integral of sin(2x)-5cos(4x)

Lösung

\int sin (2 x) - 5 cos (4 x) d x

- \frac{1}{2} cos (2 x) - \frac{5}{4} sin (4 x) + C

Schritte zur Lösung

\int sin (2 x) - 5 cos (4 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

= \int sin (2 x) d x - \int 5 cos (4 x) d x

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

\int 5 cos (4 x) d x = \frac{5}{4} sin (4 x)

= - \frac{1}{2} cos (2 x) - \frac{5}{4} sin (4 x)

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{2} cos (2 x) - \frac{5}{4} sin (4 x) + C

\int 10 x^{5} e^{x^{6}} d x

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

\frac{\partial}{\partial x} (\frac{1}{a} e^{- (\frac{x}{a})^{2}})

t an g e n t f (x) = 8 x^{2} + 3 x, at x = - 45

\frac{d y}{d x} = 2 sec^{2} (x) e^{2 y}, y (0) = 0