利用定积分：
∫[0,π/2]cosxdx
=∫[0,π/2]cosxdx
=sinx[0,π/2]
=1
对∫[0,π/2]πcos^2xdx
=∫[0,π/2]π/2(1+cos2x)dx
=π/2(x+1/2sin2x)[0,π/2]
=π/2(π/2-0)
=π^2/4
1,求曲线y=cosx（0
利用定积分：
∫[0,π/2]cosxdx
=∫[0,π/2]cosxdx
=sinx[0,π/2]
=1
对∫[0,π/2]πcos^2xdx
=∫[0,π/2]π/2(1+cos2x)dx
=π/2(x+1/2sin2x)[0,π/2]
=π/2(π/2-0)
=π^2/4