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

## Homework Statement

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

## Answers and Replies

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

HallsofIvy
Science Advisor
Homework Helper
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
Science Advisor
Homework Helper
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
Science Advisor
Homework Helper
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?

## Homework Statement

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

## 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
Science Advisor
Homework Helper
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
$$\frac{1}{4}\int_0^{625} sin(u) du= -\frac{1}{4}\left[cos(u)\right]_0^{625}$$
$$= -\frac{1}{4}(-0.984387)= 0.246097$$

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

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

HallsofIvy
Science Advisor
Homework Helper
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
Science Advisor
Homework Helper
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.

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

dx=cos(x^2)dm
Why is this true?

Char. Limit
Gold Member
Why is this true?
Well, if dx = cos(x^2) dm, then sec(x^2) = dm/dx and m=... I got lost.

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:
Char. Limit
Gold Member
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?

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