your answer will be pi by 2 minus 0 so here it gives you 4 by root 2 pi is your i so i is nothing but 2 root 2 pi absolutely correct oh sure sure so see the way i have used this property more than once i first made it half again i made it half again i made it half is that fine shira nikhil i'm sorry i'm not going to give you any break because we all we just have 15 minutes of class left and i'm sure you are habituated for a 4 hour class so 3 hour class should be nothing for you so let's move on okay i would like to do this problem because this is a very famous ncrt question and it comes every year please mark it with a star because it's going to definitely come for your school exams as well or we have to prove that 0 to pi by 2 sine log sine x is same as integral 0 to pi by 2 log cos x and this result is pi by 2 log half any idea i think the first part is pretty easy right you can prove this by using king's property there's nothing in the first part the idea is to prove from here to here it's a very important problem very important question okay not very simple one but again very important for school exams so let us start with assuming that this integration log of sine x from 0 to pi by the dx is i so using the property king's property using king's property you can write i as 0 to pi by 2 log sine pi by 2 minus x dx which is nothing but log of cos x okay so this proves the first part of the question there was nothing very hard about it very easy now proving the next step we'll add these two i's so let's add these two so let's add these two i i becomes a two i correct since the limit of integration is same for both of them 0 to pi by 2 we can add the function itself log of sine x and log of cos x so it becomes 0 to pi by 2 log of sine x cos x now here we can do a simple activity i can multiply with the two here and divide with the two please note that this is happening within the log this is happening within the log i'm multiplied with the two and divided with the two so two i gets converted into 0 to pi by 2 log sine 2x by 2 and by use of log properties i can split this sine 2x by 2 as log sine 2x minus 0 to pi by 2 log 2 so i've done nothing i've just used log a by b property log a by b is log a minus log b and i've separately applied integrals to both of them clear no doubt about it now this is pretty easy to integrate right because log 2 at the end of the day is a constant isn't it log 2 is a constant so integrating constant means that x will come out so this becomes 0 to pi by 2 log sine 2x and this will be x log 2 and you just have to put the limits 0 to pi by 2 which is nothing but 0 to pi by 2 log sine 2x minus pi by 2 log okay this is the status of 2y so far okay now see in this integral i'm going to substitute i'm going to substitute 2x as t okay i'm going to substitute 2x as t which means dx is equal to dt by 2 okay so let me rephrase everything log sine t dx is going to be dt by 2 limit of integration will be 0 to pi if i'm not mistaken why 0 to pi because when you put a 0 in place of x t will also be 0 and when you put a pi here sorry pi by 2 here t will be pi correct let me erase this else it'll clutter up the space there oh stop stop stop stop oh my god sorry so 0 to pi right and there was a half and this term i'll just copy down as such pi by 2 log 2 now here i will use the halving the limit property i'll use half the limit property because it it works here because if you take log sine t and log sine pi minus t they are same actually isn't it okay so i can use half the limit property here so if you use half the limit property it becomes 2i is equal to half 2 times 0 to pi by 2 log sine t this 2 and this 2 gets cancelled correct now there is nothing in the name there is nothing in the name means i can write this i can write this integral back in terms of x so i can write this as 0 to pi by 2 log sine x dx correct and isn't this the i integral isn't this my i so ultimately what are you trying to say you're trying to say 2i is i minus pi by 2 log 2 which means i itself is minus pi by 2 log 2 yes or no which means i is nothing but pi by 2 log half that is what we wanted to prove i understand this is a pretty lengthy problem but here the approach is pretty straightforward okay so i i just freeze the screen over here anybody who wants to ask any questions here please let me know yeah fragile fragile is asking some questions uh first part it can be only change the variable if it isn't present in other places see you just have this feeling in your mind that that is the beginning of a problem if that is the bringing of problem you can do anything with the variable okay don't have don't carry any baggage that there was an x already used before so can i use it or not think as if that is the very first step of the problem can i have the one last problem before we close the session i'm writing one integers separately and using the variable while solving that no no problem see variable name doesn't matter at all because ultimately you're putting the limits of integration okay so don't don't be worried about whether i can use the variable name which has already been used in some other integral it doesn't matter ultimately it's just a number at the end of the day we have already done these questions okay fine i'll give you from my end integrate try it out try it out try it out pretty see what works it has something to do with the previous problem hint is you can you can start you can start simplifying this is actually root 2 sign x plus 5 by 4 okay this is the hint which i'm going to give you that's it okay karbya says 5 by 4 log 2 uh just a sign mistake is there karbya the answer is exactly negative of what you said exactly the same shawak or is the sign different because the book says negative of what karbya got ah now she has got it okay check it out fine so in the interest of time i'm going to solve this so okay now substitute x plus 5 by 4 as theta so dx is d theta so it becomes log root 2 sign theta d theta okay now what about the limits of integration when you put a minus 5 by 4 you get a zero when you put a pi by 4 you'll get a pi by 2 so this is your i this is nothing but 0 to pi by 2 log root 2 0 to pi by 2 log sign theta d theta okay now this integral is something which we did just a while ago and the answer was 5 by 2 log half okay here also the answer will be half log 2 into pi by 2 so it's pi by 2 log half okay let me simplify this further so this becomes minus minus pi by 4 oh once again uh minus pi by what did the answer karbya gave pi by 4 log 2 okay okay no no that's wrong so this i'm sorry this you can write it as 5 by 4 log 2 this is minus pi by 2 log 2 remember when you are making log half as log 2 a negative sign will come in front okay now this is as good as minus pi by 4 log 2 okay so this becomes your answer now this is this problem would have been very lengthy had you not known the result for 0 to pi by 2 log sign theta okay so some results you can actually remember them because they really save a lot of your time is that fine guys all right then so we'll end the session right now here thank you so much for coming online hope to see you on an offline session next week yeah Krishna has a doubt sir i use kink's property to simplify and got the log of cost 2x kink's property hey how Krishna because one sign will become cost then other cost will actually become a sign right thank you so much pleasure is mine Nikhil thank you so much so those who have those who are fine with the session you can leave those who have some doubt you can ask me right now a krishna once again let me unmute the a krishna because one sign will become cost then other cost will actually become a sign thank you so much pleasure is mine Nikhil thank you so much to those who have those who are fine with the session you can leave those who have some doubt you can ask me right now a krishna once again let me unmute then other cost will actually become a sign okay thank you so much guys bye bye