第1个回答 2014-12-07
∫ [π/2,-π/2] (cosx-cos^3x)^(1/2)dx
=∫ [π/2,-π/2](cosx*sin^2x)^(1/2)dx
=∫ [π/2,0](cosx)^(1/2)sinxdx+∫ [0,-π/2](cosx)^(1/2)(-sinx)dx
=-∫ [π/2,0](cosx)^(1/2)d(cosx)+∫ [0,-π/2](cosx)^(1/2)d(cosx)
= -2/3(cosx)^(3/2) [π/2,0] + 2/3(cosx)^(3/2)[0,-π/2]
=4/3本回答被提问者采纳