sin^2xcos^3x

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sin^2xcos^3x的不定积分,

sin^2xcos^3x的不定积分,∫sin^2xcos^3xdx=∫sin^2x(1-sin^2x)dsinx=∫sin^2x-sin^4xdx=(1/3)sin^3x-(1/5)sin^5x+C实际非常简单∫sin^2x(1-sin^2

∫1/(sin^2xcos^2x)

∫1/(sin^2xcos^2x)∫1/(sin²xcos²x)dx=∫4/(4sin²xcos²x)dx=4∫1/sin²2xdx=2∫csc²2xd(2x)=-2cot2x+C

sin^2xcos^2x不定积分

sin^2xcos^2x不定积分∫sin^2xcos^2xdx=1/4∫(sin2x)^2dx=1/8∫(1-cos4x)dx=1/8(x-1/4sin4x)+C=x/8-(sin4x)/32+C满意请采纳,

(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x

(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x+3sin^2xsin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^

sin^2xcos^2x/(cos^3x+sin^3x)^2dx的不定积分

sin^2xcos^2x/(cos^3x+sin^3x)^2dx的不定积分积分结果:(3Cos[x]+Cos[3x]-8Sin[x]^3)/(18(Cos[x]+Sin[x])(-2+Sin[2x])),常数项自己加就可以了

1-(sin^6x+cos^6x)=3sin^2xcos^2x 如何证明?

1-(sin^6x+cos^6x)=3sin^2xcos^2x如何证明?1-(sin^6x+cos^6x)=3sin^2xcos^2x只要证明sin^6x+cos^6x=1-3sin^2xcos^2xsin^6x+cos^6x=(sin^2

∫sin^3xcos^2xdx

∫sin^3xcos^2xdx 

化简[1-(sin^4 x-sin^2 xcos^2 x+cos^4 x)]/(sin^2 x)+3

化简[1-(sin^4x-sin^2xcos^2x+cos^4x)]/(sin^2x)+3sin^2x注意到sin^2x+cos^2x=1所以1=(sin^2x+cos^2x)^2=sin^4x+2sin^2xcos^2x+cos^4x1-

若lim x趋向0,(xcosα-sinα)/2x^3拆成(xcosα/2x^3)-(sinα/2x

若limx趋向0,(xcosα-sinα)/2x^3拆成(xcosα/2x^3)-(sinα/2x^3)那分子能用无穷小量代换了吗?按理说加减不能替换,貌似拆成这样也不能替换,困扰我好久了不需要拆xcosα趋于0所以分子趋于-sinα,是个

sin^8X+cos^8X+4sin^2Xcos^2x-2sin^4xcos^4X(化简)

sin^8X+cos^8X+4sin^2Xcos^2x-2sin^4xcos^4X(化简)原式=sin^8X-2sin^4xcos^4X+cos^8X+4sin^2Xcos^2x=(sin^4x-cos^4x)^2+4sin^2Xcos^2

∫sin^3xcos^2xdx第一步∫sin^2xcos^2x*sinxdx(这部看懂了)第二步∫(

∫sin^3xcos^2xdx第一步∫sin^2xcos^2x*sinxdx(这部看懂了)第二步∫(1-cos^2x)cos^2x*(-d(cosx))为什么sinx变成(-d(cosx))了?∫sin^3xcos^2xdx=-∫sin^2

cos xsin b-sin xcos b和sin xcos x-

cosxsinb-sinxcosb和sinxcosx- 行吗?

1-2sin xcos x/cos的平方x

1-2sinxcosx/cos的平方x1-2sinxcosx/(cosx)^2=1-2sinx/cosx=1-2tanx1-2tanx

已知tanx=-3/4,求2sin^2 x+3sin xcos x-cos^2 x

已知tanx=-3/4,求2sin^2x+3sinxcosx-cos^2x原式=(2sin²x+3sinxcosx-cos²x)/(sin²x+cos²x)分子分母同除cos²x=(2tan

sin^4x-sin^2xcos^2x+cos^4x =(sin^2x+cos^2x)^2-3sin

sin^4x-sin^2xcos^2x+cos^4x=(sin^2x+cos^2x)^2-3sin^2xcos^2x那么这个依据呢?老师针对三角函数的证明,一般都是采取化简的思路来解答.对等式右边进行化简,=(sin^2x+cos^2x)^

求定积分∫(1/sin^2xcos^2x)dx上标派/3下标派/4

求定积分∫(1/sin^2xcos^2x)dx上标派/3下标派/4∫[π/4→π/3]1/(sin²xcos²x)dx=4∫[π/4→π/3]1/sin²2xdx=2∫[π/4→π/3]csc²2xd

求不定积分(1/sin^2xcos^2x)dx

求不定积分(1/sin^2xcos^2x)dx原式=∫4dx/(2sinxcosx)²=4∫dx/sin²2x=2∫csc²2xd2x=-2cot2x+C原式=∫4/4sin^2xcos^2xdx=∫4/sin

求不定积分,∫sin^2xcos^2x dx

求不定积分,∫sin^2xcos^2xdx利用半角公式如图降次计算.经济数学团队帮你解答,请及时采纳.

求∫1/sin^2xcos^2x dx

求∫1/sin^2xcos^2xdx∫dx/(sinxcosx)^2=∫4dx/(sin2x)^2=2∫d2x/(sin2x)^2=2∫(csc2x)^2d2x=-2cot2x+C∫dx/(sinxcosx)^2=∫4dx/(sin2x)^

2sinxcosx/sin²xcos²x怎么化简?

2sinxcosx/sin²xcos²x怎么化简?