What is the integral of sin(x^2) dx?

[-EquinoX-]

Homework Statement

what is the integral of sin(x^2) dx?

Homework Equations

The Attempt at a Solution

 

[Dick]

It's one of those integrals like e^(-x^2) that doesn't have an elementary antiderivative. Why are you asking?

 

[HallsofIvy]

That does not have an integral in terms of elementary functions.

 

[-EquinoX-]

as I am asked to calculate integral from y^2 to 25 of y * sin(x^2) dx and I am stuck with the sin(x^2), the y can be treated as a constant. Can you give some help?

tried to use some online help here and the result was just bizarre:
http://www.numberempire.com/integralcalculator.php

 

[Dick]

Is that REALLY the whole problem? Or is there more you aren't telling us about?

 

[-EquinoX-]

this is the whole problem:

integral from 0 to 5, integral from y^2 to 25 of y * sin(x^2) dx dy

It's a double integral

 

[HallsofIvy]

So, since you cannot integrate sin(x²) in elementary functions, reverse the order of integration, as I suggested.

 

[-EquinoX-]

ok so after reversing it I have integral from 0 to 25 , integral from 0 to sqrt(x) of y sin(x^2) dy dx. Doing the first integration results in integral from 0 to 25 of (sin(x^2)*x)/2 and I got -cos(x^2)/4 evaluated from 0 to 25. Is this correct so far?

 

[bekibless]

Homework Statement

what is the integral of sin(x^2) dx?

Homework Equations

The Attempt at a Solution

how i can evaluate it
gi me now

 

[bekibless]

please give me a solution no to me
many thanks to you.

 

[HallsofIvy]

Why have you suddenly jumped into this thread from back in February? Did you not read the previous posts? As said in the very first response, $sin(x^2)$ does NOT have an elementary anti-derivative.

After EquinoX told us that the problem was really
$$\int_{y= 0}^5\int_{x= y^2}^{25} y sin(x^2)dx dy$$
it was suggested that he reverse the order of integration. Doing that it becomes
$$\int_{x= 0}^{25}}\int_{y= 0}^{\sqrt{x}} y sin(x^2)dy dx$$
$$= \frac{1}{2}\int_{x= 0}^{25}\left[y^2\right]_{y=0}^{\sqrt{x}} sin(x^2) dx$$
$$= \frac{1}{2}\int_{x= 0}^{25} x sin(x^2) dx$$
which can be integrated by using the substitution $u= x^2$:
If $u= x^2$, du= 2x dx so x dx= (1/2)du. When x= 0, u= 0 and when x= 25, u= 625 so the integral is
$$\frac{1}{4}\int_0^{625} sin(u) du= -\frac{1}{4}\left[cos(u)\right]_0^{625}$$
$$= -\frac{1}{4}(-0.984387)= 0.246097$$

 

[g_edgar]

According to Maple, it is a Fresnel S integral...
$\int \sin(x^2)\,dx =<br /> \frac{\sqrt {2}\sqrt {\pi }}{2}\,{\rm S} \left( {\frac {\sqrt {2}x}{<br /> \sqrt {\pi }}} \right)$

 

[anushyan88]

sin x^2 = 1 - cos 2x
and we can use 1 and cos 2x seperatly and solve this problem.

 

[HallsofIvy]

sin x^2 = 1 - cos 2x
and we can use 1 and cos 2x seperatly and solve this problem.

No, "sin x^2" MEANS sin(x^2) and cannot be integrated in that way. If your function is really (sin(x))^2= sin^2(x), you should have told us that immediately.

 

[zooboodoo]

if we do a maclaurin series expansion on sin(x^2) can't we use that to find the integral of sin(x^2)dx?

 

[clamtrox]

Of course; the solution to integrals almost always exists, even if you cannot express it in terms of elementary functions. This means that the solution series won't have a nicely identifiable set of coefficients -- you'll need to leave it in the series form.

 

[jaydeepsamant]

INTsin x^2dx
=INT(1-cos2x)/2.dx
=1/2INTdx-INTcos2xdx
=x/2-sin2x/2

 

[HallsofIvy]

So you resurrected this thread from over a year ago just to say you did not understand it?

The original question was to integrate $sin(x^2)$, NOT $sin^2(x)$ for which your solution would be appropriate.

That was said back in November of 2009.

 

[Mstf_akkoc]

I think I have a solution. I hope it was not so late :).

tan(x^2)=m

dx=cos(x^2)dm integral [sin(x^2)] = integral [mdm/(m^2+1)]

m^2+1=a and 2mdm=da ...

integral [sin(x^2)] = 0.5*ln[(tan(x^2))^2+1]+c

 

[micromass]

dx=cos(x^2)dm

Why is this true?

 

[Char. Limit]

Why is this true?

Well, if dx = cos(x^2) dm, then sec(x^2) = dm/dx and m=... I got lost.

 

[Mstf_akkoc]

Actualy is not true. I have done a misteke when calculete the [tan(x^2)]'

2xdx=cos^2(x^2)dm is true

if I find a solution with this I ll write.

 

[Char. Limit]

Actualy is not true. I have done a misteke when calculete the [tan(x^2)]'

2xdx=cos(x^2)dm is true

if I find a solution with this I ll write.

No no no.

If tan(x^2) = m, then $$2x sec^2(x^2) dx = dm$$ and $$2x dx = cos^2(x^2) dm$$

Can you see why?

 

[CLouD8521]

Why have you suddenly jumped into this thread from back in February? Did you not read the previous posts? As said in the very first response, $sin(x^2)$ does NOT have an elementary anti-derivative.

After EquinoX told us that the problem was really
$$\int_{y= 0}^5\int_{x= y^2}^{25} y sin(x^2)dx dy$$
it was suggested that he reverse the order of integration. Doing that it becomes
$$\int_{x= 0}^{25}\int_{y= 0}^{\sqrt{x}} y sin(x^2)dy dx$$
$$= \frac{1}{2}\int_{x= 0}^{25}\left[y^2\right]_{y=0}^{\sqrt{x}} sin(x^2) dx$$
$$= \frac{1}{2}\int_{x= 0}^{25} x sin(x^2) dx$$
which can be integrated by using the substitution $u= x^2$:
If $u= x^2$, du= 2x dx so x dx= (1/2)du. When x= 0, u= 0 and when x= 25, u= 625 so the integral is
$$\frac{1}{4}\int_0^{625} sin(u) du= -\frac{1}{4}\left[cos(u)\right]_0^{625}$$
$$= -\frac{1}{4}(-0.984387)= 0.246097$$

I have a problem. Isn't:
$$\left[cos(u)\right]_0^{625} = cos(625)-cos(0) = (-0.984387)-(1)=-1.984387$$

Therefore;
$$\frac{1}{4}\int_0^{625} sin(u) du = -\frac{1}{4}(-1.984387)= 0.496097$$