定积分s(π/2,0)cos^3xdx
问题描述：
定积分s(π/2,0)cos^3xdx
答
∫[0,π/2] cos^3xdx
=∫[0,π/2] (1-sin^2x)dsinx
=(sinx-sin^3x/3)[0,π/2]
=2/3
答
∫(0 π/2) cos³x dx
=∫(0 π/2)cosxcos²x dx
=∫(0 π/2)(1-sin²x) d(sinx)
=sinx -sin³x/3 (0 π/2)
=(1 -1/3)-(0-0)
=2/3