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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.

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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.

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$\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} $

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