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

Lösung

\int_{0}^{π} sin (2 x) sin (x) d x

0

Schritte zur Lösung

\int_{0}^{π} sin (2 x) sin (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{0}^{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}^{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}^{π} cos (3 x) d x + \int_{0}^{π} cos (x) d x)

\int_{0}^{π} cos (3 x) d x = 0

\int_{0}^{π} cos (x) d x = 0

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

Vereinfache

= 0

\int_{0}^{1} t 1 3^{t^{2}} d t

\int_{1}^{x} \frac{5}{t ^{3} + 1} d t

\int_{1}^{2} x \frac{1}{2} x^{2} + 9 d x

\int_{0}^{2} x^{2} (1 + 2 x)^{3} d x

\int_{0}^{x} sin (t) d t