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∫sinxsin2xsin3xDx

这题利用公式求

∫sinxsin2xsin3xdx =(1/2)∫[cosx-cos(3x)]sin(3x)dx (应用三角函数的积化和差) =(1/2)∫[cosxsin(3x)-cos(3x)sin(3x)]dx =(1/4)...

积化和差∫sinxsin2xsin3xdx=1/2∫(cosx-cos3x)sin3xdx=1/2∫cosxsin3xdx-1/2∫cos3xsin3xdx=1/4∫(sin2x+sin4x)dx-1/4∫sin6xdx=-1/8cos2x-1/16cos4x+1/24cos6x+C数学软件验算:

sinxsin2xsin3x=(sinxsin3x)sin2x=(cos2x-cos4x)sin2x=cos2xsin2x-cos4xsin2x =cos2xsin2x-(1/2)(sin2x-sin6x) ∫sinxsin2xsin3xdx=∫cos2xsin2xdx-(1/2)∫sin2xdx+(1/2)∫sin6xdx =(1/2)(sin2x)^2+(1/4)cos4x-(1/12)cos6x+C

n=1时公式成立; 现在假设对n-1公式成立 那么sinx+sin2x+sin3x+……+sinnx=sinx+sin2x+sin3x+……+sin(n-1)x+sinnx =[sin((n-1)x/2)sin(nx/2)]/sin(x/2)+sinnx =[sin((n-1)x/2)sin(nx/2)+sinnxsin(x/2)]/sin(x/2) =sin(nx/2)[sin((nx/2-x/2)+2cos(nx...

利用积化和差公式-2sin((A+B)/2)*sin((A-B)/2)=cosA-cosB 2sin(x/2)*sinx=cos(x/2)-cos(3x/2) 2sin(x/2)*sin2x=cos(3x/2)-cos(5x/2) ... 2sin(x/2)*sinnx=cos((2n-1)x/2)-cos((2n+1)x/2) 裂项相消 原式就等于cos(x/2)-cos((2n+1)x/2)

∫[sinxsin(3x)]dx =∫½[cos(x-3x)-cos(x+3x)]dx =½∫[cos(-2x)-cos(4x)]dx =½∫[cos(2x)-cos(4x)]dx =½∫cos(2x)dx -½∫cos(4x)dx =¼∫cos(2x)d(2x)-⅛∫cos(4x)d(4x) =-¼sin(2x)+⅛sin(4x)+C 提示:先对

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