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求 ∫√(a^2-x^2)dx 的不定积分,
人气:263 ℃ 时间:2020-03-21 20:30:44
解答
令x=asiny,dx=acosy dy
√(a²-x²)=√(a²-a²sin²y)=acosy
原式=a²∫cos²y dy
=a²/2*∫(1+cos2y) dy
=a²/2*y+a²/2*1/2*∫cos2y d(2y)
=a²/2*y+a²/4*sin2y
=a²/2*y+a²/2*sinycosy
=a²/2*arcsin(x/a)+a²/2*(x/a)*√(a²-x²)/a
=(a²/2)arcsin(x/2)+(x/2)√(a²-x²)+C
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