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不定积分x^2dx/(a^2-x^2)^(1/2) (a>0)
人气:347 ℃ 时间:2020-06-26 23:46:00
解答
令x = a siny,dx = a cosy dy∫ x²/√(a² - x²) dx= ∫ (a² sin²y)(a cosy dy)/(a cosy)= a²∫ sin²y dy= (a²/2)∫ (1 - cos2y) dy= (a²/2)(y - 1/2 sin2y) + C= (a&#...
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