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

I have no idea how you got -0.984387. Please advise.You are correct. There was a mistake in the calculation. It should be:\frac{1}{4}\int_0^{625} sin(u) du= -\frac{1}{4}\left[cos(u)\right]_0^{625}= -\frac{1}{4}(cos(625)-cos(0))=-\frac{1}{4}(-0.984387-1)= 0.496097
  • #1
-EquinoX-
564
1

Homework Statement



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

Homework Equations





The Attempt at a Solution

 
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  • #2
It's one of those integrals like e^(-x^2) that doesn't have an elementary antiderivative. Why are you asking?
 
  • #3
That does not have an integral in terms of elementary functions.
 
  • #4
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
 
  • #5
Is that REALLY the whole problem? Or is there more you aren't telling us about?
 
  • #6
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
 
  • #7
So, since you cannot integrate sin(x2) in elementary functions, reverse the order of integration, as I suggested.
 
  • #8
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?
 
  • #9
-EquinoX- said:

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
 
  • #10
please give me a solution no to me
many thanks to you.
 
  • #11
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, [itex]sin(x^2)[/itex] does NOT have an elementary anti-derivative.

After EquinoX told us that the problem was really
[tex]\int_{y= 0}^5\int_{x= y^2}^{25} y sin(x^2)dx dy[/tex]
it was suggested that he reverse the order of integration. Doing that it becomes
[tex]\int_{x= 0}^{25}}\int_{y= 0}^{\sqrt{x}} y sin(x^2)dy dx[/tex]
[tex]= \frac{1}{2}\int_{x= 0}^{25}\left[y^2\right]_{y=0}^{\sqrt{x}} sin(x^2) dx[/tex]
[tex]= \frac{1}{2}\int_{x= 0}^{25} x sin(x^2) dx[/tex]
which can be integrated by using the substitution [itex]u= x^2[/itex]:
If [itex]u= x^2[/itex], du= 2x dx so x dx= (1/2)du. When x= 0, u= 0 and when x= 25, u= 625 so the integral is
[tex]\frac{1}{4}\int_0^{625} sin(u) du= -\frac{1}{4}\left[cos(u)\right]_0^{625}[/tex]
[tex]= -\frac{1}{4}(-0.984387)= 0.246097[/tex]
 
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  • #12
According to Maple, it is a Fresnel S integral...
[itex]\int \sin(x^2)\,dx =
\frac{\sqrt {2}\sqrt {\pi }}{2}\,{\rm S} \left( {\frac {\sqrt {2}x}{
\sqrt {\pi }}} \right)
[/itex]
 
  • #13
sin x^2 = 1 - cos 2x
and we can use 1 and cos 2x seperatly and solve this problem.
 
  • #14
anushyan88 said:
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.
 
  • #15
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?
 
  • #16
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.
 
  • #17
INTsin x^2dx
=INT(1-cos2x)/2.dx
=1/2INTdx-INTcos2xdx
=x/2-sin2x/2
 
  • #18
So you resurrected this thread from over a year ago just to say you did not understand it?

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

That was said back in November of 2009.
 
  • #19
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
 
  • #20
Mstf_akkoc said:
dx=cos(x^2)dm

Why is this true?
 
  • #21
micromass said:
Why is this true?

Well, if dx = cos(x^2) dm, then sec(x^2) = dm/dx and m=... I got lost.
 
  • #22
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.
 
Last edited:
  • #23
Mstf_akkoc said:
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 [tex]2x sec^2(x^2) dx = dm[/tex] and [tex]2x dx = cos^2(x^2) dm[/tex]

Can you see why?
 
  • #24
HallsofIvy said:
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, [itex]sin(x^2)[/itex] does NOT have an elementary anti-derivative.

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

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

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

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

The integral of sin(x^2) dx cannot be expressed in terms of elementary functions. It is an unsolved integral known as the Fresnel integral.

2. Is there a way to approximate the integral of sin(x^2) dx?

Yes, there are various methods for approximating this integral, such as Simpson's rule or the Riemann sum.

3. Can the integral of sin(x^2) dx be solved using numerical methods?

Yes, numerical methods such as the trapezoidal rule or Gaussian quadrature can be used to approximate the value of this integral.

4. What is the significance of the integral of sin(x^2) dx in mathematics?

The integral of sin(x^2) dx is a special case of the Fresnel integral, which has applications in optics, diffraction, and quantum mechanics.

5. How can I evaluate the integral of sin(x^2) dx using software?

There are various mathematical software packages, such as Mathematica or MATLAB, that have built-in functions for evaluating this integral numerically or symbolically.

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