I'm bored and am humbly seeking a little entertainment...

Might want to use Euler's formulas here.

Did anyone win?

Solved it, what now?

Post the proof to this thread with your steam ID

Use latex

OP i just finished a class on Intro to Numerical Analysis and I know a little about Monte Carlo integration scheme and Gauss Quadrature integration scheme. What do I do

Solve it however you want. OP will deliver. I myself didn't have to use any advanced mathematics to solve this integral. Not sure if it'd help.

I have added you on steam

Am I too late? I've been writing on a sheet of paper trying to figure it out. If x123z solved it then post the solution

...

The second line implies that cos^p (x) = .5(1 + cos(px)).

Take p = 1. Then you are claiming that cos(x) = .5 + cos(x). How is this true?

[math] \int \frac{sin(nx)}{sin(x)} dx [/math]

is bad enough

those shitty games aint worth the time

no thanks

No idea hahaha I did not understand that part, I have never worked with integrals or this advanced math, so i used this as a way to learn and maybe get some new games. That cos^x part i still do not understand, but i can proudly say i can solve some basic integrals now :)(p.s I saw it somewhere on the net)

Sheesh. I've spent over an hour on this problem and have the integral as the sum of a bunch of things like e^(nix) and then I made it into the sum of a bunch of trig functions but I'm definitely stuck. I think OP is trolling. He said he solved it without using "any advanced mathematics" but I think it's required to solve

I agree, I would be very surprised if there's a closed form (i.e. no arbitrary sums/products), at least one that doesn't involve some stupid obscure function that only the stackexchange integration autists know of.

Just testing out simple cases like cos^2(x)/sin(3x) or cos(x)/sin(7x) on a computer returns disgusting looking dogpiles very fast.

I suppose it might be possible to break it apart with a tan substitution but the nx is going to make a mess of the denominator and the final form is still going to be a bajillion terms long.

I would be happy to be wrong though.

It is a finite sum and doesn't have very many terms. Cos(x)/sin(7x) would have 6 terms.

You definitely only need calculus 2 and some trig formulas. No infinite sums or advanced math.

Careful on the 3rd line

What is wrong in it? We do not have this in school yet so i had to study it on my own for an hour.

[math]\int \frac{cos^p(x)}{sin(nx)}dx=\frac{ln(x)}{n}+\frac{(n^2-3p)x^2}{2(6n)}+[/math]
[math]\frac{(7n^4-30n^2p+15p(3p-2))x^4}{4(360n)}+\frac{(31n^6-147n^4p+105n^2p(3p-2)-21p(15p^2-30p+16))x^6}{6(15120n)}+...[/math]

Come on now. What trig integral doesn't have trig functions in the antiderivative?

In the 3rd line you said
[eqn] \int \cos(px) \csc(nx) dx = \int \cos(px) dx + \int \csc(nx) dx [/eqn]

This is simply not true. You aren't allowed to separate integrals like that

How do i do them then? I really wanto understand this

That's the thrill of the hunt my friend. There's always a set of tricks for doing integrals that are standard to learn, but sometimes an integral requires a nonstandard trick and you have to find it on your own.

Yea, that seems to be the case. OP is just trolling us. You need way more than calc 2 because that's what I have and there is definitely no substitution or method to integrate this. There is no tried and true way to power reduce a trigonometric function to an arbitrary power

Here's a hint and proof I'm not lying. Check it in your calculators if you want.
[math] p

N^* ?

natural numbers excluding 0

the first reply was meant for this answer

One that you write as an infinite sum

How did you get this?

wolfram alpha, looked up the series expansion for the problem, integrated the terms that I saw. ezpz

wolframalpha.com/input/?i=cos^p (x)/sin(nx)

Impressive. Practically anything can be integrated in an infinite series. Glad I made the requirement to be closed form or I'd have to give you the games

First we'll prove the following formula. This proof can be safely omitted.

