原式=4∫[0,π/2][1-(cosθ)^2][1+(cosθ)^2]dθ=4∫[0,π/2](1/4)[1-cos2θ)(3+cos2θ)dθ=∫[0,π/2][3-2cos2θ-(cos2θ)^2]dθ=[0,π/2][3θ-sin2θ]-(1/2)∫[0,π/2](1+cos4θ)dθ=3π/2-0-(1/2)θ[0,π/2]-(1/8...=4∫[0,π/2](1/4)[1-cos2θ)(3+cos2θ)dθ這里的[1-cos2θ)(3+cos2θ)是怎么做出來的？=[1-(1+cos2θ)/2][1+(1+cos2θ)/2]=(1/4)(2-1-cos2θ)(2+1+cos2θ)=(1/4)(1-cos2θ)(3+cos2θ)(cosθ)^2=(1+cos2θ)/2.