integral from 0 to pi/2 of cos(9x)cos(18x)

Lösung

\int_{0}^{\frac{π}{2}} cos (9 x) cos (18 x) d x

\frac{1}{27}

Dezimale

0.03703 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} cos (9 x) cos (18 x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{0}^{\frac{π}{2}} \frac{1}{2} (cos (27 x) + cos (- 9 x)) 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 (27 x) + cos (- 9 x) d x

Vereinfache

cos (27 x) + cos (- 9 x) : cos (9 x) + cos (27 x)

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} cos (9 x) + cos (27 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 (9 x) d x + \int_{0}^{\frac{π}{2}} cos (27 x) d x)

\int_{0}^{\frac{π}{2}} cos (9 x) d x = \frac{1}{9}

\int_{0}^{\frac{π}{2}} cos (27 x) d x = - \frac{1}{27}

= \frac{1}{2} (\frac{1}{9} - \frac{1}{27})

Vereinfache

\frac{1}{2} (\frac{1}{9} - \frac{1}{27}) : \frac{1}{27}

= \frac{1}{27}

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