what does the integral of sin(sin(x)) equals to?

[drhu...@gmail.com]

what does the integral of sin(sin(x)) equals to? wolfram falls. what is your integrator give?


DrHuang.com

[Jim Burns]

On 2/10/2017 12:05 AM, drhu...@gmail.com wrote:
> what does the integral of sin(sin(x)) equals to?
> wolfram falls. what is your integrator give?

sin(sin(x)) very likely does not have an integral
expressible with elementary functions. However, I don't
know how to prove that, or even define elementary
function, really.

However, you could integrate the power series for
sin(sin(x)). With the 1/n! factors, it should converge
very nicely.

[abu.ku...@gmail.com]

since sine of x is a lenght at most one and at least zero,
how can sine of a length be defined

[abu.ku...@gmail.com]

it's just a scanty error about arcsine,
the inverse function of the sine of x
is the arcsine of f(x) or of y

[aweso...@gmail.com]

On Thursday, February 9, 2017 at 9:05:55 PM UTC-8, drhu...@gmail.com wrote:
> what does the integral of sin(sin(x)) equals to? wolfram falls. what is your integrator give?
>
>
> DrHuang.com

I am going to give you an answer that won't satisfy you. The integral of sin(sin x) is sin(sin x). I know it sounds ridiculous, but there is a point in integral calculus in which ridiculous and insane integrals just equal itself. It's the horseshoe method.

[Adrian Farley]

On Friday, November 3, 2017 at 7:01:47 PM UTC-4, aweso...@gmail.com wrote:
> On Thursday, February 9, 2017 at 9:05:55 PM UTC-8, drhu...@gmail.com wrote:
> > what does the integral of sin(sin(x)) equals to? wolfram falls. what is your integrator give?
> >
> >
> > DrHuang.com
>
> I am going to give you an answer that won't satisfy you. The integral of sin(sin x) is sin(sin x).

If that were true, sin(sin x) would be the exponential function.

[Jim Burns]

On 11/3/2017 7:01 PM, aweso...@gmail.com wrote:
> On Thursday, February 9, 2017 at 9:05:55 PM UTC-8,
> drhu...@gmail.com wrote:

>> what does the integral of sin(sin(x)) equals to? wolfram
>> falls. what is your integrator give?

> I am going to give you an answer that won't satisfy you.
> The integral of sin(sin x) is sin(sin x).

No, it's not. This is not helpful to anyone who
might want to know the answer for some reason --
even if Dr Huang is not such a person. Everyone
potentially reads these posts.

We know that sin(sin(x)) is not the integral of
sin(sin(x)) because, if it were, it's derivative
would be sin(sin(x)), and it's not. It's
cos(sin(x))*cos(x)

> I know it sounds ridiculous, but there is a point in
> integral calculus in which ridiculous and insane
> integrals just equal itself. It's the horseshoe method.

There's nothing ridiculous or insane about about an
integral just equaling itself. But there's only one
integrand for which that's true.
INT exp(x) dx = exp(x) + C

And there's nothing ridiculous or insane about
INT sin(sin(x)) dx

It's _difficult_ but that's a different matter.

Actually, if one asks Wolfram something like
integrate sin(sin(x)) from 0 to 1
it will give you an answer
0.4306061031206909...
integrated numerically.

There might not be an integral in terms of elementary
functions. (Possibly elliptic functions? Only a guess.)
But the integral is well-defined. It exists and can be
calculated.

Depending upon what one is using the integral for,
different transformations might be useful. Some thoughts
which I had:

----
Substitute sin(x) = y

cos(x) dx = dy
dx = dy/sqrt(1 - y&6)

INT sin(sin(x)) dx = INT sin(y)/sqrt(1 - y^2) dy

----
Expand the outer sine function as an infinite series,
Exchange the integral and infinite sum, then integrate
the terms of the form sin(x)^(2k+1) dx

(We know the exchange works because that infinite
sum converges absolutely.)

INT sin(sin x) dx
= INT [ SUM_k ((-1)^k/(2*k+1)!)*(sin x)^(2*k+1) ] dx
= INT [ (sin x) - (sin x)^3/3! + (sin x)^5/5! - ... ] dx

When we exchange the order of INT and SUM we get
= SUM_k ((-1)^k/(2*k+1)!)*[ INT (sin x)^(2*k+1) dx ]

The integrals INT (sin(x))^(2*k+1) dx integrate neatly.
The substitution u = cos(x) is useful
sin(x) dx = du
(sin x)^2 = (1 - u^2)

INT (sin(x))^(2*k+1) dx
= INT (1 - u^2)^k du

And so on. I haven't finish it, though it could be
finished. It looks like we'll get some infinite series
of the odd powers of cos(x).

Whether this is a useful way to characterize
INT sin(sin x) dx depends upon what you intend
to do with it. I fully expect that there are other ways
to "solve" this problem.

[Me]

On Saturday, November 4, 2017 at 3:22:06 AM UTC+1, Jim Burns wrote:
> On 11/3/2017 7:01 PM, aweso...@gmail.com wrote:
> > On Thursday, February 9, 2017 at 9:05:55 PM UTC-8,
> > drhu...@gmail.com wrote:
> > >
> > > what does the integral of sin(sin(x)) equals to? wolfram
> > > falls. what is your integrator give?
> > >
> > I am going to give you an answer that won't satisfy you.
> > The integral of sin(sin x) is sin(sin x).
> >
> No, it's not. This is not helpful to anyone who
> might want to know the answer for some reason --
> even if Dr Huang is not such a person. Everyone
> potentially reads these posts.

Right. I just checked this one:
https://www.wolframalpha.com/input/?i=Integral+sin(sin+x)