sinx sin^3x cos^3x的不定积分-搜答案-神马作文网

f(x)=根号3cos²ωx+sinωxcosωx=根号3/2(cos²2ωx)+1/2(sin2ωx)=sin(2ωx+π/3)；y轴右侧的第一个最高点的横坐标为π/6,π/3ω

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

设倾斜角是qtan(q)=k=cosa/(-根号3)=-根号3/3cosa由于-1

sin（x-π/2）=sin(x+3π/2)=-cosx2sin²x=1-cos2x2sinxcosx=sin2xf（x）=2sin²x-2√3sinxsin（x-π/2）=1-c

∫sin^3xcos^2xdx=-∫sin^2xcos^2xdcosx=-∫(1-cos^2x)*cos^2xdcosx=-∫(cos^2x-cos^4x)dcosx=(1/5)*cos^5x-(1/

（Ι）∵函数f（x）=sinxsin（π2+x）+3cos2x=sinxcosx+32（1+cos2x）=12sin2x+32cos2x+32=sin（2x+π3）+32，…（4分）∴函数f（x）的最

设倾斜角是a斜率k=tan(a)=cosα所以-1

f(x) = xcos(3x)求导

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

最大值1,周期pai化为：-sin(2x-(pai/3))再问：能写详细点么再答：sinxsin(x+π/2)+sin2π/3cos2x=-1/2sinx+√3/2cos2x=sin(pai/3-2x

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

sin^2(a+π)Xcos(π+a)Xcot(-a-2π)/tan(π+a)Xcos^3(-a-π)=sin^2a(-cosa)(-cota)/tana(-cos^3a)=-sinacos^2a/s

看：(对不起,第一条的变数全部都是t,刚才做的时候不小心把t打错作x了)

化简得到f(x)=2sin(2x-%pi/6)+1x属于(0,2%pi/3),所以f(x)属于（0,3]已知M-2

用二倍角公式,化简成sin(2x)和cos（2x）,然后合并,再求导,求出单调增或减区间.取倒数零点,找到极值,然后找到最大值

f(x)=sin²x+√3sinxsin(x+π/2)=(1-co2x)/2+√3sinxcosx=(1-co2x)/2+√3/2*sin2x=1/2+√3/2*sin2x-cos2x/2=

（Ⅰ）函数f（x）=sin2x+2sinxsin(π2−x) +3sin2(3π2−x)=sin2x+2sinxcosx+3cos2x=1+sin2x+2cos2x=2+sin2x+cos2

∫ 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

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