integral of-9sec(x)tan(x)-2sec^2(x)

Lösung

\int - 9 sec (x) tan (x) - 2 sec^{2} (x) d x

- 9 sec (x) - 2 tan (x) + C

Schritte zur Lösung

\int - 9 sec (x) tan (x) - 2 sec^{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

= - \int 9 sec (x) tan (x) d x - \int 2 sec^{2} (x) d x

\int 9 sec (x) tan (x) d x = 9 sec (x)

\int 2 sec^{2} (x) d x = 2 tan (x)

= - 9 sec (x) - 2 tan (x)

Füge eine Konstante zur Lösung hinzu

= - 9 sec (x) - 2 tan (x) + C

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

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