作业帮sin^2x (1 cos^2x)不定积分
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/07 01:24:02
原式=-∫cos²xdcosx=-cos³x/3+C再问：第一步能讲一下为什么吗？再答：dcosx=-sinxdx采纳吧
sin^2x+cos^2x=1
f(x)=cos(3x)*cos(2x)+sin(3x)*sin(2x)=cos(3x-2x)=cosxf'(x)=-sinx
=sin^2(x)*[cos^2(x)-1]=-sin^4(x)再答：别忘了负号再问：嗯谢谢
5(sinx)^2+sin2x--(cosx)^2=15(sinx)^2+2sinxcosx--(cosx)^2=(sinx)^2+(cosx)^24(sinx)^2+2sinxcosx--2(cos
Sinx-siny=2/3cosx-cosy=1/2分别平方得(Sinx-siny)^2=(2/3)^2(cosx-cosy)^2=(1/2)^2展开相加得-2cos(x-y)+2=4/9+1/4-2
题目应该是当x逼近到0得时候,limx^2*cos(1/x)=0lim(sin(x^2*cos(1/x)))/x=lim(x^2*cos(1/x))/x=lim(x*cos(1/x))=0再问：你用罗
左边=(1-2sinxcosx)/(cos²x-sin²x)=(sin²x+cos²x-2sinxcosx)/(cos²x-sin²x)=(
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
3/2cosx-3/2(sinx)^2
∫cos2xdx/(sin^2xcos^2x)=4∫cos2xdx/(2sinxcosx)^2=4∫cos2xdx/(sin2x)^2=2∫cos2xd(2x)/(sin2x)^2=2∫d(sin2x
2cosx(sinx-cosx)+1=2sinxcosx-2cosx^2+1=sin2x+1-2cosx^2=sin2x-cos2x=√2sin(2x-π/4)
你确定第一个符号是加号不是乘号?
（1）f(x)=2cos(x/2)·(sin(x/2)+cos(x/2))-1=2cos(x/2)·sin(x/2)+2cos^2(x/2)-1=sinx+cosx(倍角公式)=√2sin(x+π/4
证明：∵cos²x-sin²x=cos2xcos⁴x+sin⁴x=1-2cos²xsin²x=1-(1-cos4x)/4=3/4+(co
sinx=2cosx,sin^2x=4cos^2xsin^2x=4-4sin^2x,sin^2x=4/5(cosx+sinx)/(cosx-sinx)+sin^2x=(1+tanx)/(1-tanx)
f(x)=2cosx*sinx-2cosx^2+1f(x)=sin2x-cos2xf(x)=根号2*sin(2x-45)周期T=π
2（√3/2sinx+1/2cosx)=2sin(x+π/6）√2*√2（√2/2sinx-√2/2cosx)=2sin(x-π/4）（3）解;2√2(1/2cosx-√3/2sinx)=2√2cos
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