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不定积分:Cos3xsin2xDx

本题要用到的积化和差公式: ∫sin2xcos3xdx=½∫[sin(3x+2x)-sin(3x-2x)]dx=½∫(sin5x-sinx)dx=(-1/10)cos5x +½cosx +C

这是基本积分,但是由于现在的高中数学把积化和差公式不做要求。因此学生感觉不熟。此题用公式sinαcosβ=[sin(α+β)+sin(α-β)]/2,就有

应该是∫sin^2xcos^3xdx吧?

∫ (x^3+2x-1) dx =(1/4)x^4+x^2 -x + C

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 =...

先积化和差,再积分! sin5xcos3x=0.5sin(8x)+0.5sin(2x) ∫sin5xcos3x dx =∫0.5sin(8x)+0.5sin(2x)dx =(1/16)∫sin(8x)d(8x) + (1/4)∫sin(2x)d(2x) =-(1/16)cos(8x) - (1/4)cos(2x) + C

cos5x=cos(3x+2x)=cos3xcos2x-sin3xsin2x cosx=cos(3x-2x)=cos3xcos2x+sin3xsin2x cosx+cos5x=2cos3xcos2x cos3xcos2x=1/2(cosx+cos5x) ∫cos3xcos2xdx =∫1/2(cosx+cos5x)dx =1/2∫cosxdx+1/2∫cos5xdx =1/2∫cosxdx+1/10∫cos5xd5x =1/2sinx+1/10sin...

u=∫e^(2x)cos3xdx =(1/3)∫e^(2x)dsin3x =(1/3)[e^(2x)sin3x-∫sin3xde^(2x)] =(1/3)[e^(2x)sin3x-2∫e^(2x)sin3xdx] =(1/3)[e^(2x)sin3x+(2/3)∫e^(2x)dcos3x] =(1/3){e^(2x)sin3x+(2/3)[e^(2x)cos3x-∫cos3xde^(2x)]} =(1/3){e^(2x)sin3x+(2/3)[e^(...

使用积化和差公式 sinAcosB=(1/2)*sin(A+B)+ (1/2)*sin(A-B) 所以得到 sin2xcos5x=(1/2)*sin(5x)- (1/2)*sin(3x) 那么积分就得到 ∫(1/2)*sin(5x)- (1/2)*sin(3x) dx =1/10 *∫sin(5x) d(5x) -1/6 *∫sin(3x) d(3x) = -1/10 *cos(5x) +1/6 *cos(3x)...

积化和差公式sinαcosβ=1/2[sin(α+β)+sin(α-β)] ∴sin2xccos3x=1/2[sin(2x+3x)+sin(2x-3x)]=1/2(sin5x-sinx)

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