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∫sinxsin2xsin3xDx

这题利用公式求

积化和差公式:sinx*siny=(1/2)[cos(x-y)-cos(x+y)] sinx*cosy=(1/2)[sin(x-y)+sin(x+y)] sinx * sin2x * sin3x = (sinx*sin2x) * sin3x = (1/2)[cos(x-2x)-cos(x+2x)] * sin3x = (1/2)(cosx-cos3x) * sin3x = (1/2)(sin3x*cosx) - (1/2)(sin3x*...

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:

被积函数是奇函数,而积分区间关于原点对称,根据”奇函数在对称区间的积分为0“可知,原式=0.

1.设u=tan(1/2*x),所以sinx=2u/(1+u^2),cosx=(1-u^2)/(1+u^2),dx=2/(1+u^2)du 代入化简得: 原式=∫1/u(3-u^2)du =∫1/3*1/udu+∫2/3*1/(3-u^2)du =-2/3*lnlu^2-3l+1/3*lnlul+C =-2/3*lnltan(1/2*x)^2-3l+1/3*lnltan(1/2*x)l+C 2.分子分母同乘sinx, ...

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)cos...

∫[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

证明:由题意,P=2xcosy-y2sinx,Q=2ycosx-x2siny,在整个平面上具有一阶连续偏导数,且?P?y=?2xsiny?2ysinx=?Q?x∴曲线积分I与积分路径无关.取路径从(0,0)到(2,0)再到(0,3),则I=∫202xdx+∫30(2ycos2?4siny)dy=4+9cos2+4cos3-4=9cos...

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