integral from 0 to pi of |sin(x)|cos(kx)

Lösung

\int_{0}^{π} ∣ sin (x) ∣ cos (k x) d x

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

Schritte zur Lösung

\int_{0}^{π} ∣ sin (x) ∣ cos (k x) d x

Entferne die Extremwerte

= \int_{0}^{π} sin (x) cos (k x) d x

Verwende die folgenden Identitäten:

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

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

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

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

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

Vereinfache

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

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

\int_{- 2}^{1} 2 x^{2} - x^{4} - x^{3} d x

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