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人気のある微分積分 >

integral of sqrt(a-x^2)

解

\int a - x^{2} d x

解

\frac{1}{2} a (arcsin (\frac{1}{a} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{a} x))) + C

解答ステップ

\int a - x^{2} d x

a - x^{2} = a (1 - \frac{x ^{2}}{a})

= \int a (1 - \frac{x ^{2}}{a}) d x

指数の規則を適用する:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{x ^{2}}{a} = (\frac{x}{a})^{2}

= \int a (1 - (\frac{x}{a})^{2}) d x

(\frac{x}{a})^{2} = \frac{x ^{2}}{a}

= \int a (1 - \frac{x ^{2}}{a}) d x

指数の規則を適用する:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

a (1 - \frac{x ^{2}}{a}) = a (1 - \frac{x ^{2}}{a})

= \int a 1 - \frac{x ^{2}}{a} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= a \cdot \int 1 - \frac{x ^{2}}{a} d x

三角関数による置換を適用する

= a \cdot \int a cos^{2} (u) d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= a a \cdot \int cos^{2} (u) d u

三角関数の公式を使用して書き換える

= a a \cdot \int \frac{1 + cos ( 2 u )}{2} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= a a \frac{1}{2} \cdot \int 1 + cos (2 u) d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= a a \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= a a \frac{1}{2} (u + \frac{1}{2} sin (2 u))

代用を戻す

u = arcsin (\frac{1}{a} x)

= a a \frac{1}{2} (arcsin (\frac{1}{a} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{a} x)))

累乗根の規則を適用する:

a a = a

a a = a

= \frac{1}{2} a (arcsin (\frac{1}{a} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{a} x)))

定数を解答に追加する

= \frac{1}{2} a (arcsin (\frac{1}{a} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{a} x))) + C

グラフ

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