integral from 0 to pi/3 of sin(3y)cos(y)

Lösung

\int_{0}^{\frac{π}{3}} sin (3 y) cos (y) d y

\frac{9}{16}

Dezimale

0.5625

Schritte zur Lösung

\int_{0}^{\frac{π}{3}} sin (3 y) cos (y) d y

Verwende die folgenden Identitäten:

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

= \int_{0}^{\frac{π}{3}} \frac{sin ( 3 y + y ) + sin ( 3 y - y )}{2} d y

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}{3}} sin (3 y + y) + sin (3 y - y) d y

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}{3}} sin (3 y + y) d y + \int_{0}^{\frac{π}{3}} sin (3 y - y) d y)

\int_{0}^{\frac{π}{3}} sin (3 y + y) d y = \frac{3}{8}

\int_{0}^{\frac{π}{3}} sin (3 y - y) d y = \frac{3}{4}

= \frac{1}{2} (\frac{3}{8} + \frac{3}{4})

Vereinfache

\frac{1}{2} (\frac{3}{8} + \frac{3}{4}) : \frac{9}{16}

= \frac{9}{16}

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