integral from 0 to xcos(x) of sqrt(25-t^2)

Lösung

\int_{0}^{x c o s (x)} 25 - t^{2} d t

\frac{25}{2} (arcsin (\frac{x}{5} cos (x)) + \frac{1}{2} sin (2 arcsin (\frac{x}{5} cos (x))))

Schritte zur Lösung

\int_{0}^{x c o s (x)} 25 - t^{2} d t

Trigonometrische Substitution anwenden

= \int_{0}^{a r c s i n (\frac{x c o s ( x )}{5})} 25 cos^{2} (u) d u

Entferne die Konstante:

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

= 25 \cdot \int_{0}^{a r c s i n (\frac{x c o s ( x )}{5})} cos^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 25 \cdot \int_{0}^{a r c s i n (\frac{x c o s ( x )}{5})} \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

= 25 \cdot \frac{1}{2} \cdot \int_{0}^{a r c s i n (\frac{x c o s ( x )}{5})} 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

= 25 \cdot \frac{1}{2} (\int_{0}^{a r c s i n (\frac{x c o s ( x )}{5})} 1 d u + \int_{0}^{a r c s i n (\frac{x c o s ( x )}{5})} cos (2 u) d u)

\int_{0}^{a r c s i n (\frac{x c o s ( x )}{5})} 1 d u = arcsin (\frac{x cos ( x )}{5})

\int_{0}^{a r c s i n (\frac{x c o s ( x )}{5})} cos (2 u) d u = \frac{1}{2} sin (2 arcsin (\frac{x cos ( x )}{5}))

= 25 \cdot \frac{1}{2} (arcsin (\frac{x cos ( x )}{5}) + \frac{1}{2} sin (2 arcsin (\frac{x cos ( x )}{5})))

Vereinfache

25 \cdot \frac{1}{2} (arcsin (\frac{x cos ( x )}{5}) + \frac{1}{2} sin (2 arcsin (\frac{x cos ( x )}{5}))) : \frac{25}{2} (arcsin (\frac{x}{5} cos (x)) + \frac{1}{2} sin (2 arcsin (\frac{x}{5} cos (x))))

= \frac{25}{2} (arcsin (\frac{x}{5} cos (x)) + \frac{1}{2} sin (2 arcsin (\frac{x}{5} cos (x))))

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