integral from 0 to pi of sin^3(t/2)

Lösung

\int_{0}^{π} sin^{3} (\frac{t}{2}) d t

\frac{4}{3}

Dezimale

1.33333 \dots

Schritte zur Lösung

\int_{0}^{π} sin^{3} (\frac{t}{2}) d t

Wende U-Substitution an

= \int_{0}^{\frac{π}{2}} sin^{3} (u) \cdot 2 d u

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{\frac{π}{2}} sin^{3} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Wende U-Substitution an

= 2 \cdot \int_{1}^{0} - 1 + v^{2} d v

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

= 2 (- \int_{0}^{1} - 1 + v^{2} d v)

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (- (- \int_{0}^{1} 1 d v + \int_{0}^{1} v^{2} d v))

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

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

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

Vereinfache

2 (- (- 1 + \frac{1}{3})) : \frac{4}{3}

= \frac{4}{3}

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