原式=∫(0到2π)(1-cos²t)(1+cost)dt
=∫(0到2π)sin²t(1+cost)dt
令t=x+π
则原式=∫(-π到π)sin²(x+π)[1+cos(x+π)]d(x+π)
=∫(-π到π)sin²x(1-cosx)dx
∵y=sin²x(1-cosx)是一个偶函数
∴原式=2∫(0到π)sin²x(1-cosx)dx
令x=s+π/2
则原式=2∫(-π/2到π/2)sin²(s+π/2)[1-cos(s+π/2)]d(s+π/2)
=2∫(-π/2到π/2)cos²s(1+sins)ds
=2[∫(-π/2到π/2)cos²sds+∫(-π/2到π/2)cos²ssinsds]
∵y=cos²x为偶函数,y=cos²ssins为奇函数
∴原式=2∫(-π/2到π/2)cos²sds=4∫(0到π/2)cos²sds
令s=π/2-u,原式=4∫(π/2到0)cos²(π/2-u)d(π/2-u)=-4∫(π/2到0)sin²udu=4∫(0到π/2)sin²udu
又因为定积分与积分变量用什么字母表示无关
∴原式=4∫(0到π/2)sin²tdt