integral from 0 to 2pi of sin(2x)sin(x)

Lösung

\int_{0}^{2 π} sin (2 x) sin (x) d x

0

Schritte zur Lösung

\int_{0}^{2 π} sin (2 x) sin (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{0}^{2 π} \frac{1}{2} (- cos (3 x) + cos (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}^{2 π} - cos (3 x) + cos (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}^{2 π} cos (3 x) d x + \int_{0}^{2 π} cos (x) d x)

\int_{0}^{2 π} cos (3 x) d x = 0

\int_{0}^{2 π} cos (x) d x = 0

= \frac{1}{2} (- 0 + 0)

Vereinfache

= 0

n = 0 \sum \infty (\frac{- 1}{2})^{n}

t an g e n t f (x) = 4 x (x^{2} - 4 x + 5)^{9}, at x = 2

\frac{d}{d x} (9 e^{- 4 x})

y^{3} \frac{d y}{d x} + x^{3} = 0

\frac{\partial}{\partial y} (x^{2} - 4 x y - 2 y^{2})