问个数学积分问题

∫e^xcos2xdx
=(1/2)∫e^xd(sin2x)=(1/2)[e^xsin2x-∫sin2xd(e^x)]
=(1/2)[e^xsin2x-∫e^xsin2xdx]=(1/2)[e^xsin2x+(1/2)∫e^xd(cos2x)]
=(1/2){e^xsin2x+(1/2)[e^xcos2x-∫e^xcos2xdx]}
=(1/4)e^x(2sin2x+cos2x)-(1/4)∫e^xcos2xdx (把最后一项移到左边)
∴∫e^xcos2xdx=(4/5)[(1/4)e^x(2sin2x+cos2x)]
=e^x(2sin2x+cos2x)/5+c

你也可以通过一起求她的对称积分:e^xsin2x
通过四则运算得到结果,不详!

∫e^x cos2xdx
=∫cos2xde^x
=e^x cos2x-∫e^xdcos2x
=e^x cos2x+∫e^x*sin2xd2x
=e^x cos2x+2∫sin2xde^x
....如此用下去
可的到

∫e^x cos2xdx =（cos2x+2sin2x）e^x/5