解：分享一种解法。∵sinθsin(θ+ωτ)dθ=(1/2)[cosωτ-cos(2θ+ωτ)]dθ，∴Rx(τ)=(x0)^2/(2π){cosωτ[θ/2-(1/4)sin(2θ+ωτ)/4]}丨(θ=0,2π)=(cosωτ)[(x0)^2]/2。供参考。
