作业帮求定积分.
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/15 11:12:05
求定积分.
求定积分.
求定积分.
∫(π/6-> π/3) (cosx)^2/[ x(π-2x)] dx
let
x = π/2 -y
dx = -dy
x= π/6,y= π/3
x= π/3,y = π/6
I=∫(π/6-> π/3) (cosx)^2/[ x(π-2x)] dx
=∫(π/3-> π/6) (siny)^2/[ y(π/2 -y)] (-dy)
=∫(π/6-> π/3) (sinx)^2/[ x(π/2 -x)] dx
2I =∫(π/6-> π/3) dx/[ x(π/2 -x)]
=(2/π)∫(π/6-> π/3) [ 1/x +1/(π/2 -x)] dx
=(2/π)ln |x/(π/2 -x) |(π/6-> π/3)
=(2/π)ln4
I =(2/π)ln2
令y=π/2-x，则x=π/2-y，当x从π/6变到π/3,时，y从π/3变到π/6
原式=∫{从π/3积到π/6}cos^2(π/2-y)/[(π/2-y)*2y] d(π/2-y)
=∫{从π/6积到π/3}sin^2 y/[y(π-2y)] dy
=∫{从π/6积到π/3}sin^2 x/[x(π-2x)] dx
所以原式*2=∫{从π...
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令y=π/2-x，则x=π/2-y，当x从π/6变到π/3,时，y从π/3变到π/6
原式=∫{从π/3积到π/6}cos^2(π/2-y)/[(π/2-y)*2y] d(π/2-y)
=∫{从π/6积到π/3}sin^2 y/[y(π-2y)] dy
=∫{从π/6积到π/3}sin^2 x/[x(π-2x)] dx
所以原式*2=∫{从π/6积到π/3}cos^2 x/[x(π-2x)] dx+∫{从π/6积到π/3}sin^2 x/[x(π-2x)] dx
=∫{从π/6积到π/3} 1/[x(π-2x)] dx
=∫{从π/6积到π/3} 1/[x(π-2x)] dx
=1/π∫{从π/6积到π/3} [1/x+1(π/2-x)] dx
=1/π[ln|x|【从π/6积到π/3】+ln|y|【从π/6积到π/3】 ]=2ln2/π
所以原式=ln2/π
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不会 哎、、、、、、