sin^3*con^2dx
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∫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=1/3*∫sin(3x+2)d(3x+2)=-1/3*cos(3x+2)
∫sin²x/cos³xdx=∫tan²xsecxdx=∫(sec²x-1)secxdx=∫sec³xdx-∫secxdx=(1/2)secxtanx
根据三角恒等式:cos2x=1-2sin²xsin²x=(1/2)(1-cos2x)sin²(3x)=(1/2)(1-cos6x)∫sin²(3x)dx=(1/
∫sin(2x+3)dx=1/2∫sin(2x+3)d(2x+3)=-1/2cos(2x+3)+c如果不明白可以令2x+3=t
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
由于dx=d(2x-3)/2所以Int(dx/sin(2x-3)^2)=Int(d(2x-3)/sin(2x-3)^2)/2=-cot(2x-3)/2+C
第一个用分部积分法即可.第二个用第一类换元法即可第三个用1的代换即1=cos^2(6x^2+2)+sin^2(6x^2+2)第一题:∫3ln^2*x+6lnx+7/xdx=3∫ln^2xdx+6∫ln
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
设sinx=a,cosxdx=da原式=a^3da=a^4/4=(sinx)^4/4=1/4
解∫x³sinx²dx=1/2∫x²sinx²dx²=1/2∫usinudu=-1/2∫ud(cosu)=-1/2[ucosu-∫cosudu]=-1
∫dx/(3+sin^2x)=∫dx/(4-cos^2x)∫dx/[2+cosx)][2-cosx]=∫(1/4)/[2-cosx]+(1/4)/[2+cosx]dx=(1/4)∫1/[2-cosx]
y=sin²x+cos²x+sin2x+2cos²x=1+sin2x+cos2x+1=sin2x+cos2x+2=√2sin(2x+π/4)+2-1
利用(sinax)^2=(1-cos2ax)/2,cos2ax你应该会积吧,然后你再去积分吧
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