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integral of sqrt(8-4x-4x^2)

Lösung

\int 8 - 4 x - 4 x^{2} d x

\frac{9}{8} (2 arcsin (\frac{2}{3} x + \frac{1}{3}) + sin (2 arcsin (\frac{2}{3} x + \frac{1}{3}))) + C

Schritte zur Lösung

\int 8 - 4 x - 4 x^{2} d x

Vervollständige das Quadrat

8 - 4 x - 4 x^{2} : - 4 (x + \frac{1}{2})^{2} + 9

= \int - 4 (x + \frac{1}{2})^{2} + 9 d x

Wende U-Substitution an

= \int - 4 u^{2} + 9 d u

Trigonometrische Substitution anwenden

= \int \frac{9}{2} cos^{2} (v) d v

Entferne die Konstante:

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

= \frac{9}{2} \cdot \int cos^{2} (v) d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \frac{9}{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{9}{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{9}{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{9}{2} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Ersetze zurück

= \frac{9}{2} \cdot \frac{1}{2} (arcsin (\frac{2}{3} (x + \frac{1}{2})) + \frac{1}{2} sin (2 arcsin (\frac{2}{3} (x + \frac{1}{2}))))

Vereinfache

\frac{9}{2} \cdot \frac{1}{2} (arcsin (\frac{2}{3} (x + \frac{1}{2})) + \frac{1}{2} sin (2 arcsin (\frac{2}{3} (x + \frac{1}{2})))) : \frac{9}{8} (2 arcsin (\frac{2}{3} x + \frac{1}{3}) + sin (2 arcsin (\frac{2}{3} x + \frac{1}{3})))

= \frac{9}{8} (2 arcsin (\frac{2}{3} x + \frac{1}{3}) + sin (2 arcsin (\frac{2}{3} x + \frac{1}{3})))

Füge eine Konstante zur Lösung hinzu

= \frac{9}{8} (2 arcsin (\frac{2}{3} x + \frac{1}{3}) + sin (2 arcsin (\frac{2}{3} x + \frac{1}{3}))) + C

x \to 0 lim (x^{2} + 4 x + 2)

n = 1 \sum \infty \frac{1}{ln ( n + 1 )}

y^{^{'}} = k (y - 9 - 10 cos (2 \cdot π \cdot t))

\frac{d ^{2}}{d x ^{2}} (2 sin (- 4 x - 3))

n = 2 \sum \infty \frac{1}{5 ^{n}}