Answer
\[\frac{1-\cos \left( {{\pi }^{2}} \right)}{2}\]
Work Step by Step
\[\begin{align}
& \int_{0}^{\pi }{\int_{x}^{\pi }{\sin {{y}^{2}}}dydx} \\
& \text{Using the graph to switch the order of integration} \\
& R=\left\{ \left( x,y \right):0\le x\le y,\text{ 0}\le y\le \pi \text{ } \right\} \\
& \text{Then}, \\
& \int_{0}^{\pi }{\int_{x}^{\pi }{\sin {{y}^{2}}}dydx}=\int_{0}^{\pi }{\int_{0}^{y}{\sin {{y}^{2}}}dxdy} \\
& \text{Integrating} \\
& =\int_{0}^{\pi }{\left[ x\sin {{y}^{2}} \right]_{0}^{y}dy} \\
& =\int_{0}^{\pi }{y\sin {{y}^{2}}dy} \\
& =\frac{1}{2}\int_{0}^{\pi }{\left( 2y \right)\sin {{y}^{2}}dy} \\
& =-\frac{1}{2}\left[ \cos {{y}^{2}} \right]_{0}^{\pi } \\
& =-\frac{1}{2}\left[ \cos \left( {{\pi }^{2}} \right)-\cos \left( {{0}^{2}} \right) \right] \\
& =-\frac{1}{2}\left[ \cos \left( {{\pi }^{2}} \right)-1 \right] \\
& =\frac{1-\cos \left( {{\pi }^{2}} \right)}{2} \\
\end{align}\]
