 and we took lot of problems, many other new types of integration, there are many types which will be exploring in today's class, so I am very funny here, but the choice of the function whom we are substituting as always will become dd, so what we will do is as t, so t to the power 4, cos square can be substituted as 1 minus sin square minus t square and cos square becomes dd, which can be integrated as c to the power 5 by 5 minus c to the power 7 by 7 plus c, substitute your t back as sin trigonometric asymptotic, we will come to that, that will be a part of this. So, in this case the one with the even power will be substituted, we are talking about this right, so if cos is to the even power then substitute cos s t, let us take another example, we are taking a type of integral of this form, where sin is h to the power n, cos is h to the power n, m and n are natural numbers, now we are taking cases here, when m is even n is, in fact you can have also the case where n is even m is odd, indirectly I am trying to say one should be even one should be odd, whichever is subjected to the even power preferably in fact you should always substitute that function s t, so here sin is subjected to the even power I will substitute sin x s t, try this one out sin to the power 3 x cos to the power 6 x, tomorrow 10 to 1 this will be there, types of papers one would be of another would be of c t level, so I would suggest the top 10 or 12 people or 30 people to write the advancement in the rest to write the c t level, you as cos square into cos. The reason cos I am on that, it behaves as dT, right, right, so it behaves as dT. So, the moment I ask for integration, right, this one. Sir, one second is cos cube, one second is ds because you can take sin x is dm, you can substitute dm. Yes, we will discuss it. Can I do this one? Yeah. What is this one? So, it becomes sin square, which I can easily substitute. So, this integral with 1 minus t square t to the power 6 into minus dT. So, you can write it as t to the power t to the power 8 minus t to the power 6 dT by 7 plus t, put your t back as, pretty simple, which has greater power, but since cos x is subjected to a higher power, I will substitute cos x as t. So, minus sin x dx becomes dT, that is sin x dx becomes minus of dT. So, what we do? I start substituting, this becomes 1 minus t square, this becomes t to the power 7, this becomes minus dT, which is the integrates, which is the inverse to t to the power 7 minus t to the power 5 dT, whose integral is minus t to the power 6 by 6 plus c. Substitute your t back as cos x, that is what would happen, you would pull out the cos x out and you would be left with unnecessarily chases, you will have to rely to higher angles. Correct? So, all you can do is just cos square x and then you have cos past 6 x, cos to the power 4 and you can take cos x as t, over them sin x dx and sin x dx. Exactly. So, what is going to happen is that, the moment you start substituting any one of the trigonometric functions as t, you will experience irrational powers coming up. So, something like cos divided by 2 will start coming up, which will make the integral more difficult to solve. So, in such cases, you will have to rely on trigonometric identities, especially the ones which help you to convert higher powers to higher angles. Right? For example, and similarly, cos to the power 4 x can be written as 1 plus cos 2 x by 2 whole square, isn't it? Yes sir. So, basically you can simplify this as 1 by 8, you will have 1 minus cos square 2 x into 1 plus cos 2 x dx, correct? Yes sir. So, this is sin square 2 x, sin square 2 x can be written as 1 minus cos 4 x by 2. So, I can have 1 by 16 times this. Yeah, if you expand this, you will have 1 plus cos 2 x minus cos 4 x and minus cos 4 x cos 2 x. Now, what is cos 4 x into cos 2 x? You can apply your transformation formula on this. Yes sir. What is the formula cos a plus b plus cos a minus b? So, we can apply the formula minus cos 4 x minus half cos 6 x plus cos 2 x. Yes sir. This station? Yes sir. See, sin square, we all know that cos 2 x is 1 minus 2 sin square x. So, from here sin square x could be written as 1 minus cos 2 x by 2, which is what I did to the first term. Second term we know is 2 cos square x minus 1. So, cos square x could be written as 1 plus cos 2 x by 2. So, that is what I did to the fourth square and 1 minus cos 2 x. I took out 1 from this and I made the formula a minus b a plus b, which became 1 minus cos square sin square 2 x and sin square 2 x is again 1 minus cos 4 x by 2. So, the idea is this to multiples of angles. The moment I start doing all these things with cos 2 x cos 2 x and minus cos 4 x minus half cos 6 x. This is what I will get. Now integrating this is a pretty easy task because we will get x. This will be sin 2 x by 4. This will be minus sin 4 x by 4. This will be minus sin 6 x by 2 plus sin to the power 5 x cos square x basically take cos x as t. The one with the even power will be taken as t. It is 1 plus half cos 2 x. What will be minus sin 2 x by 4? Cos 2 x integral is sin 2 x by 2. Cos 2 x derivative is sin 2 x. Cos 2 x is 1 minus 2 sin square x and 2 cos square x minus 1. 1 plus cos 2 x to the power 5 itself will be minus sin 2 x. Now expansion. Now 10 z to the power 8 and the last what is sin c 2? 45. So, 45 z to the power 6. 1 by z to the power 6 plus next term would be 10 c 3. 10 c 3 is 120. 10 c 3 is 10 into 9 into 8 by 6 which is 120. So, 120 z to the power 4 plus 1 by z to the power 4. Next term would be. So, you all know the expansion of a plus b to the power n and c 2 a to the power n minus 2 b square and goes on till n c n a to the power 0 or you go to b to the power n directly. Now, previous a as z and b as 1 by z 0 which is what? 1 z to the power 10. 10 c 10 1 by z to the power 10. So, what I am doing 1 which is 10 z to the power 9. Similarly, the second last term would be n c n minus 1 a to the power n minus 1. So, what will be that term? 10 c 9 z in which will be 10 in. So, what I am doing is I am collecting these two terms here. 10 z square right. 10 c 1, 10 c 2, 10 c 3, 10 c 4, 10 c 5. 10 c 5 is now 252 that is it. Now, you will think. So, what if you have an under term signed to the pi 11 x plus the pi 11 x. And this is for only cos term. This question only causes. I am not using this. If I ask you a simple question. Yes sir. If z is this, what is z to the power 10 x correct? 10 x. So, when you add them, when you add them you will get 2 cos 10 x correct. So, the benefit of doing 8 x is you do not do it. What would this 6 x? Yeah correct. Again 120 into 2 cos 4 x into x. That is the question that you have to keep on looking at k terms. So, here.