这么复杂的积分怎么求原函数呀？

如题所述

令t=2nπx，则x=t/2nπ，dx=dt/2nπ
原式=(1/2nπ)*∫[1-t^2/(2nπ)^2]*costdt
=(1/2nπ)*∫costdt-[1/(2nπ)^3]*∫t^2*costdt
=(1/2nπ)*sint-[1/(2nπ)^3]*∫t^2*d(sint)
=(1/2nπ)*sint-[1/(2nπ)^3]*(t^2*sint-2∫tsintdt)
=(1/2nπ)*sint-[1/(2nπ)^3]*t^2*sint-[2/(2nπ)^3]*∫td(cost)
=(1/2nπ)*sint-[1/(2nπ)^3]*t^2*sint-[2/(2nπ)^3]*(tcost-∫costdt)
=(1/2nπ)*sint-[1/(2nπ)^3]*t^2*sint-[2/(2nπ)^3]*(tcost-sint)+C
=(1/2nπ)*sin(2nπx)-(1/2nπ)*x^2*sin(2nπx)-[2/(2nπ)^2]*xcos(2nπx)+[2/(2nπ)^3]**sin(2nπx)+C

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