∫√xcos√x

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∫√xcos√x
已知函数f(x)=sin^2ωx+√3cosωxcos(π/2-ωx)(ω>0)

f(x)=sin^2ωx+√3cosωxcos(π/2-ωx)(ω>0)=(1-cos2ωx)/2+(√3/2)sin2ωx=sin(2ωx-π/6)+1/2∵函数y=f(x)的图像相邻两条对称轴之间

不定积分∫(sin2x)/(sin²x)dx ∫sin³xcos²xdx

∫(sin2x)/(sin²x)dx=∫(2sinxcosx)/(sin²x)dx=2∫cosx/sinxdx=2∫(1/sinx)d(sinx)=2ln|sinx|+C_____

y=xcos(1/x)

y'=cos(1/x)+x(-sin(1/x))(-1/x^2)=cos(1/x)+1/x*sin(1/x)

求不定积分:∫xcos(4x^2 +5)dx

∫xcos(4x^2+5)dxlet4x^2+5=tdt=d(4x^2+5)=4d(x^2)=4*2xdx=8xdxsodx=[dt/8x]∫xcos(4x^2+5)dx=∫xcostdt/8x=1/

计算不定积分∫xcos(x/2)dx 求过程跟答案

∫xcos(x/2)dx=2∫xcos(x/2)d(x/2)=2∫xdsin(x/2)=2xsin(x/2)-2∫sin(x/2)dx=2xsin(x/2)-4∫sin(x/2)d(x/2)=2xsi

求不定积分,∫sin^2xcos^2x dx

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.

f(x) = xcos(3x)求导

f'(x)=x'cos3x+x*(cos3x)'=cos3x+x(-sin3x)*(3x)'=cos3x-3xsin3x

∫ln(tanx)/sin xcos x dx. 不定积分的详细步骤过程和答案,拜托大神.

intln(tanx)/(sinxcosx)dx=intln(tanx)*cosx/sinx*1/cos^2xdx=intln(tanx)*1/tanxd(tanx)=intln(tanx)d[ln(

求不定积分∫xcos xdx

∫cos²xdx=∫cosxdsinx=sinxcosx-∫sinxdcosx=sinxcosx+∫sin²xdx=sinxcosx+∫(1-cos²x)dx=sinxc

已知函数f(x)=-√3sinωxcosωx+cos²ωx,x∈R,ω>0⑴求函数f(x)的值域

f(x)=-√3sinωxcosωx+cos²ωx=-(√3/2)sin(2ωx)+[1+cos(2ωx)]/2=cos(2ωx)*cos(π/3)-sin(2ωx)*sin(π/3)+1/

log10(e^-xcosπx)求导

=1/(ln10)[e^(-x)*cos(pix)]*[e^(-x)cos(pix)]'=1/(ln10)[e^(-x)*cos(pix)]*[-e^(-x)cos(pix)-e^(-x)(-pisi

∫(1/sin²xcos²x)dx怎么求,

∫(1/sin²xcos²x)dx=∫(sin2x+cos2x/sin²xcos²x)dx=∫(1/sin²x+1/cos²x)dx=-co

∫xcos(x/2)dx

用分部积分∫xcos(x/2)dx=2∫xcos(x/2)d(x/2)=2∫xdsin(x/2)=2xsin(x/2)-2∫sin(x/2)dx=2xsin(x/2)-4∫sin(x/2)d(x/2)

∫xcos(1+x^2)dx=

原式=0.5∫cos(1+x²)d(x²)=0.5sin(1+x²)+C再问:能给下过程么?3Q再答:这都是可以直接积分的,xdx=0.5d(x²)=0.5d(

求不定积分∫xcos(x^2)dx

∫xcos(x^2)dx=∫cos(x^2)(xdx)=∫cos(x^2)(d(x^2)/2)=(1/2)∫cos(x^2)d(x^2)=(1/2)sin(x^2)+C

∫ xcos(x/3) dx ...

∫xcos(x/3)dx=3∫xdsin(x/3)=3xsin(x/3)-3∫sin(x/3)dx+C=3xsin(x/3)+9cos(x/3)+CC为任意常数

求不定积分 1.∫(x√X+1/X²)dx 2.∫xe^xdx 3.∫x²1nxdx 4.∫xcos

1.∫(x√x+1/x^2)dx=∫x^(3/2)dx+∫x^(-2)dx=(2/5)x^(5/2)+(-1)x^(-1)+C=(2/5)x^(5/2)-x^(-1)+C2.∫xe^xdx=∫xd(e

∫(1/sin³xcos³x)dx 怎么解

1/[(sinx)^3(cosx)^3]=[sinx/(cosx)^3]+(2/sinxcosx)+[cosx/(sinx)^3]∫(1/sin³xcos³x)dx=[(1/2)/

∫sin²xcos³x dx

∫sin^2xcos^3xdx=∫sin^2x(1-sin^2x)dsinx=∫sin^2x-sin^4xdx=(1/3)sin^3x-(1/5)sin^5x+C不是让你求助我吗.再问:∫sin^2x