integral from 0 to 2pi of sin^2(nx)

Lösung

\int_{0}^{2 π} sin^{2} (n x) d x

π - \frac{1}{4 n} sin (4 πn)

Schritte zur Lösung

\int_{0}^{2 π} sin^{2} (n x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

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 (2 n x) d x

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 x - \int_{0}^{2 π} cos (2 n x) d x)

\int_{0}^{2 π} 1 d x = 2 π

\int_{0}^{2 π} cos (2 n x) d x = \frac{sin ( 4 πn )}{2 n}

= \frac{1}{2} (2 π - \frac{sin ( 4 πn )}{2 n})

Vereinfache

\frac{1}{2} (2 π - \frac{sin ( 4 πn )}{2 n}) : π - \frac{1}{4 n} sin (4 πn)

= π - \frac{1}{4 n} sin (4 πn)

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