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sin3xsin2xDx等于多少

由cos5x=cos(3x+2x) =cos3xcos2x-sin3xsin2x ① cosx=cos(3x-2x) =cos3xcos2x+sin3xsin2x ② ②-①得 cosx-cos5x=2sin3xsin2x 即sin3xsin2x=1/2(cosx-cos5x) 则sin3xsin2xdx=[1/2(cosx-cos5x)]dx =1/2cosxdx-1/2cos5xdx =1/2sinx-1/2*1/5sin5x =1/2sinx-1/10sin5x

sin2xsin2x+sin2xsin(2x+π/2)=sin2xsin2x-sin2xsin2x=0

sin3x=3sinx-4sin^3xsin2x=2sinxcosx

sinxsin3x=-(cos4x-cos2x)/2∫sinxsin2xsin3xdx=(1/4)∫sin4xdx - (1/2)∫sin2xcos4xdx=-(1/16)cos4x - (1/4)∫cos4xd(cos2x)=-(1/16)cos4x - (1/4)∫(2(cos2x)^2-1)d(cos2x)=-(1/16)cos4x + (1/4)cos2x - (1/2)∫(cos2x)^2d(cos2x)=-(1/16)cos4x + (1/4)cos2x - (1/6)(cos2x)^3

预备知识 cos(A-B)=cosAcosB+sinAsinB cos(A+B)=cosAcosB-sinAsinB cos(A-B)-cos(A+B)=2sinAsinB ∴ sinAsinB=1/2[cos(A-B)-cos(A+B)] ∫sin3xsinxdx=1/2 ∫(cos2x-cos4x)dx=1/4 sin2x -1/8 sin8x +C

先利用积化和差公式:sin(A)cos(B) = (1/2)[sin(A+B) + sin(A-B)]∴sin(2x)cos(3x) = (1/2)[sin(2x+3x) + sin(2x-3x)]= (1/2)[sin5x + sin(-x)]= (1/2)(sin5x - sinx)∴∫ sin(2x)cos(3x) dx= (1/2)∫ sin(5x) dx - (1/2)∫ sinx dx= (1/2)(1/5)∫ sin(5x) d(5x) - (1/2)∫ sinx dx= (1/10)(-cos(5x)] + (1/2)cosx + C= (1/10)[5cosx - cos(5x)] + C

这题用这条积化和差公式较快,sinAsinB=(-1/2)[cos(A+B)-cos(A-B)]证明的话将右边展开就行,下面入正题.原式=(-1/2)∫[cos5x-cosx]dx=(-1/10)∫cos5xd(5x)+(1/2)∫cosxdx=(-1/10)sin5x+(1/2)sinx+C

积化和差2sin3xsin2x=-(cos5x-cosx)=cosx-cos5x所以原式=∫cosxdx-∫cos5xdx=sinx-1/5*sin5x+C

∫Sin3xSinxdx 用积化和差公式 =1/2∫[cos(3x-x)-cos(3x+x)]dx =1/2∫[cos2x-cos4x]dx =(sin2x)/4-(sin4x)/8+c 上课睡觉去了?.

cos8x=cos(3x+5x)=cos3xcos5x-sin3xsin5xcos2x=cos(-2x)=cos(3x-5x)=cos3xcos5x+sin3xsin5x故sin3xsin5x=1/2*(cos2x-cos8x)∫Sin3xSin5xdx=∫1/2*(cos2x-cos8x)dx=1/4*sin2x-1/16*sin8x+C

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