 Hi, I'm Zor. Welcome to a new Zor education. I would like to continue talking about indefinite integrals. I think it's very important to devote some time to basically do the job to find integrals. And the more the merrier, actually I would like this lecture to be an illustration of the fact that integration is really more of an art. Rather than the skill because you really have to think about how to approach each individual problem based obviously on your experience. And that's why the more integrals you take, you find the more comfortable you will feel the next time. Now this lecture is part of the course of advanced mathematics which we present for teenagers and high school students. It's presented on unizor.com. All the lectures and corresponding notes to lectures are brought to this website. So if you found this particular lecture on YouTube, I would rather suggest you to go to the unizor.com and watch it from there. Because the result is very detailed explanation about whatever is going on in the text which is supporting each video. Okay, so, well, let's just do the work. We have a lot of work to do. We have six different problems which I would like to present. As usually with all the problems which I present, I do encourage you to try to do it yourself first and then listen to a lecture. Regardless whether you will or will not succeed, it's still a good idea to think about this before you listen to somebody else. Alright, so my first integral is, well, before going into each particular problem, I think it's worth to remind two major approaches which I actually explained before, but I'm going to use it in one or another capacity in these particular problems. One is the integration by parts and I would like just to remind you that integral U, dv is equal to Uv minus integral dv. So, where U and V are both functions, so this is differential of functions, U of, let's say, X and V of X. So, this is the integration by part and another thing which is kind of a substitution. If you know that integral of f of X, dx is equal to g of X, then you can substitute instead of just an independent variable X in the other function, but plus C obviously. So, you can have f of, let's say, W of X, g W of X equals to g of W of X plus C. This is substitution. So, if you, for instance, have your integral reduced to this level, to this form, then you can attempt to do the integration of function f of X and then in the solution just substitute W of X. So, I will use these two. I consider them as known and there are a couple of lectures which I did before to basically prove these two rules of integration and examples of using them. Okay, now let's go to concrete examples. Sine of X cosine cube of X dx. Okay, so what's creative about this particular example? Well, you have to notice in this particular case and it's very easy to notice that, well, sine and cosine are two different functions. So, whenever you have two different functions, you cannot really use something which you know, like integral of something is equal to something. This is, you don't know. However, what you can actually do is you have to notice that the sine, well actually minus sine is a derivative from a cosine, which means you can rewrite it as minus cosine of X dx divided by cosine cube of X, right? So, this is a derivative of cosine. Instead of a sine, I put minus derivative of cosine. Derivative of cosine is minus sine. That's why I put minus in front so it neutralizes that sine. Now, grouping this together, you obviously see that this is minus differential of cosine of X, right? What is differential of the function? It's derivative times differential of the argument divided by cosine cube of X. Now, how should I approach this? Well, there is a substitution, obviously, which we're just talking about. If you will substitute Y is equal to cosine of X, you have this integral as minus dy divided by Y cube, right? Or minus Y to the power of minus 3 dy. Now, this is a power function. So, what do we know about power functions? Well, if you integrate any power function, you have to increment the power by 1. Why? Because the derivative will subtract one from the power, right? So, basically, this is minus Y to the power of minus 2. But we need some kind of a coefficient to neutralize this. So, the derivative of Y to the power of minus 2 is what? It's minus 2 times Y to the power of minus 3, right? So, I have to put here 1 over minus 2 to neutralize the differentiation. So, when I differentiate this, well, this minus is from this one. Y to the power of minus 2, if we take derivative would be minus 2 times Y to the power of minus 3. So, we have to neutralize minus 2, and that will be the result. Well, plus C, obviously. Now, instead of Y is, we can put the substitution. So, this minus and minus, it's all plus. So, it's 1 half 1 over cosine square of X, right? Y to the power of minus 2 is 1 over cosine square. Now, what do we have to do next? Check. Now, how can I check my integration by differentiation, right? So, let's differentiate this function. It will be 1 half times cosine to the power of minus 2 derivative, right? Equals 2. 1, 2. Now, this is the power function. Well, it's actually a function of function, but outer function is the power function. The power is minus 2, so it's minus 2 times cosine to the power of minus 3 of X. Times, since it's a function of function, I have to differentiate inner function, which is cosine, which is minus sine of X. So, 2 and 2 minus and minus, and we have sine of X divided by cosine cube. The end. So, that's my first problem. 