Question

∫√(sin^3 x-sin^5 x)dx 上限π 下限0 求定积分

Answer 1

∫√(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

Answer 2

sin³ x-sin^5x=sin³x(1-sin²x) =sin³xcos²x

当0<x<0.5π时，sinx>0,cosx>0

√(sin³xcos²x)=sinxcosx√sinx

当0.5π<x<π时，sinx>0,cosx<0

√(sin³xcos²x)=-sinxcosx√sinx

sinxcosx√sinx=sin^(1.5)x*cosx

我们知道dsin^(1.5)x=1.5sin^(0.5)xcosxdx=1.5cosx√sinxdx

所以∫sinxcosx√sinxdx=(2/3)*∫sixdsin^(1.5)x

令sinx=t，得到
∫sixdsin^(1.5)x=∫tdt^(1.5)=∫1.5*t^(1.5)dt=(1.5/2.5)t^(2.5)

下略

Answer 3

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