integral from 0 to x of sqrt(36-t^2)

Lösung

\int_{0}^{x} 36 - t^{2} d t

18 (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6})))

Schritte zur Lösung

\int_{0}^{x} 36 - t^{2} d t

Trigonometrische Substitution anwenden

= \int_{0}^{a r c s i n (\frac{x}{6})} 36 cos^{2} (u) d u

Entferne die Konstante:

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

= 36 \cdot \int_{0}^{a r c s i n (\frac{x}{6})} cos^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 36 \cdot \int_{0}^{a r c s i n (\frac{x}{6})} \frac{1 + cos ( 2 u )}{2} d u

Entferne die Konstante:

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

= 36 \cdot \frac{1}{2} \cdot \int_{0}^{a r c s i n (\frac{x}{6})} 1 + cos (2 u) 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

= 36 \cdot \frac{1}{2} (\int_{0}^{a r c s i n (\frac{x}{6})} 1 d u + \int_{0}^{a r c s i n (\frac{x}{6})} cos (2 u) d u)

\int_{0}^{a r c s i n (\frac{x}{6})} 1 d u = arcsin (\frac{x}{6})

\int_{0}^{a r c s i n (\frac{x}{6})} cos (2 u) d u = \frac{1}{2} sin (2 arcsin (\frac{x}{6}))

= 36 \cdot \frac{1}{2} (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6})))

Vereinfache

36 \cdot \frac{1}{2} (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6}))) : 18 (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6})))

= 18 (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6})))

\frac{d}{d x} ((2 x^{5} - 3)^{8})

\frac{d}{d x} (4 e^{\frac{x}{2}})

d er i v a t i v e tan^{2} (3 x) + 1

y^{^{''}} + 9 y = 3 t^{2}

d er i v a t i v e π \cdot x^{\frac{1}{2}} - x^{\frac{1}{3}}