integral from 0 to pi/2 of 16cos^5(x)

Lösung

\int_{0}^{\frac{π}{2}} 16 cos^{5} (x) d x

\frac{128}{15}

Dezimale

8.53333 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} 16 cos^{5} (x) d x

Entferne die Konstante:

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

= 16 \cdot \int_{0}^{\frac{π}{2}} cos^{5} (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 16 \cdot \int_{0}^{\frac{π}{2}} (1 - sin^{2} (x))^{2} cos (x) d x

Wende U-Substitution an

= 16 \cdot \int_{0}^{1} (1 - u^{2})^{2} d u

Multipliziere aus

(1 - u^{2})^{2} : 1 - 2 u^{2} + u^{4}

= 16 \cdot \int_{0}^{1} 1 - 2 u^{2} + u^{4} 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

= 16 (\int_{0}^{1} 1 d u - \int_{0}^{1} 2 u^{2} d u + \int_{0}^{1} u^{4} d u)

\int_{0}^{1} 1 d u = 1

\int_{0}^{1} 2 u^{2} d u = \frac{2}{3}

\int_{0}^{1} u^{4} d u = \frac{1}{5}

= 16 (1 - \frac{2}{3} + \frac{1}{5})

Vereinfache

16 (1 - \frac{2}{3} + \frac{1}{5}) : \frac{128}{15}

= \frac{128}{15}

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