第1个回答 2012-02-15
See you again ^ ^
∫[-π/2->π/2] √(cosx - cos³x) dx
= 2∫[0->π/2] √[cosx (1 - cos²x)] dx，√(cosx - cos³x)是偶函数
= 2∫[0->π/2] √cosx sinx dx
= -2∫[0->π/2] √cosx d(cosx)
= (-4/3)(cosx)^(3/2)_[0->π/2]
= (-4/3)(0 - 1)
= 4/3本回答被提问者采纳
∫[-π/2->π/2] √(cosx - cos³x) dx
= 2∫[0->π/2] √[cosx (1 - cos²x)] dx，√(cosx - cos³x)是偶函数
= 2∫[0->π/2] √cosx sinx dx
= -2∫[0->π/2] √cosx d(cosx)
= (-4/3)(cosx)^(3/2)_[0->π/2]
= (-4/3)(0 - 1)
= 4/3本回答被提问者采纳
第2个回答 2020-02-26