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∫sinxsin2xsin3xDx

这题利用公式求

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:

首先,利用两次积化和差公式: sinXsin2Xsin3X =-(1/2)(cos3X-cosX)sin3X =-1/4(sin6X)+1/2(sin4X)+1/2sin(2X) 分别设u1,u2,u3为-1/4(sin6X),1/2(sin4X),1/2sin(2X) 则u1的n阶导数为-1/4(sin(6X+n(π/2))*6^(n).....这个是复合函数求导 同理u2的n阶...

利用三角函数的积化和差公式即可

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)

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...

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设S=sinx+sin2x+sin3x+……+sinnx 两边同乘以2sin(x/2)(x≠2kπ,k∈Z) 得2sin(x/2)S=cos(x/2)-cos[(2n+1)x/2]=2sin(nx/2)sin[(n+1)x/2] 所以S=sin(nx/2)sin[(n+1)x/2]÷sin(x/2)

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