integral from 0 to 2pi of sin(x)cos(nx)

Lösung

\int_{0}^{2 π} sin (x) cos (n x) d x

\frac{n cos ( 2 π + 2 πn ) - cos ( 2 π + 2 πn ) + 2 - n cos ( 2 π - 2 πn ) - cos ( 2 π - 2 πn )}{2 ( n + 1 ) ( - n + 1 )}

Schritte zur Lösung

\int_{0}^{2 π} sin (x) cos (n x) d x

Verwende die folgenden Identitäten:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

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

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

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

= \frac{1}{2} (\frac{- cos ( 2 π + 2 πn ) + 1}{1 + n} + \frac{- cos ( 2 π - 2 πn ) + 1}{1 - n})

Vereinfache

\frac{1}{2} (\frac{- cos ( 2 π + 2 πn ) + 1}{1 + n} + \frac{- cos ( 2 π - 2 πn ) + 1}{1 - n}) : \frac{n cos ( 2 π + 2 πn ) - cos ( 2 π + 2 πn ) + 2 - n cos ( 2 π - 2 πn ) - cos ( 2 π - 2 πn )}{2 ( n + 1 ) ( - n + 1 )}

= \frac{n cos ( 2 π + 2 πn ) - cos ( 2 π + 2 πn ) + 2 - n cos ( 2 π - 2 πn ) - cos ( 2 π - 2 πn )}{2 ( n + 1 ) ( - n + 1 )}

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