integral of 1/2 (1+cos(2x))

Lösung

\int \frac{1}{2} (1 + cos (2 x)) d x

\frac{1}{2} (x + \frac{1}{2} sin (2 x)) + C

Schritte zur Lösung

\int \frac{1}{2} (1 + cos (2 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 1 + cos (2 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 1 d x + \int cos (2 x) d x)

\int 1 d x = x

\int cos (2 x) d x = \frac{1}{2} sin (2 x)

= \frac{1}{2} (x + \frac{1}{2} sin (2 x))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} (x + \frac{1}{2} sin (2 x)) + C

y^{^{'}} = - 4 y, y (0) = 8

x \to 1 lim (\frac{1}{x} - \frac{1}{e ^{x} - 1})

d er i v a t i v e ln (x + x^{2} - 1)

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

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