integral from 0 to pi/2 of-8sin^2(3x)

Lösung

\int_{0}^{\frac{π}{2}} - 8 sin^{2} (3 x) d x

- 2 π

Dezimale

- 6.28318 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} - 8 sin^{2} (3 x) d x

Entferne die Konstante:

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

= - 8 \cdot \int_{0}^{\frac{π}{2}} sin^{2} (3 x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= - 8 \cdot \int_{0}^{\frac{π}{2}} \frac{1}{2} (1 - cos (6 x)) d x

Entferne die Konstante:

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

= - 8 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (6 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

= - 8 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d x - \int_{0}^{\frac{π}{2}} cos (6 x) d x)

\int_{0}^{\frac{π}{2}} 1 d x = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (6 x) d x = 0

= - 8 \cdot \frac{1}{2} (\frac{π}{2} - 0)

Vereinfache

- 8 \cdot \frac{1}{2} (\frac{π}{2} - 0) : - 2 π

= - 2 π

\frac{d y}{d x} = cos (x)

(2 y + 2 t - 7) d t + (4 y + 2 t - 1) d y = 0, y (- 1) = 2

d er i v a t i v e f (x) = 2 x - 3

d er i v a t i v e f (x) = (1 + 3 x^{4})^{2} + 3 x^{3}

y^{^{''}} + y^{^{'}} + 2 y = 0, y (0) = 2, y^{^{'}} (0) = 0