integral of 16tan^4(x)sec^6(x)

Lösung

\int 16 tan^{4} (x) sec^{6} (x) d x

16 (\frac{tan ^{5} ( x )}{5} + \frac{2 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{9}) + C

Schritte zur Lösung

\int 16 tan^{4} (x) sec^{6} (x) d x

Entferne die Konstante:

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

= 16 \cdot \int tan^{4} (x) sec^{6} (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 16 \cdot \int sec^{2} (x) (1 + tan^{2} (x))^{2} tan^{4} (x) d x

Wende U-Substitution an

= 16 \cdot \int u^{4} (1 + u^{2})^{2} d u

Multipliziere aus

u^{4} (1 + u^{2})^{2} : u^{4} + 2 u^{6} + u^{8}

= 16 \cdot \int u^{4} + 2 u^{6} + u^{8} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 16 (\int u^{4} d u + \int 2 u^{6} d u + \int u^{8} d u)

\int u^{4} d u = \frac{u ^{5}}{5}

\int 2 u^{6} d u = \frac{2 u ^{7}}{7}

\int u^{8} d u = \frac{u ^{9}}{9}

= 16 (\frac{u ^{5}}{5} + \frac{2 u ^{7}}{7} + \frac{u ^{9}}{9})

Setze in

u = tan (x)

ein

= 16 (\frac{tan ^{5} ( x )}{5} + \frac{2 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{9})

Füge eine Konstante zur Lösung hinzu

= 16 (\frac{tan ^{5} ( x )}{5} + \frac{2 tan ^{7} ( x )}{7} + \frac{tan ^{9} ( x )}{9}) + C

t an g e n t f (x) = 38 - x^{2}, (- 6, 2)

t an g e n t y = 3 - x^{3} (0.3)

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

z \to 2 i lim (i z^{4} + 3 z^{2} - 10 i)

d er i v a t i v e f (x) = x^{2} - 8 x