 Hello students, welcome to Centrum Academy YouTube channel where we bring for you every day a new problem in mathematics in our problem solving So here we are here. We are again with one of the questions on integral calculus. Okay, so guys and girls Let's read this question and let's find out the way to solve it Let I n so I with an index of n is related to a definite integral Which is x to the power n cos x integral from 0 to pi by 2 and being a non-negative integer Then we have to find which of the following option is basically give this summation to you Summation of i n by n factorial plus i n minus 2 by n minus 2 factorial n going from 2 to infinity Okay, guys and girls. Let's have a look at this question The first feeling that we get when we read this question is Can we apply? Can we apply integration by parts on this? Yes, so we have two functions here One is x to the power n which happens to be an Algebraic function and there's another function which is cos of x which happens to be a trigonometric function So can we start applying integration by parts on it by considering this to be my first function We can call it as you and this we can call it as a v, right? So integration by parts formula everybody knows that integral of u v with respect to x will be u integral of v Minus integral of The product of derivative of u with the integral of v Right, so this is our integration by parts formula integration by parts also called as ibp integration by parts Okay, so let us use this formula over here. Let's use this formula So considering u to be my x to the power n and v to be my cos of x Can I say I can write i n x to the power n? cos of x integration is sin x and Mind you when you are doing integration by parts with the limits with the limits of integration You have to put the limits of integration immediately after writing this expression minus Integral from 0 to pi by 2 Derivative of x to the power n will be n x to the power n minus 1 into sin of x dx Okay, so far so good. No issues with that Alright my friends. So here if you want to write this down you can simplify this as Pi by 2 to the power n So basically I'm putting the upper limit so pi by 2 to the power n sin pi by 2 is going to be 1 and Putting a 0 will give everything 0 The other term will become integral of n times 0 to pi by 2 x to the power n minus 1 sin x dx Right. So basically you can call this as you can call this as i n minus 1 as well Because you're just changing the index over here. You're just changing the index over here from n to n minus 1 But mind you in my question the requirement is to strike a relation with i n minus 2 Yes, my dear friends i n minus 2. So what I need to do here? Obviously, I need to apply integration by parts once again on this fellow Yes, I need to apply integration by parts once again to that and let's see What do we achieve when we apply integration by parts once again to that? Okay, let's start applying integration by parts again for doing integration by parts I will take this to be you and this to be V Right, so let's see. What is the result? So x to the power n minus 1 integral of sin is going to be negative cos x Okay, limits of integration. We all know 0 to pi by 2 Minus but if you realize there's already a minus which will come along with cos x so that I can write directly as a plus 0 to pi by 2 then you'll have n minus 1 x to the power n minus 2 sin x dx. Yes or no, is it fine? Everybody's happy so far All right, so let's move on. So when we further simplify this a rather, you know, write it in a more simple-looking format All of you please pay attention. The moment you put a pi by 2 as the upper limit in this expression it's going to become a 0 because cos x is sitting over there and The moment you put the lower limit which is 0 in this again This is going to become a 0 because x to the power n minus 1 is sitting there so this guy is anyways going to be a 0 okay, and May I say that this is going to be n minus 1 times i n minus 2? Yes or no? Yes or no? Yes or no? Okay, so if you agree with me on that you must also agree with me that this further Simplifies to this expression this further simplifies to this expression I hope nobody has any problem up till now any problem up till now. Okay, great so now now Let's bring down The nn minus 1 i n minus 2 to the left-hand side So we have pi by 2 to the power n Okay, and let's also divide it by n factorial because that is what we are expected to arrive at in the summation So as you can see here clearly you have been given that there is an n factorial which is sitting below i n Right, so in light of that if I divide it by n factorial Okay, all of you please pay attention over here because some factorial properties are going to surface out over here Correct. So now nn minus 1 will cancel out with the n factorial to a large extent giving you n minus 2 factorial Yes, okay, so we have basically simplified that expression to a term which is going to be Looking somewhat like that. Is it fine any questions any concerns so far, please Let me know in the comment box Now what we are doing we are summing this up. We are summing this up from 2 to infinity. Yes, my dear friends 2 to infinity Let's not have let's not forget this that the summation is going from 2 all the way till infinity So summation of this expression i n upon n factorial plus i n minus 2 by n minus 2 factorial from n equal to 2 to infinity which is as good as You summing 1 by n factorial pi by 2 to the power n from n equal to 2 to infinity Right. Okay. Now guys and girls I would like to draw your attention here to the McLaren series expansion for e to the power x Right. So all of you have seen the McLaren series expansion for e to the power x which runs like this 1 plus x plus x square by 2 factorial plus x cube by 3 factorial plus x 4 by 4 factorial and So on all the way till infinity isn't it? Okay, now in this expression twice substituting twice substituting your x as a pi by 2, right? So when you do so The right hand side of the expression will appear to be somewhat like this Pi by 2 square 1 by 2 factorial then we have pi by 2 cube divided by 3 factorial and Pi by 2 to the power 4 divided by 4 factorial and all the way till infinity. Okay Now what I'm going to do is I'm going to bring the 1 and the pi by 2 to the left Thereby leaving me with an expression on the right side, which will basically Look like this Right all the way till infinity Now don't you think don't you realize that this expression on the right side is the summation here that we require So if you see this summation, it is basically saying the summation of 1 by 2 factorial pi by 2 to the power 2 1 by 3 factorial pi by 2 to the power 3 1 by 4 factorial pi by 2 to the power 4 and so on and so forth till infinity So isn't that the summation that we have already got on the right side here? So this is 1 by n factorial pi by 2 to the power n n going from 2 to infinity Right, so that leaves us with actually the solution to this question So what we needed is this answer? So this becomes the answer to this given question So yes, my dear friends So since we required this expression which is equal to this expression and this expression happens to be Equal to this fellow. Can I say this is my final answer to this question which happens to be which of the options? Let's find it out. Let's find it out. So e to the power pi by 2 minus 1 minus pi by 2 Which is obviously option number 8. Thank you so much for watching. Stay safe. Stay healthy