sinxcosx+sin4x

I(2n)=∫ (sinx)^(2n) dx=-∫ (sinx)^(2n-1) dcosx=-cosx.(sinx)^(2n-1) +(2n-1).∫ (sinx)^(2n-2).(cosx)^2 dx=-cosx.(sinx)^(2n-1) +(...

不定积分sinxcosx/1+sin4x dx huamin8000

分子｜sinxcosx｜= (1/2)｜sin2x｜, 2x 周期是 π， x 周期是 π/2 分母 1+sin^4x = 1+(1/4)(1-cos2x)^2 = 5/4 - (...

解答:解析:f(x)=cos4x+2sinxcosx-sin4x=(cos2x-sin2x)(cos2x+sin2x)+2sinxcosx=cos2x+sin2x=2sin(2x+π4) (1)最小正周期T=2π2=π(2)当x∈[0,...

∫ sinxcosx/(1 + sin⁴x) dx= ∫ sinx/(1 + sin⁴x) dsinx= (1/2)∫ 1/[1 + (sin²x)²] dsin²x= (1/2)arctan(...

y=(sinx)^4+(cosx)^4=[(sinx)^2+(cosx)^2]^2-2(sinx)^2(cosx)^2=1-(1/2)[sin(2x)]^2=1-(1/4)2[sin(2x)]^2=1/4-(1/4)2[sin(2x)]^2+...

你好！利用sinx与cosx的展开公式写出sin4x与cos4x的展开式，相加即得。经济数学团队帮你解答，请及时采纳。谢谢！

乘以2sinx,积化和差就变成了 sin2x-0+sin3x-sinx+sin4x-sin2x+...+sinnx-si(n-2)x+sin(n+1)x-sin(n-1)x=sin(n+1)x+sinnx-sinx再除以2sinx,即为答案...

* sin3x = (1/2)(sin3x*cosx) - (1/2)(sin3x*cos3x) = (1/2)*(1/2)[sin(3x-x)+sin(3x+x)] - (1/2)(sin3x*cos3x) = (1/4)(sin2x+sin4x) - (...

不定积分：∫[（cosx+sinx）⁴]dx=∫[（cosx+sinx）²]²dx=∫（1+sin2x）²dx=∫[1+2sin2x+sin²（2x）]dx=∫dx+2∫sin2xdx+...