integral of tan^2(x)*sec^3(x)

Lösung

\int tan^{2} (x) \cdot sec^{3} (x) d x

- \frac{1}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{1}{4} sec^{4} (x) sin (x) + C

Schritte zur Lösung

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Multipliziere aus

(- 1 + sec^{2} (x)) sec^{3} (x) : - sec^{3} (x) + sec^{5} (x)

= \int - sec^{3} (x) + sec^{5} (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 sec^{3} (x) d x + \int sec^{5} (x) d x

\int sec^{3} (x) d x = \frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣

\int sec^{5} (x) d x = \frac{sec ^{4} ( x ) sin ( x )}{4} + \frac{3}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)

= - (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{sec ^{4} ( x ) sin ( x )}{4} + \frac{3}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)

Vereinfache

- (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{sec ^{4} ( x ) sin ( x )}{4} + \frac{3}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) : - \frac{1}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{1}{4} sec^{4} (x) sin (x)

= - \frac{1}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{1}{4} sec^{4} (x) sin (x)

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{1}{4} sec^{4} (x) sin (x) + C

\frac{\partial}{\partial x} (ln (x + 4 y + 9 z))

t an g e n t y = 6 x - x^{2}, (1, 5)

\int \frac{x + 2}{x ^{2} + 9} d x

\int \frac{6}{x - 5} d x

\frac{\partial}{\partial y} (3 x^{2} y)