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

Lösung

\int_{0}^{π} cos (2 x) sin (n x) d x

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

Schritte zur Lösung

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

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

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

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

Vereinfache

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

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

\int_{1}^{5} \frac{x}{2} d x

\int_{1}^{2} - x + 2 d x

\int_{3}^{\infty} 2 e^{- 2 x} d x

\int_{0}^{2} x e^{\frac{x}{2}} d x

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