(1-sin6x-cos6x)/(1-sin4x-cos4x)=[1-(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x)]/(1-sin^4x-cos^4x)=(1-sin^4x-cos^4x+sin^2xcos^2x)/(1-sin^4x-cos^4x)=(1-1+sin^2xcos^2x+sin^2xcos^2x)/(1-1+2sin^2xcos^2x)=3sin^2xcos^2x/2sin^2xcos^2x =3/2 ...
sin6x+cos6x=√2sin[6x+(π/4)]所以，最小正周期T=2π/6=π/3
sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x =1-2sin^2xcos^2x 1-sin6x-cos6x/1-sin4x-cos4x =[1-(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x)]/(1-sin^4x-cos^4x)=(1-sin^4x-cos^4x+sin^2xcos^2x)/(1-sin^4x-cos^4x)=(1-1+sin^2xcos^2x+sin^2xcos...
方法如下，请作参考：若有帮助，请采纳。
sin^6x+cos^6x =(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x)=sin^4x-sin^2xcos^2x+cos^4x =(sin^2x+cos^2x)^2-3sin^2xcos^2x =1-3sin^2xcos^2x 所以分子=3sin^2xcos^2x sin^4x+cos^4x =(sin^2x+cos^2x)^2-2sin^2xcos^2x =1-2sin^2xcos^2x 所以分母=2sin^2x...
要证1−sin6x−cos6x1−sin4x−cos4x=32只需证1−(sin2x+cos2x)(sin4x−sin2xcos2x+cos4x)1−sin4x−cos4x=32, 即证1−sin4x−cos4x+sin2xcos2x1−sin4x−cos4x=32,即证1+sin2xcos2x1−sin4x−cos4x=32, 只需证sin2xcos2x1−sin4x−cos4x=12,即证1−sin4x−...
利用泰勒展开sin6x=6x-36x^3因为极限为0,则分子的最高阶低于分母。故-36x^3=xf(x),推出f(x)=-36x^2。带入式2得-36 baqktdgt 小吧主 15 我的泰勒公式贴70楼 三角的方 全微分 9 这不是泰勒嘛 艾克奥特曼 数项级数 6 你这个等价在复习全书基础篇里有,是six-x的等价,扩展的,这里就...
\\=&\frac14x^2+x-\frac23(x+1)^{3/2}+\left(\frac{8}{25}(x+1)^{3/2}-\frac{1}{8}(x+1)^2-\frac{1}{5}(x+1)\right)\sin(2\log(x+1))\\ &\left(\frac{6}{25} (x+1)^{3/2}-\frac{1}{8} (x+1)^2-\frac{1}{10}(x+1)\right)\cos(2\log(x+1)) \end...
2.用换元法求下列不定积分(1) ∫sin3xdx;(2) ∫cos5xdx(3) ∫e^(2-3x)dx ;(4) ∫1/(1+x)dx ;∫√(1-2x)dx; ⑥∫1/(√(2-3x))dx) ∫(1-3x)^9dx ;(8) ∫1/((1-x)^2)dx ;(9) ∫xe^(x^2)dx ;(10) ∫x^2sinx^3dx ;(11)) ∫x/(3-2x^2)dx ;(1 2...
∫dx/[√(1-6x)]= -1/6∫d(1-6x)/[√(1-6x)]= (-1/3)√(1-6x)+C ∫cos(2t+5) dt=1/2∫cos(2t+5) d(2t+5)=(1/2)sin(2t+5)+C ∫2t/(1+t) dt=∫2-(2/(1+t)) d(1+t)=2t-2ln|t+1|+C ∫x[4次方√(4x^2)-5] dx =1/2∫[4次方√(4x^2)-5]...