求定积分∫x√[cos²x (1-cos²x)](上限为∏,下限为0)

先求一下不定积分∫x cosx sinx dx的解:
∫x cosx sinx dx
=∫ (1/4)* sin2x *xd(2x)
=-1/4∫x d(cos2x)
=-1/4*x*cos2x+1/8sin2x
∫x√[cos²x (1-cos²x)](上限为∏,下限为0)
=∫(上π,下0) x|cosx|sinx dx
=∫(上π/2,下0)x cosx sinx dx-∫(上π,下π/2)x cosx sinx dx
={-1/4*x*cos2x+1/8sin2x}|(上π/2,下0)-{-1/4*x*cos2x+1/8sin2x}|(上π,下π/2)
=π/8-(-3π/8)=π/2