作业帮∫x^3sin^2x (x^4 2x^2 1)dx
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/19 10:29:57
∫sin^3(x)cos^2(x)dx=∫sin^2(x)cos^2(x)sin(x)dx=-∫sin^2(x)cos^2(x)dcos(x)=∫[cos^2(x)-1]cos^2(x)dcos(x)
原式等于：∫[1-cos^2(x)]/cos^3(x)dx=∫dx/cos^3(x)-∫dx/cos(x)=(secxtanx+ln|secx+tanx|)/2-ln|secx+tanx|+C
∫[cos^3(x)]/[sin^2(x)]dx=积分:(cos^2x)/(sin^2x)dsinx=积分:(1-sin^2x)/sin^2x)dsinx=积分;1/sin^2xdsin^2x-积分1
把一个sin(x)拿出来∫sin^3(x)cos^2(x)dx=-∫sin^2(x)cos^2(x)d(cos(x))=-∫(1-cos^2)cos^2(x)d(cos(x))=-∫cos^2-cos
∫(cos^3x/sin^2x)dx=∫[(1-(sinx)^2]/(sinx)^2dsinx=∫[1/(sinx)^2-1]dsinx=-1/sinx-sinx+C
f(x)=sin(π-x)cos(3π/2+x)+sin(π+x)sin(3π/2-x)=(sinx)(sinx)+(-sinx)(-cosx)=sinx(sinx+cosx)f'(x)=cosx(s
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
我知道这个题是个定积分题,请追问我给出积分限.我按我以前做过的同一题给你做吧,积分限是0→π∫[0→π]√(sin^3x-sin^5x)dx=∫[0→π]√[sin³x(1-sin²
f(x)=cos(3x)*cos(2x)+sin(3x)*sin(2x)=cos(3x-2x)=cosxf'(x)=-sinx
由和差化积公式分子=2sin[(x^3+x^2)/2]cos[(x^3+x^2-2x)/2]x→0,则(x^3+x^2)/2→0,sin则(x^3+x^2)/2和(x^3+x^2)/2是等价无穷小而c
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
换元法?没必要啊显然这是个奇函数而积分限关于原点对称所以原式=0
3/2cosx-3/2(sinx)^2
解∫x³sinx²dx=1/2∫x²sinx²dx²=1/2∫usinudu=-1/2∫ud(cosu)=-1/2[ucosu-∫cosudu]=-1
sin^2(x)+cos^2(x+30)+sin(x)cos(x+30）=sin^2(x)+cos(x+30)[cos(x+30)+sinx]=sin^2(x)+cos(x+30)(cosxcos30
y=sin²x+2sinxcosx+3cos²xy=(sin²x+cos²x)+2sinxcosx+(2cos²x-1)+1=1+sin2x+cos2
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次分部积分法解用!代表积分号=!(x^3-x+1)（1-cos2x）/2dx=(x^3-x+1)（x/2-sin2x/4）-!(3x^2-1)（x/2-sin2x/4）dx+c=-!(3x^2-1)