Answer
$$\int_{0}^{\frac{\pi}{2}}sin^{2}x \, cos^{2}x \, dx=\frac{\pi}{16}$$
Work Step by Step
$$\int_{0}^{\frac{\pi}{2}}sin^{2}x \, cos^{2}x \, dx=\int_{0}^{\frac{\pi}{2}}(sinx \,cosx)^{2} \, dx$$
$$=\int_{0}^{\frac{\pi}{2}}(\frac{sin2x}{2})^{2} \, dx$$
$$=\int_{0}^{\frac{\pi}{2}}\frac{sin^{2}2x}{4} \, dx$$
$$=\int_{0}^{\frac{\pi}{2}}\frac{1-cos4x}{8} \, dx$$
$$=\left [\frac{1}{8}x -\frac{sin4x}{32} \right ]_{0}^{\frac{\pi}{2}}$$
$$=\frac{\pi}{16}$$
