sinx cos³x的积分
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fx=-√3cos2x-sin2x=-2sin(2x+π/3)所以最小正周期为πf'x=-4cos(2x+π/3),f'x>0时递增x在(π/12,π/3)上递增f'x=0,x=π/12.极小值f(π
肯定的啊,就是乘以1了啊.1/sinx=(1*sinx)/(sinx*sinx)=sinx/(sinx*sinx)要什么公式啊?
令sinxdx=-d(cosx)t^3/(1+t^2)dt=[(t^3+t)-t]/(1+t^2)*dt=t-t/(1+t^2)t^2/2-1/2*ln(1+t^2)+Ccosx^2/2-ln(1+c
原式=2sinxcos(x+π/3)+√3cos²x+sinxcosx=2sinxcos(x+π/3)+cosx(√3cosx+sinx)=2sinxcos(x+π/3)+2cosx·sin
f(x)=2(cosx)^2-2√3sinxcosx-1=(cos2x+1)-√3sin2x-1=cos2x-√3sin2x=2cos(2x+π/6)周期T=2π/│ω│=2π/2=π因为y=cosx
F(X)=5√3cos²x+√3sin²x-4sinxcosx=√3cos²x+√3sin²x+4√3cos²x-4sinxcosx=√3+2√3(c
(1)最简单的方法是用“积化和差”公式2sinαcosβ=sin(α+β)+sin(α-β)原式=2×2sinxcos(x+π/3)=2[sin(x+x+π/3)+sin(x-x-π/3)]=2[si
答:y=sin2x+2sinxcosx+2cos2x=2sin2x+2cos2x=2√2sin(2x+π/4)1)当sin(2x+π/4)=-1时,y的最小值为-2√22x+π/4=2kπ-π/2,x
y'=cosx-3sin²xcosx
单击图片可以放大!
∫(0→π/2)sinxcos³xdx=-∫(0→π/2)cos³xd(cosx)=-∫(1→0)t³dt……【将cosx用t代换,0-π/2没有产生周期重复,可以使用,
y=sin方x+sinxcos(派/6-x)=(3/2)sin²x+(√3/2)sinxcosx=(√3/2)sin(2x-π/3)+3/4周期为π增区间为[kπ-π/12,kπ+5π/12
y=2sin²x-√3sinxcosx+cos²x=sin²x-√3sinxcosx+(sin²x+cos²x)=(1-cos2x)/2-√3/2*s
y=2sinxcos^2x/(1+sinx)=2sinx﹙1-sin²x﹚/(1+sinx)=2sinx﹙1-sinx﹚=-2﹙sinx-½﹚²+½y=sin^
y=2√3*sinxcosx+2cos^2x=√3sin2x+cos2x+1=sin(2x+π/6)+1∴最小正周期:t=2π/2=π
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y=2sinxcos^2x/1-sinx=2sinx(1+sinx)=2(sinx+1/2)^2-1/2-1/2≤sinx+1/2≤3/20≤(sinx+1/2)^2≤9/4函数y=2sinxcos^
解原式=2sinxcos(x+π/3)+根号3cos的平方x+1/2sin2x=2sinxcos(x+π/3)+根号3cos的平方x+sinxcosx=2sinxcos(x+π/3)+cosx(根号3
令f'(x)=-sinx+cos^2(x)-sin^2(x)=-sinx+1-2sin^2(x)=0得sinx=1/2或sinx=-1.x=2k*pi+pi/3或2k*pi+2pi/3或x=2k*pi
f(x)=2cosxsin(x+π/6)+2sinxcos(x+π/6)=2sin(2x+π/6),(1)x∈[0,π/6],∴2x+π/6∈[π/6,π/2],∴f(x)的值域是[1,2].(2)f