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