∫√(sin^3 x-sin^5 x)dx 上限π 下限0 求定积分

sin³ x-sin^5x=sin³x(1-sin²x) =sin³xcos²x
√(sin³ x-sin^5x)=sin^(3/2)x|cosx|
∫[0,π]√(sin³ x-sin^5x)dx
=∫[0,π/2]sin^(3/2)x cosxdx-∫[π/2,π]sin^(3/2)x cosxdx
=∫[0,π/2]sin^(3/2)x d(sinx)-∫[π/2,π]sin^(3/2)x d(sinx)
=2/5 sin^(5/2)|[0,π/2]-2/5sin^(5/2)|[π/2,π]
=4/5

∫√(sin^3 x-sin^5 x)dx
=∫√(sin^3 x*cos^2 x) dx
=∫√(sin^3 x)*|cosx| dx
=∫（上限π，下限π/2）-cosx*√(sin^3 x)dx+∫(上限π/2,下限0)cosx*√(sin^3 x)dx
=∫（上限π，下限π/2）-√(sin^3 x)dsinx+∫(上限π/2,下限0)√(sin^3 x)dsinx
=-2/5*(sinx)^(5/2)|(上限π，下限π/2)+2/5*(sinx)^(5/2)|(上限π/2，下限0)
=2/5+2/5
=4/5

sin³ x-sin^5x=sin³x(1-sin²x) =sin³xcos²x
当00
√(sin³xcos²x)=sinxcosx√sinx
当0.5π