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

Lösung

\int_{0}^{π} cos (x) \cdot sin (n x) d x

\frac{n ( ( - 1 ) ^{n} + 1 )}{( n + 1 ) ( n - 1 )}

Schritte zur Lösung

\int_{0}^{π} cos (x) sin (n x) d x

Verwende die folgenden Identitäten:

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

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

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

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

= \frac{1}{2} (\frac{( - 1 ) ^{n} + 1}{n + 1} + \frac{( - 1 ) ^{n} + 1}{n - 1})

Vereinfache

\frac{1}{2} (\frac{( - 1 ) ^{n} + 1}{n + 1} + \frac{( - 1 ) ^{n} + 1}{n - 1}) : \frac{n ( ( - 1 ) ^{n} + 1 )}{( n + 1 ) ( n - 1 )}

= \frac{n ( ( - 1 ) ^{n} + 1 )}{( n + 1 ) ( n - 1 )}

\int_{0}^{1} 2 t^{2} d t

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\int_{0}^{1} (x + 1)^{2} d x