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sinx 4Cosx

sinx^4+cosx^4 =(sin²x+cos²x)²-2sin²xcos²x =1-2sin²xcos²x≤1 ∴不是恒等的! f(x)=sinx^6+cosx^6=[(sinx)^2+(cosx)^2]*[(sinx)^4-(sinx)^2*(cosx)^2+(cosx)^4] =[(sinx)^4-(sinx)^2*(cosx)^2+(cosx)^4] =...

sinx^4+cosx^4=sinx^4+cosx^4+2sinx^2cosx^2-2sinx^2cosx^2 =(sinx^2+cosx^2)^2-2sinx^2cosx^2 =1-2sinx^2cosx^2 =1-0.5*(2sinxcosx)^2 =1-0.5*(sin2x)^2 不知道化简成这种程度够不够

sinx/cosx + cosx/sinx = 4 sin²x/(sinxcosx) + cos²x/(sinxcosx) = 4 (sin²x + cos²x)/(sinxcosx) = 4 1/(sinxcosx) = 4 sinxcosx = 1/4 sin2x = 1/2 2x = 2kπ + π/6 或2kπ + 5π/6 x = kπ + π/12 或 kπ + 5π/12 ,k ∈Z

解:

y'=x^4sinx+cosx y=∫(x^4sinx+cosx)dx =∫x^4sinxdx+∫cosxdx =-∫x^4dcosx+sinx =-x^4cosx+∫cosxdx^4+sinx =-x^4cosx+4∫cosx*x^3dx+sinx =-x^4cosx+4∫x^3dsinx+sinx =-x^4cosx+4x^3sinx-4∫sinxdx^3+sinx =-x^4cosx+4x^3sinx-12∫x^2sinxdx+sinx =-x...

答案在图片上,希望得到采纳,谢谢。愿您学业进步☆⌒_⌒☆

∫sinx(cosx)四次方dx =-∫cos^4x dcosx =-1/(4+1) (cosx)^(4+1)+c =-1/5 (cosx)^5+c

同样的方法可求(cosx)^4的积分。

sinx+cosx=√2sin(x+π/4)=0 sin(x+π/4)=0 x+π/4=2kπ或2kπ+π,k∈Z x=2kπ-π/4或2kπ+3π/4,k∈Z

令 3sinx+4cosx=A(2sinx+cosx)+B(2sinx+cosx)' 解得A=2,B=1 ∫(3sinx+4cosx)/(2sinx+cosx)dx = ∫ [ 2(2sinx+cosx) + (2sinx+cosx)' ] / (2sinx+cosx) dx = ∫ 2 dx + ∫ (2sinx+cosx)' / (2sinx+cosx) dx = 2x + ln|2sinx+cosx| +C

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