integral of sin(x^2)

dx=cos(x^2)dm

Why is this true?

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.

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]