結果 : integral of sqrt 16 x 2 dx
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Integral of 1/sqrt(16 - x^2) dx using trigonometric substitution

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1/sqrt(16-x^2) dx の積分

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三角関数の置換積分 sqrt(x^2 - 16)/x

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`int_(0)^(4)sqrt(16-x^(2))dx=`

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微積分ヘルプ: ∫ x^2/√(16-x^2 ) dx の積分 - 三角関数置換による積分 - 解決済み!

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How to solve integral 1/sqrt(16-x^2) sin^-1 (x/2) dx with u sub.

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Integral of sqrt(16 - 5x^2) dx using trigonometric substitution

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integrate 1/(sqrt(16 - x ^ 2)) dx

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