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

Lösung

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

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

Schritte zur Lösung

\int \frac{1 - cos ( 2 x )}{2} 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

\int + x^{2} sec^{2} (x^{3} + 1) d x

\frac{d}{d x} (\frac{x ^{2}}{( 5 + x )})

\frac{d}{d x} (- x^{2} - 4 x - 2)

\int cosh (0)

\int_{0}^{\frac{π}{3}} tan^{2} (x) d x