integral from 0 to y of sqrt(1-y^2)

Lösung

\int_{0}^{y} 1 - y^{2} d y

\frac{1}{2} (arcsin (y) + \frac{1}{2} sin (2 arcsin (y)))

Schritte zur Lösung

\int_{0}^{y} 1 - y^{2} d y

Trigonometrische Substitution anwenden

= \int_{0}^{a r c s i n (y)} cos^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{0}^{a r c s i n (y)} \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

= \frac{1}{2} \cdot \int_{0}^{a r c s i n (y)} 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

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

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

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

= \frac{1}{2} (arcsin (y) + \frac{1}{2} sin (2 arcsin (y)))

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