5 to go. Next, integral logarithm of sine of X times cotangent of X dx. All right. Again, we have to do something about this before we can't integrate it directly, right? We don't know what it is. Well, let's just think. The first guess would be to replace cotangent with cosine of X divided by sine of X. That's basically the definition of a cotangent, right? So, I didn't really change anything. Now, again, let's just think about it. We have sine and sine, and this is the cosine. Same as in the previous example, we can use the cosine and dx as a differential of sine of X, correct? The derivative of sine of X is cosine times differential of independent argument. So, I can rewrite this as integral logarithm of sine of X 1 over sine of X times differential of sine of X. Obviously, it's easier because I can put Y is equal to sine of X and my integral is equal to logarithm of Y times 1 over Y times dy. That's easier, right? But still not exactly because we have logarithm and we have a power function. Again, another guess. Remember, logarithm X derivative is 1 over X. So, this is actually differential of logarithm of X. Well, actually, logarithm of Y, obviously. Derivative of logarithm of Y, which is 1 over Y, times differential of the argument. So, my next substitution is logarithm of Y is equal to Z, which makes my integral equal to Z times dz. Logarithm is Z and d of logarithm is dz. Now, what is this? This is a power function. So, this is 1 half z square plus c, right? The derivative of z square is 2z. 2 and 2 will cancel each other and z will be whatever we have to integrate. So, all we have to do now is go back to the substitution. So, instead of z, we will put logarithm, right? Now, instead of Y, we can use whatever substitution here. And this is the answer. Now, let me try to check it out. Let's differentiate this one. So, it's 1 half. Well, this is a power function. That's the outermost function. So, it's three functions. Outermost is power, then inside there is a logarithm and inside there is a sine. So, differentiation would be in three steps. First, I differentiate the power function, which is logarithm of sine of X. Then I differentiate inner function, which is logarithm, and that's 1 over sine of X. And then I differentiate the innermost function, which is sine. And that's the cosine derivative of the sine and cosine. So, that's my derivative. And obviously, these two are exactly cotangent. Oh, I forgot two, obviously. Differentiating two logarithms. One half of logarithm square is one half times two logarithms. So, that's why two, and the two is supposed to cancel out. All right. So, that's the answer. We've done number two. Okay, let's go on. Oh, that's complex. Integral logarithm X minus 3X square root of logarithm X. Okay. Well, hmm. Okay, we have to notice something specific, since I'm presenting this as basically an exercise. It's a development tool to develop the creativity and analytical abilities and logic, etc. There must be something which would simplify this, right? Now, what is it? 1 over X and dx is differential of logarithm, right? So, it's logarithm X divided by square root of minus 3. Divided by square root of logarithm X. D logarithm of X, right? So, differential of logarithm is 1 over X times dx. So, that's exactly what I have noticed this. Now, how can I do next one? Well, obviously, I have to substitute, all right? So, what I have is Y minus 3 divided by square root of Y dy. Now, how can I do that? Actually, it's very simple. I separate Y minus 3 into Y and minus 3. So, it's two different integrals. Y over square root of Y dy minus 3 over square root of Y dy equals 2. This is Y to the power 1 half dy minus 3 times Y to the power 1 minus 1 half, right? dy. So, now that's easy. These are just power functions. So, what do I have? I have Y to the power of 1 half. So, I need plus 1. So, it's Y to the power of 3 halves. And since when I take the derivative, 3 seconds will be out. So, I have to put 2 thirds minus. Now, Y to the power of minus 1 half. How can I get that? Well, that's Y to the power of plus 1 half. 3 times Y to the power of 1 half. And now, since if I will make derivative from this, it will be 1 half times Y to the power of minus 1 half. I have to neutralize this 1 half by multiplying by 2. Plus C. Well, and that's the answer. All I have to do is substitute back. It's 2 thirds logarithm to the power of 3 seconds of X minus 6 logarithm to the power of 1 half of X plus C. That's the answer. Okay, let me try to check it out. Well, actually, you know what? It's kind of simple. So, I have my answers here and it corresponds to my answer. So, that's why I will skip the checking out. I do suggest you to do it yourself. So, I will save some time. Okay, next. Integral X square E to the power of minus... No, plus X over 2. All right. Now, how can we approach this? Well, we know that differentiation of exponential function basically leaves it in the same exponential form, maybe with some multipliers. So, most likely, we would like to... Yes, and differentiation of the power reduces the power by 1, right? So, what we probably have to do is we have to apply the integration by parts having something like this first. Now, what am I missing? Differential of E to the power of X2 is the derivative, which is E to the power of X2 times 1 half, the inner function, right? So, I have to compensate this 1 half, so I'll put 2, right? So, differential of E to the power of X over 2 would be E to the power of X over 2, like this, times 1 half will cancel with this one. Now, I will use integration by parts, which is 2 X square E to the power X over 2 minus 2 