∫cosx X cos2x dx

如题所述

第1个回答 2020-03-02

∫cosx·cos2x dx = (1/2)∫(cos3x+cosx)dx = (1/6)sin3x + (1/2)sinx + C追问

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第2个回答 2020-03-02

∫ xcos2x.sinx dx
=-∫ xcos2x.dcosx
=-xcos2x.cosx +∫ cosx.( cos2x - 2xsin2x ) dx
=-xcos2x.cosx +∫ cosx.cos2x dx - 2∫xsin2x.cosx dx
=-xcos2x.cosx +(1/2)∫ (cosx+ cos3x) dx - 2∫xsin2x dsinx
=-xcos2x.cosx +(1/2)sinx+ (1/6)sin3x - 2[xsin2x.sinx -∫ sinx(sin2x + 2xcos2x. sinx) dx ]
-3∫ xcos2x.sinx dx =-xcos2x.cosx +(1/2)sinx+ (1/6)sin3x - 2xsin2x.sinx +2∫ sinx.sin2xdx
=-xcos2x.cosx +(1/2)sinx+ (1/6)sin3x - 2xsin2x.sinx +∫ (cosx-cos3x)dx
=-xcos2x.cosx +(1/2)sinx+ (1/6)sin3x - 2xsin2x.sinx +sinx-(1/3)sin3x
=-xcos2x.cosx +(3/2)sinx- (1/6)sin3x - 2xsin2x.sinx
∫ xcos2x.sinx dx = (-1/3)[-xcos2x.cosx +(3/2)sinx- (1/6)sin3x - 2xsin2x.sinx ] + C

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