∫cos(2x π 3)dx
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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)
∫(1+cos^2x)/cos^2xdx=∫1/cos^2x+1dx=∫1/cos^2xdx+x=∫1d(tanx)+x=tanx+x+c
∫(-π/2→π/2)(cos²2x+8)dx=∫(0→π/2)(1+cos4x)dx+8∫(-π/2→π/2)dx=(x+1/4*sin4x)|(0→π/2)+8π=π/2+8π=17π/
1/3sin(3x+2)郁闷,这是最简单的积分啊看好了,设3x+2=u则3dx=du代入积分∫cos(3x+2)dx=∫cosu(1/3du)=1/3sinu=1/3sin(3x+2)OK?
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
再答:见图
令t=tan(x/2)则cosx=[cos²(x/2)-sin²(x/2)]/[cos²(x/2)+sin²(x/2)]=[1-tan²(x/2)]/
令t=3x+2,则dt=3dx→dx=1/3·dt∫cos(3x+2)dx=∫cost·1/3·dt=1/3·∫costdt=1/3·sint+C=1/3·sin(3x+2)+C再问:则dt=3dx→
cos^2(2x)=2cos4x-1∫cos^2(2x)dx=∫(2cos4x-1)dx=1/4∫2cos4xd4x-∫dx=1/2sin4x-x+C(C为常数)
=∫(1-cos4x)/2dx=∫1/2dx-∫cos4x/8d4x=0.5x-1/8*sin4x+C(C为任意常数)再问:为什么1-cos^(2)2x=(1-cos4x)/2?是用了什么公式吗,还是
∫x/[(cos^2)x]dx=∫xdtanx=xtanx-∫tanxdx(分部积分法)=xtanx+ln|cosx|+C
分部积分法∫udv=uv-∫vduu=3x,v=sin(x/3)结果是3xsin(x/3)+9cos(x/3)
∫(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
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
原式=2∫sec²xdx=2tanx+C
∫(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