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根号下(2x-x^2) 1/根号下(2x-x^2) 这两个的不定积分
人气:100 ℃ 时间:2020-01-30 12:10:12
解答
根号下(2x-x^2) =根号下(1-(x-1)^2)
x-1=sint,x=1+sint,t=arcsin(x-1),dx=cost dt
原积分=Scost *cost dt
=S(cos2t+1)/2 *dt
=1/2*Scos2tdt+1/2*Sdt
=1/4*sin2t+1/2*t+c
=1/2*(x-1)*根号下(2x-x^2)+1/2*arcsin(x-1)+c
1/根号下(2x-x^2)=1/根号下(1-(x-1)^2)
x-1=sint,x=1+sint,t=arcsin(x-1),dx=cost dt
原积分= S1/cost *costdt
=Sdt
=t+c
=arcsin(x-1)+c
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