integral from 0 to c of sin(3x)cos(nx)

Lösung

\int_{0}^{c} sin (3 x) cos (n x) d x

\frac{- n cos ( 3 c - c n ) - 3 cos ( 3 c - c n ) + n cos ( c n + 3 c ) + 6 - 3 cos ( c n + 3 c )}{2 ( n + 3 ) ( - n + 3 )}

Schritte zur Lösung

\int_{0}^{c} sin (3 x) cos (n x) d x

Verwende die folgenden Identitäten:

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

= \int_{0}^{c} \frac{sin ( 3 x + n x ) + sin ( 3 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}^{c} sin (3 x + n x) + sin (3 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}^{c} sin (3 x + n x) d x + \int_{0}^{c} sin (3 x - n x) d x)

\int_{0}^{c} sin (3 x + n x) d x = \frac{- cos ( 3 c + c n ) + 1}{3 + n}

\int_{0}^{c} sin (3 x - n x) d x = \frac{- cos ( 3 c - c n ) + 1}{3 - n}

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

Vereinfache

\frac{1}{2} (\frac{- cos ( 3 c + c n ) + 1}{3 + n} + \frac{- cos ( 3 c - c n ) + 1}{3 - n}) : \frac{- n cos ( 3 c - c n ) - 3 cos ( 3 c - c n ) + n cos ( c n + 3 c ) + 6 - 3 cos ( c n + 3 c )}{2 ( n + 3 ) ( - n + 3 )}

= \frac{- n cos ( 3 c - c n ) - 3 cos ( 3 c - c n ) + n cos ( c n + 3 c ) + 6 - 3 cos ( c n + 3 c )}{2 ( n + 3 ) ( - n + 3 )}

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