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人気のある微分積分 >

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

解

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

解

\frac{5}{8} arcsin (x - 1) + \frac{1}{4} sin (2 arcsin (x - 1)) - \frac{1}{32} sin (4 arcsin (x - 1)) - \frac{2}{3} (- x^{2} + 2 x)^{\frac{3}{2}} + C

解答ステップ

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

平方完成する

2 x - x^{2} : - (x - 1)^{2} + 1

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

u置換積分法を適用する

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

拡張

(u + 1)^{2} - u^{2} + 1 : u^{2} - u^{2} + 1 + 2 u - u^{2} + 1 + - u^{2} + 1

= \int u^{2} - u^{2} + 1 + 2 u - u^{2} + 1 + - u^{2} + 1 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

= \int u^{2} - u^{2} + 1 d u + \int 2 u - u^{2} + 1 d u + \int - u^{2} + 1 d u

\int u^{2} - u^{2} + 1 d u = \frac{1}{8} (arcsin (u) - \frac{1}{4} sin (4 arcsin (u)))

\int 2 u - u^{2} + 1 d u = - \frac{2}{3} (- u^{2} + 1)^{\frac{3}{2}}

\int - u^{2} + 1 d u = \frac{1}{2} (arcsin (u) + \frac{1}{2} sin (2 arcsin (u)))

= \frac{1}{8} (arcsin (u) - \frac{1}{4} sin (4 arcsin (u))) - \frac{2}{3} (- u^{2} + 1)^{\frac{3}{2}} + \frac{1}{2} (arcsin (u) + \frac{1}{2} sin (2 arcsin (u)))

代用を戻す

u = x - 1

= \frac{1}{8} (arcsin (x - 1) - \frac{1}{4} sin (4 arcsin (x - 1))) - \frac{2}{3} (- (x - 1)^{2} + 1)^{\frac{3}{2}} + \frac{1}{2} (arcsin (x - 1) + \frac{1}{2} sin (2 arcsin (x - 1)))

簡素化

\frac{1}{8} (arcsin (x - 1) - \frac{1}{4} sin (4 arcsin (x - 1))) - \frac{2}{3} (- (x - 1)^{2} + 1)^{\frac{3}{2}} + \frac{1}{2} (arcsin (x - 1) + \frac{1}{2} sin (2 arcsin (x - 1))) : \frac{5}{8} arcsin (x - 1) + \frac{1}{4} sin (2 arcsin (x - 1)) - \frac{1}{32} sin (4 arcsin (x - 1)) - \frac{2}{3} (- x^{2} + 2 x)^{\frac{3}{2}}

= \frac{5}{8} arcsin (x - 1) + \frac{1}{4} sin (2 arcsin (x - 1)) - \frac{1}{32} sin (4 arcsin (x - 1)) - \frac{2}{3} (- x^{2} + 2 x)^{\frac{3}{2}}

定数を解答に追加する

= \frac{5}{8} arcsin (x - 1) + \frac{1}{4} sin (2 arcsin (x - 1)) - \frac{1}{32} sin (4 arcsin (x - 1)) - \frac{2}{3} (- x^{2} + 2 x)^{\frac{3}{2}} + C

グラフ

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