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sin 8x

sin^8x+cos^8x= =(sin^4x-cos^4x)^2+2sin^4xcos^4x =[(sin^2x-cos^2x)(sin^2x+cos^2x)]^2+2sin^4Xcos^4x =(sin^2x-cos^2x)^2+2sin^4Xcos^4x =sin^4x-2sin^2Xcos^2x+cos^4x+2sin^4Xcos^4x=sin^4x+cos^4x+2sin^4Xcos^4x-2sin^2Xcos^2x=(sin^4x+

解:∫(sinx)^8dx=∫((sinx)^2)^4dx =∫((1-cos(2x))/2)^4dx (应用倍角公式) =∫(1-4cos(2x)+6(cos(2x))^2-4(cos(2x))^3+(cos(2x))^4)dx =∫(1-4cos(2x)+6(cos(2x))^2+(cos(2x))^4)dx-4∫(cos(2x))^3dx =∫(1-4cos(2x)+3(1+cos(4x))+((1+cos(4x))/2)^2)dx-4∫(cos(2x)

∫(sinx)^8dx=-∫ (sinx)^7 dcosx= -cosx (sinx)^7 + ∫ 7(cosx)^2(sinx)^6 dx=-cosx (sinx)^7 +7∫ (1- (sinx)^2)(sinx)^6 dx8∫(sinx)^8dx= -cosx (sinx)^7 + 7∫(sinx)^6dx ∫(sinx)^8dx = (1/8) [-cosx (sinx)^7 + 7∫(sinx)^6dx]= (1/8) {-cosx (sinx)^7 + (7/6)[-cosx (sinx)^5

设t=8x,则y=cost所以y'=-sint,t'=8原函数的导数y'=-8sint=-8sin8x

∫sin(8x)cos(3x)dx=1/3∫sin(8x)dsin(3x)=1/3sin(8x)sin(3x)-1/3∫sin(3x)dsin(8x)=1/3sin(8x)sin(3x)-8/3∫sin(3x)cos(8x)dx ……① 而∫sin(8x)cos(3x)dx+∫sin(3x)cos(8x)dx=∫sin(11x)dx=-1/11 cos(11x)+C ……② 联立两式求二元一次方程组的解:①-②得

lim x→0 sin(5x)/sin(8x)=lim x→0 (5x)/(8x)=5/8

1)已知:f、g的周期都是:Tf(x+T)=f(t)g(x+T)=g(t)h(t)=f(t)+g(t)h(t+T)=f(t+T)+g(t+T)=f(t)+g(t)=h(t)那么f+g的周期还是:T.2) sinx 和 cosx 的周期均为:2π; sin^8x 和 cos^8x 的周期都是:π; 那么由1):sin^8x+cos^8x 的周期还是:π3)化简: sin^8x+

sinx=0的解为x=kπ,k ∈Z而8x^2>=0所以k ∈N令8x^2=kπ,得x=√kπ/8,k ∈N

令u=8x+5 则y=sinu 希望我的回答能帮助你,如果你认可我的回答,敬请及时采纳,在我回答的右上角点击【采纳答案】

∫(sinx)^8dx=-∫ (sinx)^7 dcosx= -cosx (sinx)^7 + ∫ 7(cosx)^2(sinx)^6 dx=-cosx (sinx)^7 +7∫ (1- (sinx)^2)(sinx)^6 dx8∫(sinx)^8dx= -cosx (sinx)^7 + 7∫(sinx)^6dx∫(sinx)^8dx = (1/8) [-cosx (sinx)^7

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