integral of xsqrt(x-x^2)

Lösung

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

\frac{1}{96} (- 32 (x (- x + 1))^{\frac{3}{2}} + 3 (2 arcsin (2 x - 1) + sin (2 arcsin (2 x - 1)))) + C

Schritte zur Lösung

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

Vervollständige das Quadrat

x - x^{2} : - (x - \frac{1}{2})^{2} + \frac{1}{4}

= \int x - (x - \frac{1}{2})^{2} + \frac{1}{4} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Multipliziere aus

(2 u + 1) - 4 u^{2} + 1 : 2 u - 4 u^{2} + 1 + - 4 u^{2} + 1

= \frac{1}{4} \cdot \int 2 u - 4 u^{2} + 1 + - 4 u^{2} + 1 d u

Wende die Summenregel an:

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

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

\int 2 u - 4 u^{2} + 1 d u = - \frac{1}{6} (- 4 u^{2} + 1)^{\frac{3}{2}}

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

= \frac{1}{4} (- \frac{1}{6} (- 4 u^{2} + 1)^{\frac{3}{2}} + \frac{1}{4} (arcsin (2 u) + \frac{1}{2} sin (2 arcsin (2 u))))

Setze in

u = x - \frac{1}{2}

ein

= \frac{1}{4} - \frac{1}{6} (- 4 (x - \frac{1}{2})^{2} + 1)^{\frac{3}{2}} + \frac{1}{4} (arcsin (2 (x - \frac{1}{2})) + \frac{1}{2} sin (2 arcsin (2 (x - \frac{1}{2}))))

Vereinfache

\frac{1}{4} - \frac{1}{6} (- 4 (x - \frac{1}{2})^{2} + 1)^{\frac{3}{2}} + \frac{1}{4} (arcsin (2 (x - \frac{1}{2})) + \frac{1}{2} sin (2 arcsin (2 (x - \frac{1}{2})))) : \frac{1}{96} (- 32 (x (- x + 1))^{\frac{3}{2}} + 3 (2 arcsin (2 x - 1) + sin (2 arcsin (2 x - 1))))

= \frac{1}{96} (- 32 (x (- x + 1))^{\frac{3}{2}} + 3 (2 arcsin (2 x - 1) + sin (2 arcsin (2 x - 1))))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{96} (- 32 (x (- x + 1))^{\frac{3}{2}} + 3 (2 arcsin (2 x - 1) + sin (2 arcsin (2 x - 1)))) + C

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