∫（π到0）x乘以根号下((cosx)^2-(cosx)^4)dx

∫[π,0] x√(cosx^2-cosx^4)dx
=∫[π,π/2] -xcosxsinxdx+∫[π/2,0] xcosxsinxdx
=(1/2)∫[π,π/2]xdcos2x+(-1/2)∫[π/2,0]xdcos2x
=(1/2)xcos2x|[π,π/2]-(1/4)sin2x|[π,π/2] +(-1/2)xcos2x|[π/2,0]+(1/4)sin2x|[π/2,0]
= -3π/4+(-1/2)(π/2)
=-π最后答案好像是π/2啊。。。∫[π,0] x√(cosx^2-cosx^4)dx=(1/4)∫[π,π/2]xdcos2x+(-1/4)∫[π/2,0]xdcos2x=(1/4)xcos2x|[π,π/2]-(1/8)sin2x|[π,π/2] +(-1/4)xcos2x|[π/2,0]+(1/8)sin2x|[π/2,0]= -3π/8+(-1/4)(π/2)=-π/2∫[0,π] x√(cosx^2-cosx^4)dx=(-1/4)∫[π,π/2]xdcos2x+(1/4)∫[π/2,0]xdcos2x=(-1/4)xcos2x|[π,π/2]+(1/8)sin2x|[π,π/2] +(1/4)xcos2x|[π/2,0]-(1/8)sin2x|[π/2,0]= 3π/8+(1/4)(π/2)=π/2