00:01
Okay, so for part a, let's start by just looking at the mclaurin series for cosine of x.
00:07
So we've got those that cosine of x is equal to the sum from k equals 0 to infinity of negative 1 to the k times x to the 2k all over 2k factorial.
00:20
Okay, so and then what we want to find is this cosine of x squared.
00:24
So what is that? okay, so first i'm really, i'm really, i'm just going to rewrite this expression for cosine of x, just using a u.
00:34
And it's not saying, i mean, it's saying the exact same thing, but i think thinking of it as a different variable that we're going to substitute in for, is a little nicer to think about.
00:46
So now if we let this u equal x squared, well, then if we just plug in, then we've got us that cosine of x squared is equal to the sum, from k equals 0 to infinity of negative 1 to the x times x squared to the 2k over 2k factorial.
01:09
So what i did is i just replaced this u here just exactly with the x squared.
01:14
I just straight substituted in.
01:16
And now we can simplify this a tiny bit, just bring this like a simple, from k equals 0 to infinity of this negative 1 to the k times x to the 4k all over 2k factorial.
01:36
And this here at the end, this is our final answer for part a.
01:39
This is the mclaurin series for cosine of x squared.
01:44
And that's that.
01:45
So, okay, and then for b, well, so there's a definite integral that we want to compute.
01:50
Well, let's start by looking at the indefinite integral.
01:52
So this is the indefinite integral of cosine of x squared, dx.
01:58
And well, okay, so we've got an expression for cosine of x squared, this mclaurin series that we just computed.
02:04
So let's use that.
02:05
Let's plug that in.
02:07
So now this is equal to the integral from k equals 0 to infinity of negative 1 to the k times x to the 4k, all of our 2k factorial dx.
02:19
Okay.
02:20
So i just replace that expression for cosine of x squared with the other one that we have.
02:24
And then, so the important factor is that we're allowed to bring this interoperable.
02:28
Sign inside of our big sum.
02:31
So this line here is an integral of a sum, but we can switch it and change it to be a sum of integrals.
02:41
So now we're looking at the sum of the integral from, yeah, so the integral of negative 1 to the k times x to the 4k over 2k factorial.
02:53
And then, so now we'd just like to compute each of those integrals.
02:57
Well, so first let's just bring out, our constants out in front of the integral.
03:02
So that's the negative 1 to the k over 2k factorial...