∫√(4-x^2)dx等于?

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∫√(4-x^2)dx等于?
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∫√(4-x^2)dx等于?
∫√(4-x^2)dx等于?

∫√(4-x^2)dx等于?
这个要转化
∫√(4-x^2)dx
设x=2cost a -->cost=x/2 -->t=arccos(x/2)
4-x^2=4sin^2t
√(4-x^2)=2sint b -->sint=√(4-x^2) /2
dx=-2sintdt c
则原代化为
∫2sint*(-2sintdt)
=2∫-2sin^2tdt
=2∫(cos2t-1)dt
=2∫cos2t dt -2∫dt
=∫cos2td2t-2t
=sin2t-2t+C
=2sintcost-2t+c
=2*x/2*√(4-x^2) /2 -2arccos(x/2)+C
=x√(4-x^2) /2-2arccos(x/2)+C