解：∫(cos²θsin²θ)dθ =¼∫(2cosθsinθ)²dθ =¼∫sin²2θdθ =¼∫½(1-cos4θ)dθ =(1/8)∫(1-cos4θ)dθ =(1/8)(θ-¼sin4θ)+C
=(1/4)∫(sin2x)^2dx=(1/8)∫(1-cos4x)dx=(1/8)(x-sin4x/4)+C
