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

  • Thread starter -EquinoX-
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1. The problem statement, all variables and given/known data

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

2. Relevant equations



3. The attempt at a solution
 

Dick

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

HallsofIvy

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

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

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So, since you cannot integrate sin(x2) in elementary functions, reverse the order of integration, as I suggested.
 
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?
 
1. The problem statement, all variables and given/known data

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

2. Relevant equations



3. The attempt at a solution
how i can evaluate it
gi me now
 
please give me a solution no to me
many thanks to you.
 

HallsofIvy

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

HallsofIvy

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

HallsofIvy

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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.
 
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
 

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