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人気のある微分積分 >

integral of xsqrt(2-\sqrt{1-x^2)}

解

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

解

- \frac{2}{5} (2 - 1 - x^{2})^{\frac{5}{2}} + \frac{4}{3} (2 - 1 - x^{2})^{\frac{3}{2}} + C

解答ステップ

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

u置換積分法を適用する

= \int - u 2 - u d u

定数を除く:

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

= - \int u 2 - u d u

u置換積分法を適用する

= - \int - (- v + 2) v d v

定数を除く:

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

= - (- \int (- v + 2) v d v)

拡張

(- v + 2) v : - v^{\frac{3}{2}} + 2 v

= - (- \int - v^{\frac{3}{2}} + 2 v d v)

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

= - (- (- \int v^{\frac{3}{2}} d v + \int 2 v d v))

\int v^{\frac{3}{2}} d v = \frac{2}{5} v^{\frac{5}{2}}

\int 2 v d v = \frac{4}{3} v^{\frac{3}{2}}

= - (- (- \frac{2}{5} v^{\frac{5}{2}} + \frac{4}{3} v^{\frac{3}{2}}))

代用を戻す

= - (- (- \frac{2}{5} (2 - 1 - x^{2})^{\frac{5}{2}} + \frac{4}{3} (2 - 1 - x^{2})^{\frac{3}{2}}))

簡素化

= - \frac{2}{5} (2 - 1 - x^{2})^{\frac{5}{2}} + \frac{4}{3} (2 - 1 - x^{2})^{\frac{3}{2}}

定数を解答に追加する

= - \frac{2}{5} (2 - 1 - x^{2})^{\frac{5}{2}} + \frac{4}{3} (2 - 1 - x^{2})^{\frac{3}{2}} + C

グラフ

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