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求不定积分sinxCosx/(sin^4x+Cos^4x)

分子分母同除以 cos^4x

y=sin^4x+cos^4x=[(sinx)^2+(cosx)^2]^2-2(sinxcosx)^2 =1-2(sinxcosx)^2 =1-[(sin2x)^2]/2 =1-[1-(cos4x)]/4 =(3+cos4x)/4 最大值=1 最小值=1/2 值域[1/2,1].

你都知道怎么变换了,还用别人帮求积分吗?

配方sin^4x+cos^4x=[(sinx)^2+(cosx)^2]^2-2(sinx)^2(cosx)^2 由已知,(sinx)^2(cosx)^2=0 而,(sinx)^2+(cosx)^2=1 所以,答案为1

f(x)=(sinx)⁴-sinxcosx+(cosx)⁴ =(sin²x+cos²x)²-2sin²xcos²x -sinxcosx =1 -(1/2)sin²(2x)-(1/2)sin(2x) =-(1/2)[sin(2x)+1/2]² +9/8 -1≤sin(2x)≤1 -1/2≤sin(2x)+1/2≤3/2 0≤[sin(2x)+1/2]²...

将(1)式两边平方可得:(sinx+cosx)^2=2,化简得:sinxcosx=1/2sin^4x+cos^4x=(sin^2x+cos^2x)^2 -2sin^2xcos^2x=1-2*(1/2)^2=1/2

sinx+cosx=√2 (sinx+cosx)^2=2 1+2sinx.cosx =2 sinx.cosx = 1/2 (sinx+cosx)^4 = 4 (sinx)^4+ (cosx)^4 + 4sinx.cosx((sinx)^2+(cosx)^2) + 6(sinxcosx)^2=4 (sinx)^4+ (cosx)^4 + 4sinx.cosx + 6(sinxcosx)^2=4 (sinx)^4+ (cosx)^4 +4(1/2) +6...

∵(sinx)^4(cosx)^2 =(1-cos2x)^2(1+cos2x)/8 =[1-(cos2x)^2](1-cos2x)/8 =(sin2x)^2(1-cos2x)/8 =[1-(cos4x)]/16-(sin2x)^2(cos2x)/8 ∴原积分=∫[1-(cos4x)]/16*dx-∫(sin2x)^2(cos2x)/8*dx =x/16-(sin4x)]/64-1/16*∫(sin2x)^2(dsin2x) =x/16-(sin...

解f(x)=cos^4x+2sinxcosx-sin^4x =cos^4x-sin^4x+2sinxcosx =(cos^2x-sin^2x)(cos^2x+sin^2x)+2sinxcosx =(cos^2x-sin^2x)+2sinxcosx =cos2x+sin2x =√2sin(2x+π/4) 函数的周期T=2π/2=π 由0≤x≤π/2 即0≤2x≤π 即π/4≤2x+π/4≤5π/4 即当2x+π/...

解:f(x)=(cos²x+sin²x)(cos²x-sin²x)-sin2x=cos2x-sin2x=-(√2)sin[2x-(π/4).即f(x)=-(√2)sin[2x-(π/4)].∴(1)T=π。(2)2kπ-(π/2)≤2x-(π/4)≤2kπ+(π/2).===>kπ-(π/8)≤x≤kπ+(3π/8).∴x∈[kπ-(π/8),kπ+(3π/8)]时,该函数递减。在其...

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