integral from 0 to pi/8 of sin^2(2x)

Lösung

\int_{0}^{\frac{π}{8}} sin^{2} (2 x) d x

\frac{π - 2}{16}

Dezimale

0.07134 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{8}} sin^{2} (2 x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{0}^{\frac{π}{8}} \frac{1}{2} (1 - cos (4 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}{8}} 1 - cos (4 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}{8}} 1 d x - \int_{0}^{\frac{π}{8}} cos (4 x) d x)

\int_{0}^{\frac{π}{8}} 1 d x = \frac{π}{8}

\int_{0}^{\frac{π}{8}} cos (4 x) d x = \frac{1}{4}

= \frac{1}{2} (\frac{π}{8} - \frac{1}{4})

Vereinfache

\frac{1}{2} (\frac{π}{8} - \frac{1}{4}) : \frac{π - 2}{16}

= \frac{π - 2}{16}