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sinxsin2x

因为cosX- cos3X =cos(2x-x)-cos(2x+x) =cos2xcosx+sin2xsinx -(cos2xcosx-sin2xsinx) =2sin2xsinx

∫sinxsin2xdx =-1/2∫(cos3x-cosx)dx =-1/2[1/3sin3x-sinx]+C =-1/6sin3x+1/2sinx+C

∫(sinxsin2x)dx =2∫sin²xcosxdx =2∫sin²xdsinx =2sin³x/3+C

其实一般都是2π,特殊抵消的才会是其他

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)

y=sinxsin2x y^2=sin^2x*4*sin^2x*cos^2x=2sin^2x*sin^2x*(2cos^2x)

sinx+sinx=2sinx sin(2x)=2sinx·cosx

这题利用公式求

t=sinx; sin2x=2sinx*cosx=2t*sqrt(1-t^2); 设y=sin2x 4t^2(1-t^2)=y^2 t^4-t^2+y^2/4=0 一元二次方程 t^2=1/2±sqrt(1/4-y^2/4) t=±sqrt(1/2±sqrt(1/4-y^2/4)) 注意,sinx应该有四个值。

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