how do you integrate sin(x^2-3)??? i really need this answer!!!?

Answer 1

This integral is a variant of the integral of sin[(Pi/2)*x^2], which cannot be integrated in closed form using elementary functions. The integral of this function from 0 to t is defined to be FresnelS(t), a special function called the Fresnel sine integral. Similar holds for cosine.
Thus, your integral is Int[sin(x^2 - 3)]dx = Int[sin(x^2)cos(3) - cos(x^2)sin(3)]dx = Sqrt[Pi/2]*(FresnelS(Sqrt[2/Pi]*x)*cos(3) - FresnelC(Sqrt[2/Pi]*x)*sin(3)).
More about the Fresnel integrals can be found in texts on special functions, physics texts on optics, or http://mathworld.wolfram.com/FresnelIntegrals.html .

Answer 2

This can be done using power series.
Int[sin(x^2-3)]
=Int[sin(x^2)cos(3)-cos(x^2)sin(3)]
=cos(3)*Int[sin(x^2)] - sin(3)*Int[cos(x^2)]

Then evaluate the indefinite integarals
Int[sin(x^2)] and Int[cos(x^2)]
Their convergent power series are respectively:
x^3 / (3*1!) - x^7 / (7*3!) + x^11 / (11*5!) ...
and
x / (1*0!) - x^5 / (5*2!) + x^9 / (9*4!)...

Answer 3

chain rule.