integral from 0 to pi/2 of sin^5(θ)

Lösung

\int_{0}^{\frac{π}{2}} sin^{5} (θ) d θ

\frac{8}{15}

Dezimale

0.53333 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} sin^{5} (θ) d θ

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Wende U-Substitution an

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \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

= - (- \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}

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

Vereinfache

- (- 1 + \frac{2}{3} - \frac{1}{5}) : \frac{8}{15}

= \frac{8}{15}

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