cosxcos2xcos3x可化为什么
答:cosxcos2xcos3x的不定积分为x/4+1/8sin2x+1/16sin4x+1/24sin6x+C。解:∫cosxcos2xcos3xdx =1/2∫cosx*(cos(3x+2x)+cos(3x-2x))dx =1/2∫cosx*(cos5x+cosx)dx =1/2∫cosxcos5xdx+1/2∫(cosx)^2dx =1/4∫(cos(5x+x)+cos(5x-x))dx+1/4∫(1+cos2x)dx =1/4∫1d...
答:推导过程如下:上述过程使用了三角函数的“积化和差公式”和“降幂公式”,不论先算哪两个相乘,结果都是一样的。
答:∴ ∫ cosx cos2x cos3x dx = x/4 + 1/24 sin6x + 1/16 sin4x + 1/8 sin2x +C
答:(abc)'=a'bc+ab'c+abc'(cosxcos2xcos3x)'=-sinxcos2xcos3x-2cosxsin2xcos3x-3cosxcox2xsin3x 最后算出的结果是14吧
答:cosxcos2xcos3x = cos2x * cosxcos3x = cos2x * (1/2)[cos(x + 3x) + cos(x - 3x)]= cos2x * (1/2)[cos4x + cos2x]= (1/2)cos2xcos4x + (1/2)cos²2x = (1/2)(1/2)[cos(2x + 4x) + cos(2x - 4x)] + (1/2)(1/2)(1 + cos4x)= (1/4)cos6x...
答:∫cosxcos2xcos3xdx=(1/2)∫(cosx+cos3x)cos3xdx=(1/2)∫cosxcos3xdx+(1/2)∫(cos3x)^2dx=(1/4)∫(cos2x+cos4x)dx+(1/4)∫(1+cos6x)dx=(1/4)∫dx+(1/4)∫cos2xdx+(1/4)∫cos4xdx...
答:4cosxcos2xcos3x=2cos2x[cos4x+cos2x]=cos6x+cos2x+cos4x+1 【2】可设原式=y 4y=∫[cos6x+cos4x+cos2x+1]dx =(1/6)∫cos6xd(6x)+(1/4)∫cos4xd(4x)+(1/2)∫cos2xd(2x)+∫dx =[(sin6x)/6]+[(sin4x)/4]+[(sin2x)/2]+x+C ∴原式=【。。。]/4 ...
答:呃,这个函数应该是正正交函数。
答:cosx*cos2x*cos4x = 2 sinx*cosx*cos2x*cos4x / (2sinx)= sin2x * cos2x *cos4x /(2sinx) = sin8x / (8sinx)cos3x*cos5x =(1/2) ( cos8x +cos2x)原式= (1/16) (1/sinx) [ sin8x cos8x + sin8xcos2x ]= (1/32) (1/sinx) [ sin16x + sin10x +sin6x ]...
答:∫cosxcos2xcos3xdx =(1/2)∫(cosx+cos3x)cos3xdx=(1/2)∫cosxcos3xdx+(1/2)∫(cos3x)^2dx =(1/4)∫(cos2x+cos4x)dx+(1/4)∫(1+cos6x)dx =(1/4)∫dx+(1/4)∫cos2xdx+(1/4)∫cos4xdx+(1/4)∫cos6xdx =(1/4)x+(1/8)sin2x...
网友评论:
干种18426133070:
cosxcos2xcos3x的不定积分 -
60426荀德
:[答案] 你好!数学之美团为你解答积化和差公式:cosαcosβ = 1/2 [ cos(α+β) + cos(α - β) ]cosx cos2x cos3x= 1/2 ( cos3x + cosx ) cos3x= 1/2 cos²(3x) + 1/2 cosx cos3x= 1/4 ( 1 + cos6x ) + 1/4 ( cos4x + c...
