# integral of sin(x^2)

by -EquinoX-
Tags: integral, sinx2
 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
Mentor
P: 16,698
 Quote by Mstf_akkoc dx=cos(x^2)dm
Why is this true?
PF Gold
P: 1,930
 Quote by micromass Why is this true?
Well, if dx = cos(x^2) dm, then sec(x^2) = dm/dx and m=... I got lost.
 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.
PF Gold
P: 1,930
 Quote by Mstf_akkoc 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?
P: 1
 Quote by HallsofIvy 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;
$$\frac{1}{4}\int_0^{625} sin(u) du = -\frac{1}{4}(-1.984387)= 0.496097$$

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