integral of sin(x^2)


by -EquinoX-
Tags: integral, sinx2
Mstf_akkoc
Mstf_akkoc is offline
#19
Dec8-10, 03:44 PM
P: 2
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
micromass
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#20
Dec8-10, 04:08 PM
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Quote Quote by Mstf_akkoc View Post
dx=cos(x^2)dm
Why is this true?
Char. Limit
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#21
Dec8-10, 04:16 PM
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Quote Quote by micromass View Post
Why is this true?
Well, if dx = cos(x^2) dm, then sec(x^2) = dm/dx and m=... I got lost.
Mstf_akkoc
Mstf_akkoc is offline
#22
Dec8-10, 04:40 PM
P: 2
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.
Char. Limit
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#23
Dec8-10, 04:42 PM
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Quote Quote by Mstf_akkoc View Post
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?
CLouD8521
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#24
May8-12, 11:57 AM
P: 1
Quote Quote by HallsofIvy View Post
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]


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