integral E to the power of X over 2 dx square, right? Now, what is differential of X square? That's a derivative, which is 2X times dx. Now, is it better? Of course, it is better, because here we have a second degree, here we have the first degree, and everything seems to be more or less the same. So, I will do again the same thing exactly, equals to 2X square E to the power of X over 2 minus... Now, 2 and 2 is 4, and I will do again integral of X d of E to the power of X over 2, and I'm missing, again I'm missing 1 half, right? Because if I will differentiate this, I will have E to the power of X over 2 times 1 half, which I have to neutralize, which means I have to multiply it by 2. Now, again, I will use differentiation by parts. So, this thing stays minus 8X E to the power of X over 2 minus, so it will be plus 8 integral E to the power of X over 2 dx, right? Equals. Now, this is easy, because we know that E to the power of X2 is a derivative of E to the power of X over 2, and we need a multiplier 2. So, it's 2X square E to the power of X over 2 minus 8X E to the power of X over 2 plus... I need 8 times 2, so it's 16, E to the power of X over 2 and plus C, right? Derivative of this would be E to the power of X over 2 times 1 half, and that would make it 8. And this is the final answer, which we can obviously simplify with this. So, that's the final answer. And, again, it corresponds to whatever answers I have, so I will skip the checking procedure, the differentiation of this. I do recommend you to do it yourself and check if you will get exactly the original function which we integrate. Okay, next. Integral dx divided by square root of X square plus 4X plus 5 minus X square, yeah. Okay, now, here again we are at loss. It doesn't really show any obvious solutions in this case. So, we have to think about certain functions which, if differentiated, will give us something like this. 1 over square root of some quadratic polynomial with minus at X square. Well, there is one, and now you just have to remember it. I mean, if you recall that the arc sine of X derivative is equal to 1 over 1 minus X square, then that would help you. Because, you see, this is also some kind of a quadratic polynomial with a minus in front of the X square, and this probably can be somehow related. Well, by the way, why do I remember it? Actually, I don't remember it. I was thinking about this, and then I kind of flip-flopped through my previous lectures and found that this is exactly what it is. Just in case if you don't know how to derive this, think about this this way. Sine of arc sine of X is equal to X, right? So, if we differentiate both parts, we will have cosine of arc sine X. This is the derivative of the sine times derivative of the inner function arc sine of X derivative. So, derivative of the left part is this, derivative of the right part is this, 1. So, what is cosine of arc sine? Well, in most cases, if you have the sine of X is equal to X, arc sine of X, and you take the cosine, it's a cosine of the angle sine of which is equal to X, and this is obviously, in most cases, this. Obviously, we can talk about whether this particular angle, this particular argument is in one part of the domain of this function or another, but let's take the simple case, right? When the sine square plus cosine square is equal to 1, so that's why cosine of arc sine is equal to this. From which follows that arc sine derivative is equal to 1 over, like I said? So, how did I come up with this? How did I guess that this might actually be useful? I don't know how to explain it. It's just based on experience. If you are doing this exercise many different times with many different functions, eventually you will run out of different possibilities. So, you will probably have it in your mind that if you have in the denominator, square root of quadratic polynomial with a minus, that's somewhere around arc sine. Now, so what do I have to do? Well, obviously, I have to somehow relate this to this. Well, the best way to do it is to have the full square, right? So, let's just think about minus x square plus 4x plus 5. I would like it to be in the form of 1 minus and have some linear combination, something like this, maybe with another coefficient. That would probably do the trick, right? Because then all I have to do is this coefficient will go out and this would be just a substitution of the argument and I will reduce my problem to this one. Now, what is exactly this? Well, let's just think about it. The full square we need from this is x minus 2 square is equal to x square minus 4x plus 4, right? So, minus would be minus here plus here and minus here. What difference between this and this? So, we need 9. Then we will have plus 5, right? Okay, so we have transformed this into this. Well, but we need 1 here, right? So, we have to divide it by 9 and here you will have here, right? So, 3 to the power of 2 would be 9 and it will cancel with this one and this would be 9. So, basically this is something which I might use in this case. So, it's equal to integral dx divided by 9, the square root of 9 times 1 minus x minus 2 divided by 3 square and now we are very close to our arc sign. So, what do I have to do? Here it is. Well, first of all, 9 goes out of the square root, it would be 3. Now, if I would like to make this substitution, now what is differential, right? If I will differentiate both sides. So, the derivative of this is one-third of x, right? Times dx, so I can do this type of a substitution very handily, I have these 3 exactly here, right? So, I can actually do this substitution and