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求不定积分sinxCosx/(sin^4x+Cos^4x)

分子分母同除以 cos^4x

你都知道怎么变换了,还用别人帮求积分吗?

完全没有错误。这两个答案是相等的。

答案二分之一,sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-2(sinx*cosx)^2 也就是求出sinx*cosx就行 把已知条件两边同时平方,可得到sinx*cosx=1/2 ,带入上式可得答案为1/2,如果是选择题,一眼可看出x=45°。。。答案1/2

(sinx)^2(cosx)^4=[1-(cosx)^2](cosx)^4=(cosx)^4-(cosx)^6 =[(1+cos2x)/2]^2-[(1+cos2x)/2]^3 =[(cos2x)^2+2cosx+1]/4-[(cos2x)^3+3(cos2x)^2+3cos2x+1]/8 ={[(1+cos4x)/2]+2cosx+1]/4-{(cos6x+3cos2x)/4+3[(1+cos4x)/2]+3cos2x+1}/8 =(cos4x+4...

(Ⅰ) f(x)=(cos4x-sin4x)-2sinx?cosx=(cos2x-sin2x)-sin2x=cos2x-sin2x=2cos(2x+π4). …(4分)∴f(x)的最小正周期T=2π2=π. …(6分)(Ⅱ)令2kπ≤2x+π4≤2kπ+π,…(8分)则kπ-π8≤x≤kπ+3π8,k∈Z.故f(x)的单调递减区间为[kπ-π8,kπ+3π...

f(x)=cos^4x-2sinxcosx-sin^4x =(cos^2x+sin^2x)(cos^2x-sin^2x)-sin2x =1*cos2x-sin2x =cos2x-sin2x =cos(2x+π/4) 则f(x)的最小正周期 T=2π/2=π

f(x)=sinxcosx-(1-2sin^2x)(sin^4x-cos^4x) =1/2sin2x-[1-(1-cos2x)](sin^2x-cos^2x)(sin^2x+cos^2x) =1/2sin2x+cos2x(cos^2x-sin^2x) =1/2sin2x+cos2xcos2x =1/2sin2x+(cos2x)^2 =1/2sin2x+1-(sin2x)^2 =-(sin2x)^2+1/2sin2x+1 =-(sin2x-...

y=sin∧4x+2√3sinxcosx-cos∧4x (3前面是根号,不是积分??) =sin∧4x-cos∧4x+√3sin2x = (sin²x-cos²x)(sin²x+cos²x) + √3sin2x =√3sin2x - cos2x =2*sin(2x-π/6) 周期T=2π/w=2π/2=π 最小值:y=-2

f(x)=cos^4x-2sinxcosx-sin^4x = cos^4x-sin^4x-2sinxcosx =( cos^2x+sin^2x)( cos^2x-sin^2x)-sin2x =( cos^2x-sin^2x)-sin2x =cos2x-sin2x =√2cos(2x+π/4) 对称轴过图像的最高点或最低点。 所以2x+π/4=kπ,k∈Z. X= kπ/2-π/8,k∈Z.这就是所求的...

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