∫π -π sinx 1+cos²x
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∫(0到π)|cosx|dx=∫(0到π/2)cosxdx+∫(π/2到π)-cosxdx=sinx(0到π/2)-sinx(π/2到π)=(1-0)-(0-1)=1+1=2
1)f(x)=1+cos(2x+π/3)-(1+cos2x)/2=1/2-sin2x根号3/2最小值1/2-根号3/2最小正周期π2)c带入得sinC=根号3/2C=π/3A=π-B-C=2π/3-a
原式=sinx(-sinx)(-cotx)/(-cosx)(-cosx)tanx=sin^2xcos^2x/sin2^xcos^2x=1
(Ⅰ)当x∈(0,π2)时,f′(x)=π+πsinx-2cosx>0,∴f(x)在(0,π2)上为增函数,又f(0)=-π-2<0,f(π2)=π22-4>0,∴存在唯一x0∈(0,π2),使f(x
∫(0~π)根号(cos^2x-cos^4x)dx=2∫(0~π/2)根号(cos^2x(1-cos^2x))dx=2∫(0~π/2)cosxsinxdx=2∫(0~π/2)sinxdsinx=(si
∵cos(x-π/6)=-√3/3∴cosx+cos(x-π/3)=cosx+cosxcosπ/3+sinxsinπ/3=cosx+(1/2)cosx+(√3/2)sinx=(3/2)cosx+(√3
再问:cos(-x)=cosx是因为-x在第四象限符号和第一象限相同,所以就化正了吗?再答:是的
cos(x-π/2)=cos(π/2-x)=-sin(-x)=sinx所以是奇函数奇变偶不变,符号看象限.
但是如何作变换呢?怎样才能反函数也成功换成的简单函数式?请回答的具体些!多谢!
f(x)=cos(2x-π/3)-(cos^2x-sin^2x)=cos(2x-π/3)-cos2x=2sin(2x-π/6)sinπ/6=sin(2x-π/6)因为y=sinx的单减区间为[π/2+
f(x)=cos(2x-π\3)+sin²x-cos²x=1/2cos2x+√3/2sin2x-cos2x=√3/2sin2x-1/2cos2x=-cos(2x+π\3)-1
∫(π/2,-π/2)√(cos^2x-cos^4x)dx=∫(π/2,-π/2)√[cos^2x(1-cos^2x)]dx=∫(π/2,-π/2)√[cos^2x*sin^2x]dx=∫(π/2,-
∫cos(5x-π/6)dx=1/5∫cos(5x-π/6)d(5x-π/6)=1/5*sin(5x-π/6)+C
∫[0-->π](1-cos³x)dx=∫[0-->π]dx-∫[0-->π]cos^2xcosxdx=x|(0-->π)-∫[0-->π](1-sin^2x)dsinx=π-∫[0-->π
cos(2pai-2x)=cos2x你的结论是错的哦~不懂可追问~满意请采纳~再问:是不是cos括号的前面有个数字就要把它带进去用诱导公式,不能直接用,是吗再答:对的必须把括号打开才能使用诱导公式~~
∫(2/π,0)cos(2x/3)dx=3/2∫(2/π,0)cos(2x/3)d(2x/3)=3/2sin(2x/3)](2π,0)3/2[sin(2*2π/3)-sin(2*0/3)]=-(3*3
√(1-cosx)(1+cosx)+√(1+cosx)(1-cosx)=2√(1-cosx)(1+cosx)=2√(1-cosx^2)=2√(Sinx^2)x∈(π/2,π)=2Sinx
由x-π3∈[2kπ,2kπ+π],可得x∈[π3+2kπ , 4π3+2kπ](k∈Z),∴函数y=cos(x-π3)的单调递减区间是[π3+2kπ , 4π