integral of 17tan^2(x)sec^3(x)

Lösung

\int 17 tan^{2} (x) sec^{3} (x) d x

17 (- \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 17 tan^{2} (x) sec^{3} (x) d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 17 \cdot \int tan^{2} (x) sec^{3} (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 17 \cdot \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)

= 17 \cdot \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

= 17 (- \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) ∣)

= 17 (- (\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

17 (- (\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) ∣)) : 17 (- \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))

= 17 (- \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

= 17 (- \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{d}{d x} (\frac{1 + sec ( x )}{1 - sec ( x )})

d er i v a t i v e ln (t^{3})

y^{^{'}} = \frac{y ^{2} + x ^{2}}{2 x y}

cos (x) y^{^{'}} + sin (x) y = 0

d er i v a t i v e x^{2} + 24 x + c