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积分号(2到-2)根号下4-x^2(sinx+1)dx
人气:128 ℃ 时间:2020-02-04 23:04:27
解答
∫[2,-2]√(4-x^2)(sinx+1)dx=∫[2,-2]√(4-x^2)sinxdx+∫[2,-2]√(4-x^2)dx=0+x√(4-x^2)|[2,-2] +∫[2,-2]dx/√(4-x^2)=∫[2,-2]d(x/2)√(1-(x/2)^2)=arcsin(x/2)|[2,-2]=-π/2-π/2=-π ∫[2,-2]√(4-x^2)sinxdx ...
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