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Integral of sin(x^2) 
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#1
Feb2809, 03:46 PM

P: 577

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 


#2
Feb2809, 03:55 PM

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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?



#3
Feb2809, 03:56 PM

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



#4
Feb2809, 04:00 PM

P: 577

Integral of sin(x^2)
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
Feb2809, 04:14 PM

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



#6
Feb2809, 04:17 PM

P: 577

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
Feb2809, 04:34 PM

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So, since you cannot integrate sin(x^{2}) in elementary functions, reverse the order of integration, as I suggested.



#8
Feb2809, 04:43 PM

P: 577

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
Jul709, 03:53 AM

P: 2

gi me now 


#10
Jul709, 03:55 AM

P: 2

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


#11
Jul709, 05:22 AM

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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 antiderivative.
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] 


#12
Jul709, 08:33 AM

P: 607

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
Nov2109, 04:20 AM

P: 1

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


#14
Nov2109, 04:40 AM

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P: 39,683




#15
Dec409, 03:59 AM

P: 29

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
Dec409, 04:24 AM

P: 939

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
Oct1210, 03:13 AM

P: 1

INTsin x^2dx
=INT(1cos2x)/2.dx =1/2INTdxINTcos2xdx =x/2sin2x/2 


#18
Oct1210, 05:29 AM

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P: 39,683

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. 


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