第1个回答 2018-12-31
原式=∫[0,π]√[sinx (1-sin²x)] dx
=∫[0,π]√(sinxcos²x) dx
=∫[0,π]√(sinx) |cosx| dx
=∫[0,π/2]√(sinx)cosx dx - ∫[π/2,π]√(sinx) cosx dx
=∫[0,π/2]√(sinx) d(sinx) - ∫[π/2,π]√(sinx) d(sinx)
=2/3 (sinx)^(3/2) |[0,π/2] - 2/3 (sinx)^(3/2) |[π/2,π]
=2/3 (1-0) - 2/3 (0-1)
=4/3本回答被网友采纳