∫sinx cosx根号(1 sin2x)dx
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/24 22:25:50
3sinxcosx=3/2sin2x,cos平方x化为cos2x形式,再通过公式合并前面两项.
(cosX)^2=(1+cos2X)/2sinXcosX=1/2sin2X代入原式中=(1/2)cos2X+(√3/2)sin2X+3/2=sin(∏/6)cos2X+cos(∏/6)sin2X+3/
y=√2/2*sin2x+(1+cos2x)/2-1/2=√2/2*sin2x+1/2*cos2x=√3/2*sin(2x+z)其中tanz=(1/2)/(√2/2)=√2/2所以T=2π/2=π
y=√3sin2x-cos2x=2sin(2x-30°)ymin=-2
f(x)=(1-cos2x)/2+(√3/2)sin2x+1/2=(√3/2)sin2x-(1/2)cos2x+1=sin(2x-π/6)+1(1)最小正周期T=2π/2=π(2)f(x)max=2,
有sin(x-45°)=√2/4=sinxcos45°-cosxsin45°,得sinx-cosx=0.5,两边平方得1-2sinxcosx=0.25.sinxcosx=3/8.tanx+1/tanx
f(x)=√3sinxcosx+cos?x=√3/2sin2x+(cos2x+1)/2=√3/2sin2x+1/2cos2x+1/2=sin(2x+π/6)+1/2f(x)的最小正周期为2π/2=π∵
f(x)=cos^2x-sin^2x+2(根号3)sinxcosx+1=cos2x+(根号3)sin2x+1=2{(1/2)cos2x+[(根号3)/2]sin2x}+1=2sin(2x+派/6)+1
f(x)=(√3)sinxcosx+cos2x+1f(x)=(√3)(2sinxcosx)/2+cos2x+1f(x)=(√3/2)sin2x+cos2x+1f(x)=(√7/2)[(√3/2)(2/
答:f(x)=(1/2)*(cosx)^2+(√3/2)sinxcosx+1=(1/2)*(cos2x+1)/2+(√3/4)sin2x+1=(1/2)[sin2xcosπ/6+cos2xsinπ/6
Letu=1+sin(x)cos(x)=1+(1/2)sin(2x)anddu=cos(2x)dx→dx=du/cos(2x)So∫cos(2x)/(1+sin(x)cos(x))dx=∫1/udu=
y=1/2*cos²x+√3/2*sinxcosx+1=1/4*(cos2x-1)+√3/4*sin2x+1=1/2*(1/2*cos2x+√3/2*sin2x)+3/4=1/2*(sin3
由sin²x+cos²x=1得出的再问:���Ƕ��˸�2��ϵ��再答:��Ϊ֮ǰ��3cos²x再答:sin²x+3cos²x=sin²
y=1/2cos²x+√3/2sinxcosx+1=1/2(1/2+1/2cos2x+√3/2sin2x)+1=1/2(sin2xcosπ/6+cos2xsinπ/6)+1+1/4=1/2s
怎么感觉cosx应该是平方啊再问:嗯的,打错了再答:(cosx)^2=(1+cos2x)/2sinxcosx=1/2*sin2x所以原式=(1+cos2x)/2-根号3/2*sin2x+1=1/2*c
由(1+tanX)/(1-tanX)=3+2√2得tanX=√2/2((sinx)*2+√2sinxcosx-(cosx)*2)/((sinx)*2+2(cosx)*2)【分子分母同除以(cosx)*
f(x)=√3sinxcosx+cos2x+1=(√3/2)sin2x+cos2x+1=[(√7)/2][(√3/√7)sin2x+(2/√7)cos2x]+1=[(√7)/2]sin(2x+α)+1
解:原式=√3sin2x+cos2x+1=2(√3/2sin2x+1/2cos2x+1=2cos(2x-pai/3)+1.
∫sinxcosx/(1+sin^4x)dx=∫sinx/(1+sin^4x)d(sinx)=1/2*∫1/(1+(sin^2x)^2)d(sin^2x)=1/2*arctan(sin^2x)+C