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

  • Thread starter -EquinoX-
  • Start date
  • #1
564
1

Homework Statement



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

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619
It's one of those integrals like e^(-x^2) that doesn't have an elementary antiderivative. Why are you asking?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
956
That does not have an integral in terms of elementary functions.
 
  • #4
564
1
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
Dick
Science Advisor
Homework Helper
26,260
619
Is that REALLY the whole problem? Or is there more you aren't telling us about?
 
  • #6
564
1
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
HallsofIvy
Science Advisor
Homework Helper
41,833
956
So, since you cannot integrate sin(x2) in elementary functions, reverse the order of integration, as I suggested.
 
  • #8
564
1
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
2
0

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
2
0
please give me a solution no to me
many thanks to you.
 
  • #11
HallsofIvy
Science Advisor
Homework Helper
41,833
956
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]
 
Last edited by a moderator:
  • Like
Likes edilson
  • #12
607
0
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
HallsofIvy
Science Advisor
Homework Helper
41,833
956
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
29
0
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
938
9
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
HallsofIvy
Science Advisor
Homework Helper
41,833
956
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
 
  • #21
Char. Limit
Gold Member
1,204
14
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
Char. Limit
Gold Member
1,204
14
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
1
0
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]
 

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

Replies
1
Views
4K
  • Last Post
Replies
4
Views
8K
Replies
5
Views
38K
  • Last Post
Replies
5
Views
2K
Replies
3
Views
2K
Replies
6
Views
10K
  • Last Post
Replies
2
Views
1K
Replies
2
Views
2K
  • Last Post
Replies
7
Views
57K
Replies
7
Views
3K
Top