sin∧4x

∫(sinx)^4/(cosx)^2dx=∫(1-(cosx)^2)^2/(cosx)^2dx=∫(1+(cosx)^4-2(cosx)^2)/(cosx)^2dx=∫1/(cosx)^2+(cosx

见图,我觉得应该是对的,你自己再看看过程哈,我敢保证方法是对的

那个前半括号里面相加等于一

用公式a³+b³=(a+b)(a²-ab+b²)cos^6x+sin^6x=(cos²x)³+(sin²x)³=(cos

∫sinx/(1+sin^4x)dx=∫dcosx/(1+(1-cos^2x)^2)=∫dcosx/(2-2cos^2x+cos^4x)=∫du/(2-2u^2+u^4)=.查不定积分表吧再问：积分表

d/dxsin^8(cos(4x))=8sin^7(cos(4x)dsin(cos(4x)/dx=8sin^7(cos(4x)*cos(cos(4x))dcos(4x)/dx=8sin^7(cos(4

√[(sinx)^4+4(cosx)^2]-√[(cosx)^4+4(sinx)^2]=√[((sinx)^2-2)^2]-√[((cosx)^2-2)^2]=(sinx)^2-2-[(cosx)^2

sin(π/4+x/2)sin(π/4-x/2)=sin(π/4+x/2)sin[π/2-(π/4+x/2)]∵π/4=π/2-π/4∴sin(π/4-x/2)=sin(π/2-π/4-x)=sin[

x=0:0.1:2*pi;s=2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x);plot(x,s)

sinx+siny+sinz-sin(x+y+z)=4sin[(x+y)/2]sin[(x+z)/2]sin[(y+z)/2]sinx+siny+sinz-sin(x+y+z)=2sin[(x+y)/

∫1/sin⁴xdx=∫csc⁴xdx=∫csc²xd(-cotx)=-cotxcsc²x+∫cotxd(csc²x)=-cotxcsc²

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

∫cosx/(4-sin^2x)dx=∫1/(4-sin^2x)dsinx=1/4∫[1/(2-sinx)+1/(2+sinx)]dsinx=1/4ln(2+sinx)-1/4ln(2-sinx)+C

f(x)=(1+1/tanx)*(sinx)^2-2sin(x+π/2)sin(x-π/4)=(1+cosx/sinx)*(sinx)^2+2sin(x+π/4)cos[(x-π/4)+π/2]=(s

求什么再问：请问我现在是提问到网上了吗？再答：是再问：我想用的是搜索，不好意思再问：第一次用APP再答：恩再答：没事再答：查到了不，没有告诉我问题再问：可以删除吗？再答：不用删，没关系的再问：求cos

[cos(4-5x)]/5+C

证明：∵cos²x-sin²x=cos2xcos⁴x+sin⁴x=1-2cos²xsin²x=1-(1-cos4x)/4=3/4+(co

sin^2(x)+cos^2(x+30)+sin(x)cos(x+30）=sin^2(x)+cos(x+30)[cos(x+30)+sinx]=sin^2(x)+cos(x+30)(cosxcos30

sin^4x-sin^2x+cos^2x=sin^2x*(sin^2x-1)+cos^2x=-sin^2x*cos^2x+cos^2x=cos^2x*(1-sin^2x)=cos^2x*cos^2x=

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