sin^3xcosx
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解y'=(xcosx²)'=cosx²+x(cosx²)'=cosx²+x(-sinx²)(2x)=cosx²-2x²sinx
y=xcosx+sinxf(-x)=-xcos(-x)+sin(-x)=-xcosx-sinx=-(xcosx+sinx)=-f(x)所以是奇函数
y=sin^2x+√3sin^2xcosx+2cos^2x你确定那边+√3sin^2xcosx如果是+√3sinxcosx那么y=sin^2x+√3sinxcosx+2cos^2x=1/2(2cos^
你错在“原式=lim(1/(sinx)^2)-lim[(x/sinx)*(cosx/(sinx)^2)]”!∵当x->0时,lim(1/(sinx)^2)=不存在lim[(x/sinx)*(cosx/
利用换元法与分部积分法求不定积分∫(xcosx/sin³x)dx求高手破解∫(xcosx/sin³x)dx=-(1/2)∫[xd(1/sin²x)]=-(1/2)[x/s
f(x)=sin³xcosx+cos³xsinx+√3sin²x=sinxcosx(sin²x+cos²x)+√3(1-cos2x)/2=½
∫[xcosx/(sinx)^2]dx=∫[x/(sinx)^2]d(sinx)=-∫xd(1/sinx)=-x/sinx+∫(1/sinx)dx=-x/sinx+∫[sinx/(sinx)^2]dx
Y=cos^4x-2sinxcosx-sin^4x=cos^4x-sin^4x-2sinxcosx=(cos^2x-sin^2x)*1-sin2x=cos2x-sin2x=根号2倍的cos(x+4分之
∫1/(sin³xcosx)dx分子分母同除以(cosx)^4=∫(secx)^4/(tan³x)dx=∫sec²x/(tan³x)d(tanx)=∫(tan&
设sinx=a,cosxdx=da原式=a^3da=a^4/4=(sinx)^4/4=1/4
罗比达法则解法.原式=lim(x->0)[(sinx-xcosx)/(sinx)^3]=lim(x->0)[(cosx-cosx+xsinx)/(3sin²x)](0/0型极限,应用罗比达法
y=log(sinxcosx)=log(sin2x/2)/log(0.5)=1-logsin(2x)π/2+2kπ
(1)∵y=[x/(1+x)]^5∴y'=5[x/(1+x)]^4/(1+x)²=5x^4/(1+x)^6(2)∵y=(2x+1)^n∴y'=2n(2x+1)^(n-1)(3)∵y=sin&
有公式sin(a+b)=sinacosb+cosasinb所以sin(x+π/3)=sinxcosπ/3+cosxsinπ/3=1/2sinx+√3/2cosx
1.y'=[2x/(2x+1)^3]-6[x^2/(2x+1)^4]2.y'=12sin^2(4x+3)*cos(4x+3)3.y'=cosx^2+2x^2*sinx^2
tanx=-1/2sinx/cosx=-1/2cosx=-2sinx代入:sin²x+cos²x=15sin²x=1sin²x=1/5sin²x+3s