1. The problem statement, all variables and given/known data what is the integral of sin(x^2) dx? 2. Relevant equations 3. The attempt at a solution
It's one of those integrals like e^(-x^2) that doesn't have an elementary antiderivative. Why are you asking?
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
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
So, since you cannot integrate sin(x^{2}) 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?
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]
According to Maple, it is a Fresnel S integral... [itex]\int \sin(x^2)\,dx = \frac{\sqrt {2}\sqrt {\pi }}{2}\,{\rm S} \left( {\frac {\sqrt {2}x}{ \sqrt {\pi }}} \right) [/itex]
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.
So you resurrected this thread from over a year ago just to say you did not understand it? 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.
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