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人気のある微分積分 >

integral of sqrt(1-x^2-y^2)

解

\int 1 - x^{2} - y^{2} d y

解

- \frac{1}{2} (1 - x^{2}) (arcsin (cos (arcsin (\frac{y}{1 - x ^{2}}))) - \frac{1}{2} sin (2 arcsin (cos (arcsin (\frac{y}{1 - x ^{2}}))))) + C

解答ステップ

\int 1 - x^{2} - y^{2} d y

三角関数による置換を適用する

= \int cos (u) - x^{2} + x^{2} sin^{2} (u) + cos^{2} (u) 1 - x^{2} d u

定数を除く:

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

= 1 - x^{2} \cdot \int cos (u) - x^{2} + x^{2} sin^{2} (u) + cos^{2} (u) d u

u置換積分法を適用する

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

定数を除く:

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

= 1 - x^{2} (- - x^{2} + 1 \cdot \int \frac{v ^{2}}{1 - v ^{2}} d v)

三角関数による置換を適用する

= 1 - x^{2} (- - x^{2} + 1 \cdot \int sin^{2} (w) d w)

三角関数の公式を使用して書き換える

= 1 - x^{2} (- - x^{2} + 1 \cdot \int \frac{1 - cos ( 2 w )}{2} d w)

定数を除く:

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

= 1 - x^{2} (- - x^{2} + 1 \frac{1}{2} \cdot \int 1 - cos (2 w) d w)

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

= 1 - x^{2} (- - x^{2} + 1 \frac{1}{2} (\int 1 d w - \int cos (2 w) d w))

\int 1 d w = w

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

= 1 - x^{2} (- - x^{2} + 1 \frac{1}{2} (w - \frac{1}{2} sin (2 w)))

代用を戻す

= 1 - x^{2} (- - x^{2} + 1 \frac{1}{2} (arcsin (cos (arcsin (\frac{1}{1 - x ^{2}} y))) - \frac{1}{2} sin (2 arcsin (cos (arcsin (\frac{1}{1 - x ^{2}} y))))))

簡素化

1 - x^{2} (- - x^{2} + 1 \frac{1}{2} (arcsin (cos (arcsin (\frac{1}{1 - x ^{2}} y))) - \frac{1}{2} sin (2 arcsin (cos (arcsin (\frac{1}{1 - x ^{2}} y)))))) : - \frac{1}{2} (1 - x^{2}) (arcsin (cos (arcsin (\frac{y}{1 - x ^{2}}))) - \frac{1}{2} sin (2 arcsin (cos (arcsin (\frac{y}{1 - x ^{2}})))))

= - \frac{1}{2} (1 - x^{2}) (arcsin (cos (arcsin (\frac{y}{1 - x ^{2}}))) - \frac{1}{2} sin (2 arcsin (cos (arcsin (\frac{y}{1 - x ^{2}})))))

定数を解答に追加する

= - \frac{1}{2} (1 - x^{2}) (arcsin (cos (arcsin (\frac{y}{1 - x ^{2}}))) - \frac{1}{2} sin (2 arcsin (cos (arcsin (\frac{y}{1 - x ^{2}}))))) + C

グラフ

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