∫(1~派/2) (x+sinx)/(1+cosx) dx 怎么求了 范围应该是 (0~派/2)

∫(1~派/2) (x+sinx)/(1+cosx) dx 怎么求了 范围应该是 (0~派/2)
∫(1~派/2) (x+sinx)/(1+cosx) dx 怎么求了
范围应该是 (0~派/2)

∫(1~派/2) (x+sinx)/(1+cosx) dx 怎么求了 范围应该是 (0~派/2)
注意：sinx/(1 + cosx) = 2sin(x/2)cos(x/2)/[2cos²(x/2)] = sin(x/2)/cos(x/2) = tan(x/2)
∫(0~π/2) (x + sinx)/(1 + cosx) dx
= ∫(0~π/2) x/(1 + cosx) dx + ∫(0~π/2) sinx/(1 + cosx) dx
= ∫(0~π/2) x/(1 + cosx) dx + x * sinx/(1 + cosx) |(0~π/2) - ∫(0~π/2) x d[sinx/(1 + cosx)],分部积分
= ∫(0~π/2) x/(1 + cosx) dx + x * tan(x/2) |(0~π/2) - ∫(0~π/2) x * [cosx(1 + cosx) - sinx(- sinx)]/(1 + cosx)² dx
= ∫(0~π/2) x/(1 + cosx) dx + (π/2)tan(π/4) - ∫(0~π/2) x * (cosx + cos²x + sin²x)/(1 + cosx)² dx
= ∫(0~π/2) x/(1 + cosx) dx + π/2 - ∫(0~π/2) x/(1 + cosx) dx,前後一项抵消
= π/2