Answer
\[\frac{{{\pi }^{2}}}{8}\]
Work Step by Step
\[\begin{align}
& \int_{0}^{\pi }{\int_{0}^{\pi /2}{{{\sin }^{2}}x}{{\cos }^{2}}y}dydx \\
& =\int_{0}^{\pi }{{{\sin }^{2}}x\left[ \int_{0}^{\pi /2}{{{\cos }^{2}}y}dy \right]}dx \\
& \text{Integrating }\int_{0}^{\pi /2}{{{\cos }^{2}}y}dy \\
& \int_{0}^{\pi /2}{{{\cos }^{2}}y}dy=\int_{0}^{\pi /2}{\left( \frac{1+\cos 2y}{2} \right)}dy \\
& =\left[ \frac{1}{2}y+\frac{1}{4}\sin 2y \right]_{0}^{\pi /2} \\
& =\left[ \frac{1}{2}\left( \frac{\pi }{2} \right)+\frac{1}{4}\sin 2\left( \frac{\pi }{2} \right) \right]-\left[ \frac{1}{2}\left( 0 \right)+\frac{1}{4}\sin 2\left( 0 \right) \right] \\
& =\frac{\pi }{4} \\
& \text{Then,} \\
& \int_{0}^{\pi }{{{\sin }^{2}}x\left[ \int_{0}^{\pi /2}{{{\cos }^{2}}y}dy \right]}dx=\frac{\pi }{4}\int_{0}^{\pi }{{{\sin }^{2}}x}dx \\
& =\frac{\pi }{4}\int_{0}^{\pi }{\left( \frac{1-\cos 2x}{2} \right)}dx \\
& =\frac{\pi }{4}\left[ \frac{1}{2}x-\frac{1}{4}\sin 2x \right]_{0}^{\pi } \\
& =\frac{\pi }{4}\left[ \frac{1}{2}\left( \pi \right)-\frac{1}{4}\sin 2\left( \pi \right) \right]-\frac{\pi }{4}\left[ \frac{1}{2}\left( 0 \right)-\frac{1}{4}\sin 2\left( 0 \right) \right] \\
& =\frac{{{\pi }^{2}}}{8} \\
\end{align}\]
