 The heat itself is the theory which is cos x by 2 sin x plus 1 p which is cos x by 2 sin x plus 1. Is there a particular formula? This is a particular way. Now find dp by dx minus sin x minus cos x derivative by 2 sin x plus 1 d whole squared. Just the purpose of simplicity so that you have to write 2 sin x plus 1 as y. So what is sin x then? y minus 1 sin x can be written as minus y minus 1 by 2 minus 2 by y squared. Won't it become minus 1 by 2 y minus 1 by y squared? Minus 2 by y squared plus half minus 2. So it's minus 3 by dx. After differentiating I got this. I took y as 2 sin x plus 1 which is the stuff itself. So when I put it back sin x will be y minus 1 by 2. So this sin x will be minus 5 minus 1 by 2 and minus 2 and I divide it by y squared. This is y squared. Now put back your 2 sin x plus 1 and minus 3 by 2 2 sin x plus 1 whole squared. Now take this dx on the right side. Let's integrate it. Integration of dp would be which I know. This is what I think. This I can use half angle problem. So make I see a sin squared else on the top. That's why it's an algorithmous device on the basis of that. Most of these are substitute sin x as that. We'll come on this as well. Both of them and start solving it. Next class onwards. So we'll continue with the indefinite integrals. So today we'll start with discussing the type having a degree grade. We'll explain this because if I try to explain it in a generic way you might get confused. Let's take a small example to explain the algorithm for it. Let's take this as an example. x squared plus x plus 1. Now just listen to the approach very clearly and then we'll start solving it. The approach here is that the first slide which is b0 plus b1x multiplied with this quadratic expression that is x squared plus x plus 1 plus another constant b2 times integral of plus b2 times integral of the same quadratic expression with random sign. Now this has been found out through observation that whenever integral of this type is being solved with the same function that is the denominator function multiplied to a polynomial whose degree is one less than the degree of this. x plus b2x squared times this particular expression plus b3 times integral of dx is the same denominator function because b1 and b2 very simple. For that you differentiate both sides with respect to x. So differentiate both sides with respect to x. So what is the derivative? It's the function itself. It's the integrand itself. This is called the integrand. So write down the integrand which is x squared plus 2x plus 3 by under root of x squared plus x plus 1. Now what is the derivative of this term? You can apply product rule on it. So first let us differentiate the first linear term keeping the other term as such. Then keeping the first term as such differentiate the other term which is going to be 2 by 2x plus 1 by 2 root of x squared. Correct? This is by the product rule on the first term. What is the derivative of this term? It's a function. b2 by the function itself, the integrand itself. Okay. Now coefficients but before we do that we will multiply both sides with 2 times under root of x squared plus x plus 1. So this will become 2 times x squared plus 2x plus 3. This one will become 2b1 x squared plus x plus 1. This one will become b0 plus b1x 2x plus 1. And this will become 2b2. Is that okay? Now let's start comparing the coefficients of x squared first. What is the coefficient of x squared? Left side what is the coefficient of x squared? 2. Hence it's 2b1. You can say I have multiplied both sides with 2 under root of x squared plus x plus 1. So this implies b1. So on this left side your coefficient of x is going to be 4. That's 2 into 2. On the right side of plus 2b0 is equal to 4. And b1 is already known to us as half. So 4 minus 3 by 2 divided which is going to be which are constants of the right side. So this will be 1 b0 plus which implies b2 is going to be 5 minus 5 by 4 into half. Which if I am not b1 is going to be half. Yes or no? Yes or no? x squared plus a squared. Nikhil what was the integral of dx1 under root x squared plus a squared? This is going to be plus under root of? Is that okay? Plus c. So this is going to be question of that also. Meanwhile which is this? But I would write times this quadratic plus another constant. Let's say b2 integral of 1 by this function under root of x squared plus 1 dx. Now how do I get b0 b1 b2 by differentiating both sides with respect to x? When I differentiate this with respect to x I get the same function back. And here I would apply chain rule, product rule. Integrating both sides. And then we start comparing the coefficients of x squared x and constant for both the sides to get my value of b0 b1 and b2. Place it back here. It's just that one thing that you need to evaluate is the integral of dx by under root x squared plus x squared. Which you can easily do by completing the square and following the standard integral formula discuss the last class. Can you have one more example? So what if the root wasn't there in the last class? Then this approach will not be able to do it. Then you divide it first. And what about if it's like both sides? Okay, you use fact that. I think you got confused. I am not sure if it is root of x squared plus dx. And it's your face. No, it's not. Then you would follow the standard integrals or the approach which I discussed. So please be watchful in the exam. Do not get confused within the types itself. This type should have an under root of a quarter take in the denominator and numerator should be a polynomial of degree greater than equal to 2. Next problem we'll take up on this. Can I clear this up? Yes. Yes, this is going to be a long method but this is, I mean, this is the only method. The answer should be the same. They may ask. The reason why we are studying it is somewhere they've asked. Thank you.