If [math] n [/math] is even the following formula holds. If [math] n [/math] is odd omit [math] \cos\phi [/math]
[eqn] \sin n\phi = 2^{n-1} \ \ \ \sin \phi \cos \phi \Big(\cos\phi + \cos\frac{\pi}{n}\Big) \Big(\cos\phi + \cos\frac{2\pi}{n}\Big)\cdots\Big(\cos\phi + \cos\frac{(n-1)\pi}{n}\Big) [/eqn]
[math] \textbf{Proof}: [/math] [math] P = x^{2n} - 2x^{n}y^{n}\cos n\theta + y^{2n}\ \ [/math] has roots given by [math] x^n = \frac{1}{2}\Big(2y^n\cos n\theta \pm \sqrt{4y^{2n}\cos^2 n\theta - 4y^{2n}}\ \Big)
\implies x = y(\cos n\phi \pm i\sin n\phi)^{1/n} [/math]
Hence our [math] 2n [/math] roots are given by [math] x = \exp\big(\phi + \frac{2\pi}{n}\big)[/math]. Similarly it can be shown [math] P_{k} = x^2 - 2xy \cos\Big(\phi + \frac{2k}{n}\Big) + y^2 \ [/math] has roots [math] x = exp\Big(\theta + \frac{2k}{n}\Big) [/math]. Because these polynomials share roots
[eqn] x^{2n} - 2x^{n}y^{n} \cos n\theta +y^{2n} = \Big\{x^2 - 2xy\cos \theta +y^2\Big\}\Big\{ x^2-2xy\cos\Big(\theta + \frac{2\pi}{n}\Big) + y^2 \Big\} \cdots [/eqn] to [math] n [/math] terms. Setting [math] y=0 [/math] shows there is no scaling factor. Setting [math] x=y=1,\ \theta = 2\phi [/math] gives [math] 2(1-\cos 2n\phi) = 2^n \Big\{1-\cos 2\phi\Big\}\Big\{1-\cos 2\big(\phi+\frac{\pi}{n}\big)\Big\}\cdots [/math]
where applying [math] \cos 2\phi = 1 - 2\sin^2 \phi [/math] gives
[eqn] \sin n\phi = 2^{n-1} \ \ \sin\phi\sin\Big(\phi +\frac{\pi}{n}\Big)\sin\Big(\phi+\frac{2\pi}{n}\Big)\cdots \sin\Big(\frac{(n-1)\pi}{n}\Big)[/eqn]
Now apply [math] \sin(x+y)\sin(x-y) = \cos^2 y - \cos^2 x [/math] on the first and last term, 2nd first and 2nd last terms, 3rd first and 3rd last terms, and so on. Lastly expand each [math] \cos^2 y - cos^2 x [/math] as [math] (\cos y + \cos x)(\cos y - \cos x) [/math] to get the desired formula. Moving on.

I'm gonna take a bit of a break. If someone solves it before I'm finished so be it.

You don't. It's an ability completely determined by genetics.

First we'll prove the following formula.

If [math] n [/math] is even the following formula holds. If [math] n [/math] is odd omit [math] \cos\phi [/math]
[eqn] \sin n\phi = 2^{n-1} \ \ \ \sin \phi \cos \phi \Big(\cos\phi + \cos\frac{\pi}{n}\Big) \Big(\cos\phi + \cos\frac{2\pi}{n}\Big)\cdots\Big(\cos\phi + \cos\frac{(n-1)\pi}{n}\Big) [/eqn]
[math] \textbf{Proof}: [/math] [math] P = x^{2n} - 2x^{n}y^{n}\cos n\theta + y^{2n}\ \ [/math] has roots given by [math] x^n = \frac{1}{2}\Big(2y^n\cos n\theta \pm \sqrt{4y^{2n}\cos^2 n\theta - 4y^{2n}}\ \Big)
\implies x = y(\cos n\phi \pm i\sin n\phi)^{1/n} [/math]
Hence our roots are given by [math] x = y\exp i\big(\phi + \frac{2\pi r}{n}\ \big),\ r =1,\dots,n. [/math] Similarly it can be shown [math] P_{k} = x^2 - 2xy \cos\Big(\phi + \frac{2\pi k}{n}\Big) + y^2 \ [/math] has roots [math] x = y \exp i\Big(\theta \pm \frac{2\pi k}{n}\Big) [/math]. Because these polynomials share roots
[eqn] x^{2n} - 2x^{n}y^{n} \cos n\theta +y^{2n} = \Big\{x^2 - 2xy\cos \theta +y^2\Big\}\Big\{ x^2-2xy\cos\Big(\theta + \frac{2\pi}{n}\Big) + y^2 \Big\} \cdots [/eqn] to [math] n [/math] terms. Setting [math] y=0 [/math] shows there is no scaling factor. Setting [math] x=y=1,\ \theta = 2\phi [/math] gives [math] 2(1-\cos 2n\phi) = 2^n \Big\{1-\cos 2\phi\Big\}\Big\{1-\cos 2\big(\phi+\frac{\pi}{n}\big)\Big\}\cdots [/math]
where applying [math] \cos 2\phi = 1 - 2\sin^2 \phi [/math] gives
[eqn] \sin n\phi = 2^{n-1} \ \ \sin\phi\sin\Big(\phi +\frac{\pi}{n}\Big)\sin\Big(\phi+\frac{2\pi}{n}\Big)\cdots \sin\Big(\frac{(n-1)\pi}{n}\Big)[/eqn]
Now apply [math] \sin(x+y)\sin(x-y) = \cos^2 y - \cos^2 x [/math] on the first and last term, 2nd first and 2nd last terms, 3rd first and 3rd last terms, and so on. Lastly expand each [math] \cos^2 y - cos^2 x [/math] as [math] (\cos y + \cos x)(\cos y - \cos x) [/math] to get the desired formula.

I'm gonna take a break. If someone solves it in the mean time so be it.

In the last formula line the last term needs a [math] \phi [/math] in it. Also in that line we know that the sign is positive when we took the square root because for small [math] \phi [/math] both sides are positive.

Use that formula to decompose the integral into partial fractions and then integrate the fractions (which are admittedly not easy to integrate. Try tan x/2 substitutions).

I'm abandoning this thread. If anyone wants to reach me you can reach me on steam.