干种18426133070:
求cosxcos2xcos3x对x的不定积分,及这种很多三角相乘除的方法. -
60426荀德
:[答案] 先用积化和差公式化简得1/4(1+cos6x+cos4x+cos2x)再分部积之得1/4x+1/24sin6x+1/16sin4x+1/8sin2x+C
干种18426133070:
cosxcos2xcos3x的导数原题1 - cosxcos2xcos3x/1 - cosx 在x趋近于0的极限 -
60426荀德
:[答案] 这个是倒数的四则运算和符合函数的导数 (abc)'=a'bc+ab'c+abc' (cosxcos2xcos3x)'=-sinxcos2xcos3x-2cosxsin2xcos3x-3cosxcox2xsin3x 最后算出的结果是14吧
干种18426133070:
(1 - cosxcos2xcos3x)/(1 - cosx)当x趋近于0时的极限 -
60426荀德
: 由三角积化和差公式 cosxcos2xcos3x =(1/2)(cosx+cos3x)xos3x =(1/4)cos2x+(1/4)cos4x+1/4+(1/4)cos6x原极限化为(x->0) (1-(1/4)cos2x-(1/4)cos4x-1/4-(1/4)cos6x)/(1-cosx) x->0 1-cosx~(1/2)x^2 上式=(1-(1/4)cos2x-(1/4)cos4x-1/4-(1/4...
干种18426133070:
求cosxcos2xcos3x对x的不定积分,及这种很多三角相乘除的方法. -
60426荀德
: 先用积化和差公式化简得1/4(1+cos6x+cos4x+cos2x)再分部积之得1/4x+1/24sin6x+1/16sin4x+1/8sin2x+C
干种18426133070:
计算cosxcos2xcos3x….cosnx
60426荀德
: cosx+cos2x+cos3x+......+cosnx =1/(2sin(x/2))*[2cosxsin(x/2)+2cos2xsin(x/2)+......+2cosnxsin(x/2)] 括号中的数列的和等于“ [sin(3x/2)-sin(x/2)]+[sin(5x/2)-sin(3x/2)]+[sin(7x/2)-sin(5x/2)]+...... +{sin[(2n+1)x/2]-sin[(2n-1)x/2]} =sin[(2n+1)x/2]-sin(x/2) =2cos(n+1)xsin(nx/2) 所以,原式=cos(n+1)xsinnx/sin(x/2).
干种18426133070:
当x→0时,1 - cosxcos2xcos3x与axn是等价无穷小,求常数a,n. -
60426荀德
:[答案] 当x→0时余弦函数在x=0的带佩亚诺余项的泰勒展开式: cosx= n+1 k=1 (−1)k−1x2k−2 (2k−2)!+o(x2n) 则当x→0时函数在x=0的带佩亚诺余项的二阶泰勒展开式分别为: cosx=1− 1 2x2+o(x2) cos(2x)=1− 1 2(2x)2+o(x2)=1-2x2+o(x2) cos(3x)=1− 1 ...
干种18426133070:
∫cosxcos2xcos3xdx -
60426荀德
:[答案] cosxcos2xcos3x = cos2x * cosxcos3x = cos2x * (1/2)[cos(x + 3x) + cos(x - 3x)] = cos2x * (1/2)[cos4x + cos2x] = (1/2)cos2xcos4x + (1/2)cos²2x = (1/2)(1/2)[cos(2x + 4x) + cos(2x - 4x)] + (1/2)(1/2)(1 + cos4x) = (1/4)cos6x + (1/4)cos2x + 1/4 + (1/4)cos4x ∫ ...
干种18426133070:
求极限的 x→∞ (1 - cosxcos2xcos3x)/(1 - cosx) -
60426荀德
:[答案] 你确定不是x->0吗?如果是趋于无穷,基本上应该是极限不存在.
干种18426133070:
∫cosxcos2xcos3xdx -
60426荀德
: cosxcos2xcos3x = cos2x * cosxcos3x = cos2x * (1/2)[cos(x + 3x) + cos(x - 3x)] = cos2x * (1/2)[cos4x + cos2x] = (1/2)cos2xcos4x + (1/2)cos²2x = (1/2)(1/2)[cos(2x + 4x) + cos(2x - 4x)] + (1/2)(1/2)(1 + cos4x) = (1/4)cos6x + (1/4)cos2x + 1/4 + (1/4)cos4...