# Solve the definite integral of $\sin(\pi x^2)$

Compute $$\int_{y=0}^{y=1} \int_{x=y}^{x=1} \sin(\pi x^2) \;dx \;dy.$$

Wolfram returns this definite integral as $1/{\pi}$. However, I am struggling to figure out the solution to this integral, as I am unaware of any trig identity, substitution, or integration by parts that would apply to $\sin(\pi x^2)$. In trying to applying Fubini's Theorem, you would still end up with the same integrand, so it does not seem to help me in this case. Any help or guidance appreciated.

Using Fubini's theorem, $$\int_0^1\int_y^1\sin(\pi x^2)\;dxdy=\int_0^1\int_0^x\sin(\pi x^2)\;dydx=\int_0^1x\sin(\pi x^2)\;dx$$ and this integral can now be handled with a substitution.
$\begin{array}\\ \int_{0}^{1} \int_{y}^{1} \sin(\pi x^2) dx dy &= \int_{0}^{1} \int_{0}^{x}\sin(\pi x^2) dy dx\\ &= \int_{0}^{1} \sin(\pi x^2) \int_{0}^{x} dy dx\\ &= \int_{0}^{1} x \sin(\pi x^2) dx\\ &= \dfrac12 \int_{0}^{1} \sin(\pi z) dz\\ &= \dfrac1{2\pi} \int_{0}^{\pi} \sin( z) dz\\ &= \dfrac1{2\pi} (-\cos(z)|_{0}^{\pi} )\\ &=\dfrac1{\pi} \end{array}$