have dy divided by square root of 1 minus y square, right? dx divided by 3 is dy and under the square root I have 1 minus y square. Now, this is arc sign, right? So, this is arc sign of y plus c, which is equal to arc sign of x minus 2 divided by 3 plus c. If I substitute back my y. Okay, so, I'm not doing any checking, you have to. And by the way, I'm not doing any checking right now because I already did it before when I was preparing for this lecture and the checking is in the notes for this lecture. If you are watching it from theunisor.com, you will see how the checking is done. Alright, my last example, which looks innocently and simply. However, that's the most difficult of them all because I don't know what to do with this. I cannot really use sine as a derivative of a cosine or a cosine derivative of a sine because they're both kind of represented here, right? However, however, I have one lucky guess. If you remember the formula for cosine of a plus b, it's equal to cosine a cosine b minus sine a sine b. Okay, now, what if I have some kind of a number which is both cosine of some angle b and sine of this angle b? Well, I know this angle. Angle is 45 degrees, cosine of 45 degrees equals 2. This is pi over 4 by the way. It's equal to square root of 2 divided by 2 and sine of 45 degrees equal exactly the same. Which means that if I will multiply this denominator by square root of 2 and I will put square root of 2 on the top as well. Okay, what that will give me? Well, it will give me that this is the cosine of x times cosine of pi over 4. And this is the sine of x times sine of pi over 4. So I can convert it into square root over 2 dx. And here I will have cosine of x plus pi over 4, right? Cosine x times cosine of pi over 4, which is square root of 2 over 2 minus sine x times sine of pi over 4, which is also square root of 2 over 2. So that would be my... Now, this is simpler, right? Because I can obviously have here, doesn't really matter, right? Because differential of x plus pi over 4 is the same as differential of x because the derivative is equal to 1. So now I can make this substitution. And I will have that this integral is equal to square root of 2 over 2 integral dy divided by cosine of y, where y is equal to x plus pi over 4. That's easier, right? But I still don't know what to do with this one. Well, let's just think about what can we do about this? Well, 1 over something always resembles the logarithm, right? Because logarithm of x is 1 over x. So it might be related to some kind of a logarithm of something. But I have this cosine, not y. If I had y, this problem would be solved, but I have a cosine of y. So I have to neutralize the cosine somehow. How can I do that? Well, here is another wild guess, if you wish. What I can do right now, I can multiply and divide this thing by cosine. What will it give me? This is what it gives me. So that's what I have on the top. Cosine y times dy is differential of d sine, right? Now cosine square is also very easy to convert into sine, right? Cosine square is 1 minus sine square. So now I rationalize this, basically. I convert it into a rational. So z is equal to sine of y. And now I have square root of 2 over 2 integral dz over 1 minus z square. Okay? Easier, because this is now power functions, etc., so it's kind of easy. Now, how to deal with this one? Well, the simplest way is just have it as a sum of two integrals, very easy. Since 1 minus z square is 1 minus z times 1 plus z, I can say that this integral is equal to square root of 2 over 2, integral 1 over 1 minus z dz plus integral of dz 1 plus z, right? And we need some multiplier. Because if I will take this into common denominator, what I will have is 1 minus z times 1 plus z. Again, 1 over 1 minus z square is equal to 1 minus z plus 1 plus z. But this will be 1 plus z plus 1 minus z, so it will be 2. So I need actually times 1 half, right? So I have to have 1 half here, which means I will take this to 4. Now, this is easy. Now, how easy is this one? Well, this is almost d of 1 minus z, right? What's the difference? Differential of z and differential of 1 minus z. Well, this is just sine, right? Because derivative of 1 minus z is minus 1, and that would give me n times dz, so that would be this one. So, and if this is true, now this is basically logarithm, right? Plus, now this one, I don't really need anything with a sine. Differential of z or z plus 1 is exactly the same, okay? So this is minus, this is plus, this is logarithm of 1 plus z, and this is with a minus, logarithm 1 minus z plus z. Well, that's the answer. All we have to do now is substitute sine instead of z. So, that's equal to square root of 2 over 4, logarithm of 1 plus z. z is a sine, and y is x plus, so that would be sine of x plus pi over 4. Minus, logarithm of 1 minus sine of x plus pi over 4, and that's my answer. So, as you see, this innocently looking integral actually was a little bit more complex than the previous ones, so you never know what you're getting into. All right, now this is actually a very important integral because I will probably have another lecture on integration, and I will suggest a couple of other maybe methods to integrate this particular function. It's a good kind of learning function to basically go through all the different aspects of integration and different approaches which you can take. Well, that's it for today. Thanks very much. I do suggest you to go through all these exercises yourself. It's all on unizord.com and check your answers. That's mandatory. Well, other than that, that's it. Thanks very much, and good luck.