integral from 1 to 2 of cos(2θ)cos(3θ)

Lösung

\int_{1}^{2} cos (2 θ) cos (3 θ) d θ

\frac{1}{2} (sin (2) - sin (1) + \frac{1}{5} (sin (10) - sin (5)))

Dezimale

0.07540 \dots

Schritte zur Lösung

\int_{1}^{2} cos (2 θ) cos (3 θ) d θ

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{1}^{2} \frac{1}{2} (cos (5 θ) + cos (- θ)) d θ

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{1}^{2} cos (5 θ) + cos (- θ) d θ

Vereinfache

cos (5 θ) + cos (- θ) : cos (θ) + cos (5 θ)

= \frac{1}{2} \cdot \int_{1}^{2} cos (θ) + cos (5 θ) d θ

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_{1}^{2} cos (θ) d θ + \int_{1}^{2} cos (5 θ) d θ)

\int_{1}^{2} cos (θ) d θ = sin (2) - sin (1)

\int_{1}^{2} cos (5 θ) d θ = \frac{1}{5} (sin (10) - sin (5))

= \frac{1}{2} (sin (2) - sin (1) + \frac{1}{5} (sin (10) - sin (5)))

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