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

Lösung

\int_{0}^{\frac{π}{2}} cos (x) cos (n x) d x

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

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} cos (x) cos (n x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

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

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

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

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