∫sin2x/sin^4x+cos^4xdx?
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 05:18:59
∫sin2x/sin^4x+cos^4xdx?
∫sin2x/sin^4x+cos^4xdx?
∫sin2x/sin^4x+cos^4xdx?
∫sin2x/sin^4x+cos^4xdx
=∫sin2x/[(sin²x+cos²x)²-2sin²xcos²x]dx
=∫sin2x/[1-1/2sin²2x]dx
=2∫sin2x/[2-sin²2x]dx
=2∫sin2x/(1+cos²2x)dx
=-∫1/(1+cos²2x)dcos2x
=-arctancos2x+c
题目是∫sin2x/(sin^4x+cos^4x)dx 这样的吗。晕
∫sin2x/sin^4x+cos^4xdx?
化简1+sin x/cos x·sin2x/2cos²(π/4-x/2)
化简1/4sin2x shin4x-sin^8 X+cos^8 X化简1/4sin2x X shin4x-sin^8 X+cos^8 X
y=sin^2x-3sinxcosx+4cos^2x是否等于y=sin^2x+cos^2x+3cos^2x-3/2*sin2x?y=sin^2x-3sinxcosx+4cos^2xy=sin^2x+cos^2x+3cos^2x-3/2*sin2x=1+3cos^2x-3/2*sin2x=1+3(1-sin^2x)-3/2*sin2x=-3sin^2x-3/2*sin2x+4
化简(sin^4+cos^4+sin^2xcos^2x)/(2-sin2x)
sin^4 x+cos^4 x=?以(i)cos2x表示.(ii)sin2x表示
sin^4x+cos^4x=5/9 求sin2X解答过程
1-cos^2x+sin2x怎么变成根号2sin[2X-π/4]+1
化简1/4sin2x·sin4x-sin^8x+cos^8x
已知cos((派/4)+x)=3/5 求(sin2x-2sin平方x)/(1-tanx)
化简1/4sin2x*sin4x-sin^8x+cos^8x
tanx=-1/2,sin2x+cos2x/4cos^2x-3sin^2x+1=
已知sin2x=(sinθ+cosθ)/2,cos^2x=sinθcosθ,那么cos2x的值是?2sin2x=sinθ+cosθ,平方得:4(sin2x)²=1+2 sinθcosθ,将cos²x=sinθcosθ代入得:4(sin2x)²=1+2cos²x,4(1- cos²2x) =1+2cos²x,4(1- cos²2x) =1+(1+co
1.∫sin^2(2x) 就是sin2x的平方2.∫(sin^x cos^x)
已知sin(x-π/4)=-1/3,求sin2x的值为什么sin2x=cos(2-π/4)?
求函数fx=(sin^4x+cos^4x+sin^2x*cos^2x)/(2-sin2x)最小正周期 .最大值 .最小值.
求函数Y=sin^4x +cos^4x +sin^2x cos^2x 除以2-sin2x的最小正周期,最大值最小值
(sin^4 x+cos^4 x+sin^2 x * cos^2 x)/2-sin2x的最小正周期,最大值和最小值..