integral of 13tan^2(x)sec^3(x)

Lösung

\int 13 tan^{2} (x) sec^{3} (x) d x

13 (- \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 13 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

= 13 \cdot \int tan^{2} (x) sec^{3} (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 13 \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)

= 13 \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

= 13 (- \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) ∣)

= 13 (- (\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

13 (- (\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) ∣)) : 13 (- \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))

= 13 (- \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

= 13 (- \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

\int - 2 x^{2} + 8 x d x

y^{^{''}} - 3 y^{^{'}} + y = 0

t an g e n t y = 3 x^{3} - 3 x

\frac{d y}{d x} = 5 e^{x}

\frac{d}{d x} (\frac{1}{x ^{3} + 1})