Evaluate $\int(x^{91}+x^{327})\cos(x)\mathrm{d}x \quad .$

Question

Evaluate $$\int\left(x^{91}+x^{327}\right)\cos(x)\mathrm{d}x \quad .$$

It's my first time to face integration like that. I just need a clue to start because I tried, but it's not working

Thanks in advance

Answer 1

If you're looking for a primitive, I wish you good luck, but a start could be:

two integration by parts (integrate twice the cos) to get something that I hope will be a recurrence between $I_n$ and $I_{n-2}$

Answer 2

Integrating by parts: $$ I_n = \int x^n\cos x dx = x^n\sin x + n\int x^{n-1}(-\sin x) dx $$ and $$ \int x^{n-1}(-\sin x) dx = x^{n-1}\cos x - (n-1)\int x^{n-2} \cos x dx =x^{n-1}\cos x - (n-1)I_{n-2} $$ So, $$ I_n = x^n\sin x + nx^{n-1}\cos x - n(n-1)I_{n-2}. $$

Answer 3

{option 1} There is a sequence in the answer. You expect something of the form (for the $x^{327}$ term):

$ \sum^{327}_{k=0} a_k x^k \cos{x}$.

If you can postulate the sequence, you can show it fits the things you've written out so far. That should give you enough to complete a proof by induction of the postulated answer.

{option 2} What I, personally, would do here is look at the taylor expansion of $x^n \cos{x}$. That's a power series, so it is easily integrable; use that.

E.g. $f(x) = x^{327} \cos{x}$

$f_1(x) = 327 x^{326} \cos{x} - x^{327} \sin{x}$

$f_2(x) = 327 \cdot 326 x^{325} \cos{x} -327x^{326} \sin{x} - 327x^{326}\sin{x} - x^{327} \cos{x}$

From this alone, I would surmise something along the lines (NB: I've just given you some sort of form that seems valid. It's not fitted to what $f_1$ and $f_2$ look like.): $f_k(x) = \sum^k_{n=1} {k \choose n} x^{327-n} \cos{x} + g_k(x)$ where $g_k(x)$ is a sum similar to $f_k(x) - g_k(x)$.

Just write out a few terms, and try to see some kind of sequence in it. If you can show that the terms you've written out do indeed follow the postulated sequence, and then take a random $k$ and differentiate it, it should equal $f_{k+1}$. If it does, you've met the requirements for a proof by induction and you have found a series $f_k(x)$.

You can then succintly write the taylor expansion of $f(x)$ around C and thus integrate your function (because the taylor expansion is a power series). Fill in the sequence for $f_k(x)$ and you have a full expression for the integral. It might not be pretty, but as long as you take infinitely many terms for the taylor series, it is the function it is expanding; so it is an analytic solution.

That's the only way I see to go. Keep in mind that this kind of exercise is a lot of work, and that my own teachers weren't mean enough to give this to us. On the other hand, it's a very good exercise to do because you have to be very careful/tidy.

Answer 4

By brute force and with the help of previous answers, I think the answer is:

If $n$ is even, then:

$$I_n = \sum_{k=0}^{n-2} (-1)^k (2k)!\, x^{n-2k} \sin x + \sum_{k=0}^{n-2} (-1)^k k!\, x^{n-(2k+1)} \cos x + (-1)^{n/2} n!\, I_0,$$

with $I_n=\int x^n \cos x\, dx$ and $I_0=\sin x$. If $n$ is odd, then:

$$I_n = \sum_{k=0}^{n-1} (-1)^k (2k)!\, x^{n-2k} \sin x + \sum_{k=0}^{n-1} (-1)^k k!\, x^{n-(2k+1)} \cos x + (-1)^{(n-1)/2}(n-1)!\,(x \sin x+\cos x)$$