Question

∫0到2π 2π(1-cosx)^2dx 详解过程 谢谢

Answer 1

∫(0→2π) 2π(1 - cosx)² dx
= 2π∫(0→2π) (1 - 2cosx + cos²x) dx
= 2π[∫(0→2π) (1 - 2cosx) dx] + 2π∫(0→2π) (1 + cos2x)/2 dx
= 2π[x - 2sinx] |(0→2π) + π[x + 1/2 * sin2x] |(0→2π)
= 2π(2π - 0) + π(2π - 0)
= 4π² + 2π²
= 6π²

Answer 2

∫(0→2π) 2π(1 - cosx)² dx
= 2π∫(0→2π) (1 - 2cosx + cos²x) dx
=2π∫(0→2π) (1 - 2cosx + (1+cos2x)/2) dx
=2π∫(0→2π) (3/2 - 2cosx + cos2x/2) dx
=2π(3/2x - 2sinx + sin2x/4) |(0→2π)
=2π*(3/2)*2π
=6π²