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## Homework Statement

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

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

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Dick

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HallsofIvy

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

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tried to use some online help here and the result was just bizarre:

http://www.numberempire.com/integralcalculator.php

- #5

Dick

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

- #6

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integral from 0 to 5, integral from y^2 to 25 of y * sin(x^2) dx dy

It's a double integral

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HallsofIvy

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how i can evaluate it## Homework Statement

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

## Homework Equations

## The Attempt at a Solution

gi me now

- #10

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please give me a solution no to me

many thanks to you.

many thanks to you.

- #11

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]

After EquinoX told us that the problem was

[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

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[itex]\int \sin(x^2)\,dx =

\frac{\sqrt {2}\sqrt {\pi }}{2}\,{\rm S} \left( {\frac {\sqrt {2}x}{

\sqrt {\pi }}} \right)

[/itex]

- #13

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sin x^2 = 1 - cos 2x

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

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

- #14

HallsofIvy

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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.sin x^2 = 1 - cos 2x

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

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- #17

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INTsin x^2dx

=INT(1-cos2x)/2.dx

=1/2INTdx-INTcos2xdx

=x/2-sin2x/2

=INT(1-cos2x)/2.dx

=1/2INTdx-INTcos2xdx

=x/2-sin2x/2

- #18

HallsofIvy

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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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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

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

- #21

Char. Limit

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

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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.

2xdx=cos^2(x^2)dm is true

if I find a solution with this I ll write.

Last edited:

- #23

Char. Limit

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No no no.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.

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

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I have a problem. Isn't: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 wasreally

[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]

[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]

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