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

Lösung

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

\frac{5}{8} (arcsin (\frac{1}{5} (2 x + 1)) + \frac{1}{2} sin (2 arcsin (\frac{2 x + 1}{5}))) + C

Schritte zur Lösung

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

= \int \frac{- 4 u ^{2} + 5}{2} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int - 4 u^{2} + 5 d u

Trigonometrische Substitution anwenden

= \frac{1}{2} \cdot \int \frac{5}{2} cos^{2} (v) d v

Entferne die Konstante:

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

= \frac{1}{2} \cdot \frac{5}{2} \cdot \int cos^{2} (v) d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \frac{1}{2} \cdot \frac{5}{2} \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Entferne die Konstante:

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

= \frac{1}{2} \cdot \frac{5}{2} \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

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}{2} \cdot \frac{5}{2} \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

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

= \frac{1}{2} \cdot \frac{5}{2} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Ersetze zurück

= \frac{1}{2} \cdot \frac{5}{2} \cdot \frac{1}{2} (arcsin (\frac{2}{5} (x + \frac{1}{2})) + \frac{1}{2} sin (2 arcsin (\frac{2}{5} (x + \frac{1}{2}))))

Vereinfache

\frac{1}{2} \cdot \frac{5}{2} \cdot \frac{1}{2} (arcsin (\frac{2}{5} (x + \frac{1}{2})) + \frac{1}{2} sin (2 arcsin (\frac{2}{5} (x + \frac{1}{2})))) : \frac{5}{8} (arcsin (\frac{1}{5} (2 x + 1)) + \frac{1}{2} sin (2 arcsin (\frac{2 x + 1}{5})))

= \frac{5}{8} (arcsin (\frac{1}{5} (2 x + 1)) + \frac{1}{2} sin (2 arcsin (\frac{2 x + 1}{5})))

Füge eine Konstante zur Lösung hinzu

= \frac{5}{8} (arcsin (\frac{1}{5} (2 x + 1)) + \frac{1}{2} sin (2 arcsin (\frac{2 x + 1}{5}))) + C

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