integral from 0 to 2pi of sin^2(1/3 θ)

Lösung

\int_{0}^{2 π} sin^{2} (\frac{1}{3} θ) d θ

\frac{8 π + 3 3}{8}

Dezimale

3.79111 \dots

Schritte zur Lösung

\int_{0}^{2 π} sin^{2} (\frac{1}{3} θ) d θ

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{0}^{2 π} \frac{1}{2} (1 - cos (\frac{2}{3} θ)) d θ

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{0}^{2 π} 1 - cos (\frac{2}{3} θ) d θ

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}^{2 π} 1 d θ - \int_{0}^{2 π} cos (\frac{2}{3} θ) d θ)

\int_{0}^{2 π} 1 d θ = 2 π

\int_{0}^{2 π} cos (\frac{2}{3} θ) d θ = - \frac{3 3}{4}

= \frac{1}{2} (2 π - (- \frac{3 3}{4}))

Vereinfache

\frac{1}{2} (2 π - (- \frac{3 3}{4})) : \frac{8 π + 3 3}{8}

= \frac{8 π + 3 3}{8}

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