计算∫(π,0)根号sinx-sin^3xdx

∫[0--->π]√(sinx-sin³x)dx
=∫[0--->π]√[sinx(1-sin²x)]dx
=∫[0--->π]√[sinxcos²x]dx
=∫[0--->π/2] cosx√(sinx)dx-∫[π/2--->π] cosx√sinx 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-(-2/3)
=4/3

∫[π,0] √(sinx-sinx^3)dx
=∫[π,π/2]-√sinxdsinx+∫[π/2,0] √sinxdsinx
=(-2/3)-(2/3)
=-4/3

∫(π,0)根号sinx-sin^3xdx
=∫(π,0)根号sinx(cos²x）dx
=∫(π,0)|cosx|根号sinxdx
=∫(0，π/2)cosx根号sinxdx-∫(π/2,π)cosx根号sinxdx
=∫(0，π/2)根号sinxdsinx-∫(π/2,π)根号sinxdsinx
=2/3 (sinx)^(3/2)|(0，π/2)-2/3 (sinx)^(3/2)|(π/2，π)
=2/3+2/3
=4/3