sin^4xcos^2x的不定积分

函数f(x)＝2sinωxcos(ωx+π4)+12=sinωxcosωx-sin2ωx+12=22sin（2ωx+π4）（1）因为函数的周期为2π，所以T=2π|2ω|＝2π，ω=±12；（2）由（

f(x)=sin^4x+cos^4x+sin^2xcos^2x/2-2sinxcosx=(sin²x+cos²x)²-3sin²xcos²x/2-2s

那个前半括号里面相加等于一

cos^4x+sin^2xcos^2x+sin^2x=cos^4x+(1-cos²x)cos²x+sin²x=cos^4x+cos²x-cos^4x+sin&#

sin^4x+cos^4x+sin^2x*cos^2x=sin^4x+cos^4x+2sin^2x*cos^2x-sin^2x*cos^2x=(sin^2x+cos^2x)^2-sin^2x*cos^

y=(sin^2x+cos^2x)^2+2sin^2xcos^2x-1=1+2sin^2xcos^2x-1=2sin^2xcos^2x=sin^2(2x)/2=(1-cos4x)/4周期显然是pi/2

y=sin^4x+cos^4x+4sin^2xcos^2x-1=(sin^2x+cos^2x)^2+2sin^2xcos^2x-1=1+2sin^2xcos^2x-1=2sin^2xcos^2x=si

∵[xcos(x+y)+sin(x+y)]dx+xcos(x+y)dy=0==>xcos(x+y)dx+xcos(x+y)dy+sin(x+y)dx=0==>xcos(x+y)(dx+dy)+sin(

用基本不等式sin^2xcos^2x+2/sin^2xcos^2x-2≥2√2-2公式没有错,但是等号无法成立,若成立,则sin²x*cos²x=√2但是sin²x*co

原式=∫4dx/(2sinxcosx)²=4∫dx/sin²2x=2∫csc²2xd2x=-2cot2x+C

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.

y=sin⁴3xcos³4xdy/dx=cos³4x*d(sin⁴3x)/dx+sin⁴3x*d(cos³4x)/dx=cos

f(x)=[(sin^2x+cos^2x)^2-sin^2xcos^2x]/(2-2sinxcosx)=(1-sinxcosx)(1+sinxcosx)/2(1-sinxcosx)=1/2sinxco

f(x)=a(sin²x+cos²x)(sin^4x-sin²xcos²x+cos^4x)+b(sin^4x+cos^4x)+6sin^2xcos^2x=a(s

t=sinx+cosx=√2sin(x+π/4)-√2=再问：上面那个颠倒的V是什么再答：那是根号呀，√2表示根号2.再问：sin^2x这个颠倒的＾也是根号？再答：这个是次方符号呀，sin^2x表示的

y=(sin^2x+cos^2x)^2+2sin^2xcos^2x-1=1+2sin^2xcos^2x-1=2sin^2xcos^2x=sin^2(2x)/2=(1-cos4x)/4周期显然是pi/2

∫e^sinx(xcosx-sinx/cosx^2)dx=∫e^xsinx*xcosxdx-∫e^sinxsinxdx/(cosx)^2=∫xe^sinxdsinx-∫e^sinxd(1/cosx)=

2sinxcosx=sin2x那么sin^2xcos^2x=sin^22x/4另外sinx等价于x,所以sin^2x等价于x^2,也即x^2sin^2x变成了x^4不知您是否明白,若有不明还可问(⊙o

证明：因为左边=sin²X(sin²X+cos²X)+cos²X=sin²X+cos²X=1=右边,所以：(sinX)^4+sin²

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