两次分部积分∫x�0�5cos2xdx=(1/2)∫x�0�5d(sin2x)=(1/2)[x�0�5sin2x-∫sin2xd(x�0�5)]=(1/2)x�0�5sin2x-∫xsin2xdx=(1/2)x�0�5sin2x+(1/2)∫xd(cos2x)=(1/2)x�0�5sin2x+(1/2)[xcos2x-∫cos2xdx]=(1/2)x�0�5sin2x+(1/2)xcos2x-(1/2)∫cos2xdx=(1/2)x�0�5sin2x+(1/2)xcos2x-(1/4)∫d(sin2x)=(1/2)x�0�5sin2x+(1/2)xcos2x-(1/4)sin2x+CC为常数
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024